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P & C, Binomial Theorem CONCEPT NO TES NOTES

01.

Fundamental Principle of Counting

01 - 04

02.

Introduction To Permutations

05 - 07

03.

Introduction To Combinations

08 - 09

04.

Applications of Basics

10 - 26

05.

More Applications

27 - 59

06.

Appendix : Binomial Theorem

60 - 86

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P & C, Binomial Theorem This chapter is one of the most interesting chapters that we’ll study at this level. The beauty and challenge of this branch of mathematics lies in the innumerous tricks and mathematical artifices that abound in this subject. Clarity of thought, more than any thing else, is what is required to understand the subject properly. Also, you’d do best if you refrain from memorising any formulae or particular cases here; concentrate on building a logical approach, solving everything from first principles. The main objective of this chapter is to count. Given a set U of things or objects or persons ( or whatever), we need to arrange a subset S of U (according to some constraints) or select a subset S of U (again, according to certain criteria). In fact, we are actually interested in counting the number of such arrangements or selections. Read the following questions: “From a team of 15 cricket players, how can we select a playing team of 11players? “There are 20 people whom we need to seat in 2 rows of 10 seats each. How many ways exist of doing so? “From a deck of 52 playing cards, in how many ways can we select two red cards and three black cards?” “How many rectangles exist on a standard 8×8 chessboard?” “How many factors does 144000 have? In general, how many factors does a natural number N have?” These are some of the many types of questions which we’ll learn to solve in this chapter. We’ll build a systematic approach to deal with such counting issues. To really appreciate the beauty of the solving techniques that we’ll develop, you are urged to try out each and every question that we solve here, on your own first, and only then look at the solution. Only this approach will help you solve counting questions elegantly. Section - 1

Section - 1 OF COUNTING FUNDAMENTAL PRINCIPLE

The fundamental principle of counting is so fundamental that you already must have used it practically a lot many times without realising it. In other words, this principle is already programmed into our minds. A logical, step-by-step application of this principle gives rise to the entire subject of permutations and combinations. Suppose you have 4 boys and 3 girls. From this group of 7, you want to select a couple ( a boy and a girl). How many ways exist of forming this couple? Let us label the boys as B1 , B2 , B3 and B4 and the girls as G1 , G2 and G3 . Mathematics / P & C, Binomial Theorem

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There are 4 ways to choose a boy. Once you’ve chosen a boy, say B2 , there are now 3 ways to choose a girl. In other words, the boy B2 can form the couple ( B2 G1 ), ( B2 G2 ) or ( B2 G3 ) . Similarly, for every other boy, there exist 3 girls with whom that boy can be paired. Thus, the total number of ways in which a pair can be formed is:

N = ( No. of ways of selecting a boy ) × ( No. of ways of selecting a girl ) = 4×3 = 12

The most crucial aspect in this calculation is that you must realise that the task of selecting a boy is independent of the task of selecting a girl. This means that which boy you select has no effect what so ever on which girl you select; the selection of a boy and that of a girl are independent of each other. Another subtle point must be made here. There are 4 ways to select a boy. These 4 ways are mutually exclusive. This means that a selection of any particular boy, once made, rules out the selection of the other 3 boys. Similarly, there are 3 mutually exclusive ways to select a girl. Let us consider another example now. We need to travel from New Delhi, India to Fiji Islands (in the South Pacific Ocean). We must change flights, first at Singapore and then at Sydney, Australia. There are 6 different flights available from New Delhi to Singapore, 5 from Singapore to Sydney and 3 from Sydney to Fiji.

Singapore

New Delhi

6 flights

Sydney

5 flights

Fiji Islands

3 flights

Fig - 1

How may ways exist of making a flight plan from New Delhi to Fiji? Assume that we are in Singapore. From Singapore, we have 5 available flights to Sydney (5 mutually exclusive ways). For each of these 5 flights, we have 3 further ways (flights) from Sydney to Fiji. Thus, we have a total of 5 × 3 = 15 ways of travelling from Singapore to Fiji. Now, how many ways do we have to reach Singapore from New Delhi in the first place? 6 flights. For each of these 6 flights, we have 15 further ways of reaching Fiji From Singapore (as we just calculated in the preceding paragraph). Thus, the total number of ways N of travelling from New Delhi to Fiji is 6 × 15 = 90. In other words,

(

)(

)(

N = No. of flights from New × No. of flights from × No. of flights from Delhi to Singapore Singapore to Sydney Sydney to Fiji

)

Intuitively easy, isn’t it? Let us now generalise the results of these two examples into our fundamental principle of counting. Consider the set of tasks {T1 , T2 , T3 ... Tn } which are all independent of each other. This means that task Ti has “no relation ” to task T j if i ≠ j; the choice of how to accomplish task Ti has thus “no effect” on the choice of how to accomplish talk T j for i ≠ j , (you’ll realise the meaning of “no relation” and “no effect” more specifically later). Task Ti can be accomplished in ki mutually exclusive ways. Mathematics / P & C, Binomial Theorem

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The fundamental principle of counting says that the set of tasks {T1 , T2 , T3 ... Tn } can be accomplished in k1 × k2 × k3 × ... × kn ways. Here are more examples that illustrate this simple yet extremely powerful principle. A proper understanding of this principle is absolutely essential for the subject of counting to be fully comprehensible. • Think of a standard six-faced die, with the markings 1, 2, 3, 4, 5 and 6. Suppose that you have two such dice. When these two dice are thrown call the number that shows up on the first die x and the one on the second die y. How many pairs ( x, y ) are possible? There are 6 mutually exclusive ways in which a number can show up on the first die. Similarly, 6 such ways exist for the second die. The fundamental principle of counting says that the total number of possible pairs is 6 × 6 = 36. •

Suppose that we have a (unlimited) supply of the letters A, B, C , D, E , F available with us. How many 5-letter chains can we form using these letters? Imagine 5 blank spaces for the 5-letter chain that we are supposed to form (numbered 1 to 5):

1

2

3 Fig - 2

4

5

•

We now have 5 tasks at hand; each task correspond to filling a space in Fig. 1. Realise that these 5 tasks are independent of each other. For example, what you fill in place 2 has no effect on what you fill in place 5 since you’ ve been assured an unlimited supply of letters. Each task can be accomplished in 6 possible ways. For example, you can fill place -1 in 6 possible ways : with A, B , C , D , E , or F . Thus, by the fundamental principle of counting, the total number of ways to form the 5-letter chain would be 6 × 6 × 6 × 6 × 6 = 65 .

•

Now consider the scenario when you don’t have an unlimited supply of letters available. Suppose that you have only one of each of the 5 letters available. How many 5-letter chains can be formed now? You must realise that this limited-supply-of-letters situation is very different from the previous one. In this situation, once you fill a particular place with a particular letter, you are left with one letter less to choose from, from the remaining places. Let us start with filling the places from left to right. To fill place -1, w e have 6 letters to choose from, but once we’ve filled place-1 , we now have only 5 letters to choose from, to fill place-2. Continuing in this way, place-3, place - 4 and place -5 can then be filled in 4, 3 and 2 ways respectively. 1

2

3

! 6 ways to fill this place

! 5 ways to fill this place

! 4 ways to fill this place

4 ! 3 ways to fill this place

5 ! 2 ways to fill this place

Fig - 3 The fundamental principle of counting tells us that the total number of chains in this case will be 6 × 5 × 4 × 3 × 2 = 720 . You might think that in this case filling up place-i and place-j are not independent tasks. In other words, how we fill place-i has a certain relation to how we Mathematics / P & C, Binomial Theorem

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fill place-j . For example, if we fill place-1 with A, we cannot fill any other place with A and thus, how we fill place -1 has a definite effect on how we are able to fill the other places. This is definitely true. But when we talk about the independence of two events, it is in a different sense: Let event X be filling place-1 and event Y be filling place-2 after event X has been accomplished, i.e. after place-1 has been filled. There are 6 ways of accomplishing X and 5 ways of accomplishing Y. Which 5 letters contribute to event Y (i.e which of the 5 remaining letters can we choose for place -2) definitely depends on how event X was accomplished, but the event Y is itself the choice of one of the 5 symbols contributing to event Y. The number of ways in which this choice can be made is still independent of how event X was accomplished. Whatever selection was made for event X, the number of ways in which Y can be accomplished still remains 5. This is the sense that should be attached to the phrase “independent events”. ____________________________________________________________________________________ These three examples should have given you a pretty good idea about three concepts : mutually exclusive events, independent events and the fundamental principle of counting. Ponder over these examples for a bit longer and think up examples of you own till your feel very comfortable with these new concepts.

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Section 1 INTRODUCTION TO -PERMUTATIONS

Section - 2

We now formally study the concept of permutations, by generalising the last example in the preceding section. The fundamental issue in Permutations is the arrangement of things. In the last example, we had 6 letters and 5 places where we could arrange (5 of) those 6 letters, We calculated that there are 720 ways of arranging those 6 letters, taken 5 at a time. In mathematical terminology, we calculated as 720 the number of permutations (arrangements) of 6 letters taken 5 at a time. Let us generalise this: Suppose we have n people. It we had n seats available to seat these people, the total number of ways to do so would be (by the logic discussed in the preceding section) n × ( n − 1) × ( n − 2) × ... × 1. This quantity is denoted by n!. n ! = n × ( n − 1) × ( n − 2) × ... × 1

Suppose now that we have only r seats, where r < n. The total number of ways now would be n × ( n − 1) × (n − 2) × ... × ( n − r + 1) . This is the number of ways of permuting n things, taken r at a time, and the notation used for this number is n Pr . Thus : n

Pr = n × (n − 1) × (n − 2) × ... × (n − r + 1)

=

n × (n − 1) × ( n − 2) × ... × (n − r + 1) × (n − r ) × ...1 (n − r ) × ... × 1

=

n! (n − r )!

Thus, using the fundamental principle of counting, you see that we’ve been able to calculate the value of n Pr . Finally, note that n Pn = n ! Lets apply this discussion to an example: Example – 1

Find the number of permutations of the word EDUCATION which (a) consist of all the letters of this word

(b) start with E and end with N

(c) contain the string “CAT”

(d) have the letters C, A and T occurring together

(e) start and end with vowels

(f) have no two vowels occurring together.

Solution: (a) We have 9 letters and we want to permute all of them. The required number of arrangements would be 9 P9 = 9! Mathematics / P & C, Binomial Theorem

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(b) Fix E at the start and N at the end of the word

E

N Fig - 4

We now have 7 places which we need to fill using the remaining 7 letters. The number of such permutations will be 7 P7 = 7!. (c) We want all those permutations in which the string “CAT” occurs. Let us treat “CAT” as a single letter/object since this is what we want - we want “CAT” to appear as a single entity. We now have the following objects (letters) which we need to permute: E

D

U

“CAT ”

I

O

N

These are in total 7 objects. Thus, they can be permuted in 7 P7 = 7! ways. This is the number of permutations that contain the string “CAT ”. (d) We now need the permutations which contain the letters C, A and T occurring together. This means that they can occur together in any order. There are 3! = 6 ways in which C, A and T can be arranged among themselves, namely CAT, CTA, ACT, ATC, TAC, and TCA. Now consider from the previous part all those permutations in which the string “CAT” occurs. These are 7! in number. Corresponding to each such permutation, we can have 6 permutations in which the constraint is that C, A and T occur together. This is because the string “CAT” can itself be permuted in 6 ways as described above. For example, consider the permutation “EDUCATION”; to this permutation will correspond 6 permutations in which C, A and T occur together: EDU EDU EDU CAT ION → EDU EDU EDU EDU

CAT ION CTA ION ACT ION ATC ION TAC ION TCA ION

The required number of permutations is therefore 7!×3! . (e) In the word EDUCATION, there are a total of 5 vowels. We want to have the starting and ending letters of our permutations as vowels. Let us first arrange the starting and ending letters and then fill in the rest of the 7 places. Out of the 5 vowels available to us, the first and the last place can be filled in 5 P2 places. Once we’ve fixed the first and the last letters, the remaining 7 places can be filled in 7 P7 = 7! ways. Thus, the total number of required permutations is 5 P2 × 7! .

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(f) There are 5 vowels and 4 consonants in the word EDUCATION. We require permutations in which no two vowels occur together. We can ensure such permutations by first fixing the 4 consonants and then arranging the 5 vowels in the 5 possible places that arise as depicted in the following figure. D

C

T

N

Fig - 5 Convince yourself that any arrangement of the 5 vowels in the 5 blank spaces above will correspond to a permutation with no two vowels together. The number of ways of arranging the 4 consonants is 4 P4 = 4!. After the consonants have been arranged, the number of ways of arranging the 5 vowels in the 5 blank spaces as depicted in Fig.5 is 5!. Thus, the required number of permutations is 4! × 5!

____________________________________________________________________________________ It would be a good exercise for you to construct more of such examples on your own and solve them. You can take an arbitrary word and find the number of permutations of that word, that satisfy any particular condition that you can think of.

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Section 1 INTRODUCTION TO -COMBINATIONS

Section - 3

In the previous section, we encountered permutations, which correspond to arrangement of objects /things/ entities. In this section, we will encounter combinations which correspond to selection of things (and not their arrangement). We do no intend to arrange things. We intend to select them. For example, suppose we have a team of 15 cricket players. We intend to select a playing team of 11 out of these 15 players. Thus, we want the number of ways in which we can select 11 players out of 15 players. (We are not interested in arranging those 11 players in a row - only the group/ combination of those 11 players matters). Let us make this concept more specific. Suppose we have a set of 6 letters { A, B, C , D, E , F } . In how many ways can we select a group of 3 letters from this set? Suppose we had to find the number of arrangements of 3 letters possible from those 6 letters. That number would be 6 P3 . Consider the permutations that contain the letters A, B and C. These are 3! = 6 in number, namely ABC, ACB, BAC, BCA, CAB and CBA. Now, what we want is the number of combinations and not the number of arrangements. In other words, the 6 permutations listed above would correspond to a single combination. Differently put, the order of things is not important; only the group/combination matters. This means that the total number of combinations of 3 letters from the set of 6 letters available to us would be 6 P3 / 3! since each combination is counted 3! times in the list of permutations. Thus, if we denote the number of combinations of 6 things taken 3 at a time by 6C3 , we have C3 =

6

6

P3 3!

In general, suppose we have n things available to us, and we want to find the number of ways in which we can select r things out of these n things. We first find the number of all the permutations of these n things taken r at a time. That number would be n Pr . Now, in this list of n Pr permutations, each combination will be counted r! times since r things can be permuted amongst themselves in r! ways. Thus, the total number of combinations of these n things, taken r at a time, denoted by nCr , will be

Cr =

n

n

Pr n! = r ! r !(n − r )!

You should now be able to appreciate the utility of the fundamental principle of counting. Using only a step-bystep application of this principle, we have been able to obtain an expression for n Cr . As we progress through the chapter, you’ll slowly realise that each and every concept that we discuss and each and every expression that we obtain follows logically as a consequence of this simple principle.

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Example – 2

Consider the word EDUCATION. (a) In how many ways can we select a pair of letters consisting of a vowel and a consonant ? Both vowels? Both consonants? (b) In how many ways can a 5-letter selection be made with more than 2 vowels? Solution: (a) We have 5 vowels and 4 consonants available in the word EDUCATION. A vowel can be selected in 5 ways while a consonant can be selected in 4 ways. Thus, a pair consisting of a vowel and a consonant can be selected in 5 × 4 = 20 ways. Two vowels can be selected in 5 C2 ways while two consonants can be selected in 4 C2 ways. (b) Since we want more than 2 vowels, we could have 3, 4 or 5 vowels in our 5-letter selection. Suppose we select 3 vowels for our 5-letter group. This can be done in 5 C3 ways. The two remaining letters (consonants) can be selected in 4 C2 ways. Thus, 5-letter groups containing 3 vowels can be formed in 5C3 × 4C2 ways. Similarly, if we had 4 vowels, the possible number of groups would be 5C4 × 4C1 . There’s only 1 group possible containing (all) the 5 vowels. Thus, the number of required selections is 5

C3 × 4C2 +5 C4 × 4C1 + 1

____________________________________________________________________________________ Now that we’re done with the introductions, lets move on and see some really interesting and diverse applications of the basic concepts covered till now.

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Section - OF 1 BASICS APPLICATIONS

Section - 4

Our approach in the applications of the basic concepts of counting will be to keep everything really basic - and try to do everything from a first principles approach. Memorization of formulae will do no good! A good deal of thinking is required. Lets start with proving certain combinatorial assertions.

Example – 3

Prove that n Cr =n Cn −r Solution: We can easily prove this relation using the expression we obtained for n Cr : n

Cr =

n! n! = = n Cn − r r !(n − r )! (n − r )!(n − (n − r ))!

However, what we would really like is not an analytical justification like the one above but a logical justification, that involves no mathematical manipulations and is instead purely based on the interpretation of n Cr . For this example, let us discuss such a logical justification in a good amount of detail. Suppose you have a group of 6 letters, say {A, B, C, D, E, F}. Out of this group, you want to count the number of subsets containing 4 letters. Consider a particular selection of 4 letters, say {B, D, E, F}. This selection can equivalently be obtained if we say that we exclude the group {A, C} from our original group of 6 letters. This means that each selected group of 4 letters corresponds to an excluded group of 2 letters. The number of (selected) 4-letter groups will therefore equal the number of (excluded) 2-letter groups, or in other words, to count the number of 4 letter groups, we can equivalently count the number of 2-letter groups. Thus, 6

C4 = 6 C 2

Generalising this logic, we obtain n Cr =n C n−r

Example – 4

Prove that n Cr =n−1 Cr + n−1 Cr −1 Solution:

You can prove this assertion yourself very easily analytically. Let us discuss a logical justification. The left hand side of this assertion says that we have a group of n people out of which we want to select a subset of r people; more precisely, we want to count the number of such r-subsets. Fix a particular person in this group of n people, say person X. Now, all the r-groups that we form will either contain X or not contain X. These are the only two options possible.

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To count the number of groups that contain X, we proceed as follows: we already have X; we need (r – 1) more people from amongst (n – 1) people still available for selection. Thus, such groups will be n−1Cr −1 in number. For the number of groups that do not contain X we need to select r people from amongst (n –1) options available. Therefore, such groups are n −1Cr in number. The total number of r-groups are hence

n −1

Cr −1 + n−1 Cr in number, which is the same as n Cr . With this example, you should begin

to realise the beauty of the logic and skill required in this subject.

Example – 5

Prove that n Pr = n−1 Pr + r n−1Pr −1 Solution: The analytical justification is again very straightforward and is left to you as an exercise. The left hand side of this assertion says that we need to count the number of arrangements of n people, taken r at a time. We again fix a particular person, say person X. All the possible r-arrangements will either contain X or not contain X. These are the only two (mutually exclusive) cases possible. If we do not keep X in our permutation, we have r people to select from a potential group of (n – 1) people. The number of arrangements not containing X will therefore be n −1 Pr . To count the number of permutations containing X, we first seat X in one of the r seats available. This can be done in r ways. The remaining (r – 1) seats can be filled by (n –1) people in n −1 Pr −1 ways. Thus, the number of arrangements containing X is r ⋅ n−1Pr −1 . These arguments prove that n Pr = n−1 Pr + r ⋅n −1 Pr −1

Example – 6 n (a) Prove that Cr =

n r

n −1

Cr −1

n (b) Prove that Cr =

n − r +1 n Cr −1 r

Solution: (a) Let us consider this assertion in a particular example, say with n = 6 and r = 4. This will make things easier to understand. Our purpose is to select 4 people out of 6 people, say the set {A, B, C, D, E, F}. To select a group of 4, we can first select a single person : this can be done in 6 ways. The rest of the 3 people can now be selected in 5C3 ways. The total number of groups possible would thus be

6 × 5C3 . But some careful thought will show that we have ‘overcounted’, doing the calculation this way. Mathematics / P & C, Binomial Theorem

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Suppose that in our first step, we select A. While selecting the remaining 3 persons, we then select B, C, D thus forming the group {A, B, C, D}. But this same group would have been formed had we selected B in our first step and A, C, D in the second step, or C in the first step and A, B, D in the second step, or D in the first step and A, B, C in the second step. We have thus counted the group {A, B, C, D} 4 times in the figure 6 × 5C3. The actual number of groups will hence be

6 × 5C3 . 4

We now generalise this: to select r people out of a group of n, we first select one person; this can be done in n ways. The remaining (r – 1) persons can be selected in n−1Cr −1 ways. The total number of r-groups thus becomes n × n –1Cr −1 . However, as described earlier, in this figure each group has been counted r times. The actual number of r-groups is therefore n

n × n −1Cr −1 Cr = r

(b) The logic for this part is similar to that of part - (a) To select r people out of n, we first select (r – 1) people out of n. This can be done in n Cr −1 ways. We now have n − (r − 1) = (n − r + 1) persons remaining for selection out of which we have to choose 1 more person. This can therefore be done in (n – r + 1)ways. The total number of r-groups thus becomes (n − r + 1) × nCr −1. However, each r-group has again been counted r times in this figure (convince yourself about this by thinking of a particular example). The actual number of r-group is thus n

Cr =

(n − r + 1) × nCr −1 r

Example – 7

Prove that n C0 + nC1 + nC2 + .......... + nCn = 2n Solution: This looks like a tough one! Lets first interpret what the left side means. n C0 is the number of ways in which we can select ‘nothing’ out of n things (there will obviously be only one such way: that we do nothing!). n C1 is the number of ways in which we can select 1 thing out of n. n Cr is the n

n number of ways of selecting r things out of n. We want the value of the sum ∑ Cr , which is the r =0

number of all groups possible of any size what so ever. Thus, our selection could be any size from 0 to n (both inclusive); what we want is the total number of selections possible. For example, consider the set {A, B, C}. The set of all possible selections that can be made from this set is {φ , {A}, {B}, {C}, {AB}, {AC} , {BC}, {ABC}}. Thus, 8 total different selections are possible (note that 8 = 23 ) Mathematics / P & C, Binomial Theorem

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To count the total number of selections possible if we have n persons, we adopt an individual’s perspective. An individual can either be or not be in our selection. Thus, we have two choices with respect to any individual; we either put him in our group or do not put him in our group. These two choices apply to every individual. Also, choosing or not choosing any individual is independent of choosing or not choosing another. Thus, the total number of ways in which an × 2 ×# × ........ ×%2 = 2n 2$### arbitrary number of individuals can be selected from n people is 2"## n times

This proves that n

∑ nCi = 2n i =0

Example – 8

Prove that

n+ m

Cr = nC0 mCr + nC1 mCr −1 + nC2 nCr −2 + ......... + nCr mC0

Solution: We interpret the left hand side as the number of ways of select r people out of a group of (n + m) people. Let this group of (n + m) people consist of n boys and m girls. A group of r people can be made in the following ways: Sl.

Group contains

1 2

0 boys, r girls 1 boy, (r − 1) girls 2 boys, (r − 2) girls

3 . . . r. (r − 1) boys, 1 girl (r + 1) r boys, 0 girsl

No. of ways possible n

C0 × mCr

n

C1 × mCr −1

n

C2 × mCr −2

n

Cr −1 × mC1

n

Cr × mC0

This table is self-explanatory. The ( r + 1) types of groups that have been listed are mutually exclusive. Thus, the total number of r-groups is n+ m

Cr = nC0 mCr + nC1mCr −1 + ........ + nCr mC0

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Example – 9

(a) Consider the set of letters {a, a, a, b, c, d }. How many permutations of this set exist? (b) Consider the set of letters {a, a, a, b, b, c, d , e}. How many permutations of this set exist? (c) Generalise the results of the previous two parts. If we have a set of ( m + n + p + .....) things where a particular thing, say X, is repeated m times, Y is repeated n times, Z is repeated p times, and so on, find the number of permutations of this set. Solution: The main issue now is that we have a repetition of things. If the things we have were all different, the number of permutations would have been easy to calculate. But now, with repetition of things, the number of permutations will change (it will actually decrease, if you think about it carefully). Let us calculate the permutations in this case from first principles (a) We have 3 (repeated) “a” letters, a “b”, a “c” and a “d”. In all, 6 letters. Consider, for a moment, that the 3 “a” s are all different. Let us denote the 3 different “a”s by a1, a2 and a3. Our set of letters is now {a1, a2 , a3, b, c, d }. The number of permutations of this set is simply 6

P6 = 6!

If we list down all these 6! permutations, we will see that 6 permutations from this list will correspond to only one permutation, had the “a”s been all the same. Why? Consider any particular permutation with the “a” s all different, say {b, a1, a2 , c, a3, d }. If we fix the letters “b”, “c” and “d”, the 3 different “a” s can be permuted amongst themselves in 3! = 6 ways. We now list down all the 6 permutations so generated on the left hand side in the figure below, and see that these 6 permutations correspond to a single permutation if the “a” s were all the same: b a1 a2 c a3 d b a1 a3 c a2 d b a2 a1 c a3 d baacad b a2 a3 c a1 d b a3 a1 c a2 d b a3 a2 c a1 d The 6 permutations on the left with different “a”s as correspond to a single permutation with the “a” s all same. Fig - 6

Thus, the actual number of permutation with the “a”s all same will be 6! 6! = = 120 3! 6

(b) We now have 3 repeated “a” s and 2 repeated “b” s, and a total of 8 letters. If we for a moment take the “a” s and “b” s as all different, the total number of permutations of this set of 8 letters would be 8 P8 = 8! Mathematics / P & C, Binomial Theorem

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However, once we list down these 8! permutations, we will see that (as in the previous part) 3! = 6 permutations in this list will correspond to a single permutation if the “a”s were all the same. Similarly, 2! = 2 permutations in this list will correspond to a single permutation if both the “b”s were the same. Thus, the actual number of permutations if “a”s and “b”s were the 8! same would be 3!2! (c) These results can now easily be generalised for this general set and the number of permutations will be (m + n + p + .....)! m !n ! p !...... ____________________________________________________________________________________ From this example once again, the power of the fundamental principle of counting should be quite evident. Using a logical development/extension of this principle, we see that we’ve been able to solve non-trivial questions like the one above. Example – 10

Consider the integral equation x1 + x2 = 4 where x1, x2 ∈ & . The non-negative solutions to this equation

can be listed down as {0, 4} , {1,3} , {2, 2} , {3,1} and {4, 0} . Thus, 5 non-negative integral solutions exist for this equation. We would like to solve the general case. How many non-negative, integral solutions exist for the equation x1 + x2 + ..... + xn = r

Solution: You might be surprised to know that this question can be solved using the general result obtained in the previous example. Can you think how? Let us consider an arbitrary integral equation, say x1 + x2 + x3 = 8. Consider any particular nonnegative integral solution to this equation, say {2,3,3} . We some how need to “tag” this solution in a new form; a form which is easily countable. This is how we do it. We break up the solution 2 + 3 + 3 = 8 as shown below: 11

111 = 8

...(1)

1=8

...(2)

1111111 = 8

...(3)

111

Similarly, 1 + 6 + 1=8 would be written as 1

111111

and 0 + 1 + 7 = 8 would be written as 1

and 0 + 0 + 8 = 8 would be written as 11111111 = 8 Mathematics / P & C, Binomial Theorem

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An alert reader must have realised the ‘trick’ by now. In each of (1), (2), (3) and (4), we have on the left hand side 8 “1” symbols and 2 “

” symbols, in different orders. Any non-negative integral ” symbols.

solution can thus be represented by a unique permutation of 8 “1” symbols and 2 “ Conversely, every permutation of 8 “1” symbols and 2 “ negative integral solution to the equation .

” symbols represents a unique non-

Thus, the set of non-negative integral solutions to the equation and the set of permutations of 8 “1” symbols and 2 “

” symbols are in one-to-one-correspondence. To count the required number

of solutions, we simply count the permutations of 8 “1” symbols and 2 “ described in the last example would be

” symbols, which as

(8 + 2)! 10! = = 45 8!2! 8!2!

This beautiful artifice described about should make it clear to you the significance of (and the challenge of producing!) elegant proofs/solutions. We now generalise this result. Any non-negative integral solutions to the equation x1 + x2 + ..... + xn = r can be represented using r “1” symbols and n – 1 “ number of permutation of these symbols will be

” symbols. The total

(n + r − 1)! n + r −1 = Cr r !(n − 1)! and hence, this is the required number of solutions.

Example – 11

Consider a rectangular integral grid of size m × n. For example, a 4 × 5 integral grid is drawn alongside: A person has to travel from point A(0, 0) to the diagonally opposite point C(m, n). He moves one step at a time, towards the east or towards the north (that is, never moves towards the west or south at any time). How many distinct paths exist from the point A to the point C?

D

C

A

B Fig - 7

Solution: Let us draw a random path on our 4 × 5 grid in Fig. 6 and think of some way to mathematically specify/describe this path D

C

A

B

A random path across the 4 × 5 grid that our travelling person can follow. The question now is: How do we mathematically characterise this path?

Fig - 8 Mathematics / P & C, Binomial Theorem

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Suppose you had to describe this path to a blind person. If you use E for a step towards the east and N for a step towards the north, you’d tell the blind person that the travelling person took the following path. “EENENEENN” This string that we just formed should immediately make you realise how to calculate the number all the possible paths. We have 5 “E” steps and 4 “N” steps. Any permutation of these 9 steps gives rise to a different unique path. For example, the string “ E E E E E N N N N ” is the path that goes straight east from A to B and then straight north from B to C. Thus, any path can be uniquely characterised by a permutation of these 9 steps. The number of permutations of these 9 9! . This is therefore the number of different paths 5!4! that the travelling person can take from A to C. For an m × n grid we will have (m + n) total steps, m of them being “E” s and the remaining n being “N” s Thus, the number of possible paths

letters, 5 of which are “E”s and 4 are “N”s, is

is

(m + n)! . m !× n !

Example – 12

(a) We have m apples, n oranges and p bananas. In how many ways can we make a non-zero selection of fruit from this assortment? (b) How many factors does 144000 have? In general, how many factors does a natural number N have? Solution: These two seemingly unrelated questions have exactly the same approach to their solutions! Before reading the solution, can you imagine how? (a) The most important point to realise in this question is the nature of objects to be selected. We have m apples. These m apples are exactly identical to each other. You cannot make out one apple from another. This means that if you have to choose r apples out of n, there’s only one way of doing it: you just pick (any of the) r apples. It doesn’t matter which apples you choose, because all the apples are identical. Thus, only 1 r-selection is possible. So how many total selections are possible? We either select 0 apples, 1 apple, 2 apples, .... r apples, ... or m apples. Thus (m + 1) ways exist to select apples. Make sure you properly understand the essence of this discussion and why n choosing r apples out of n can be done in only 1 and not Cr ways.

We therefore have (m + 1) ways to select apples, (n +1) ways to select oranges and (p + 1) ways to select bananas. The selection of a particular fruit is independent of the selection of another fruit. Hence, we have (m +1) (n +1) (p +1) ways to select a group of fruit.

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But wait! In the (m +1) ways of selecting apples, there’s also one way in which we select no apple. Similarly, in the (n + 1) ways of selecting oranges, there’s one in which we select no orange, and in the (p + 1) ways of selecting bananas, there’s one in which we select no banana. Thus, in the product (m +1) (n +1) (p +1), there’ll be one case involving no fruit of any type. We have to exclude this case if we want a non-zero selection of fruit. Therefore, the number of ways of making a non-zero selection is (m +1)(n +1)(p +1) —1. (b) You might be wondering how the factors problem is related to the first part! Read on. Lets consider a smaller number first. Take 60, for example, and list down all its factors (including 1 and 60):

{1, 2,3, 4,5, 6,10,12,15, 20,30, 60} From elementary mathematics, you know that any number can be factorized into a product of primes. For example, 60 can be written in its prime factorization from as:

60 = 2 × 2 × 3 × 5 = 22 ⋅ 31 ⋅ 51

Such a representation exists for every natural number N. Can we somehow use this representation to find the number of factors? Consider any factor of 60, say 12, The prime factorization from of 12 is 22 ⋅ 31 . Similarly, this representation for 15, for example, is 31 ⋅ 51 and for 30 would be 21 ⋅ 31 ⋅ 51 . With discussion in our mind, we rephrase our problem: We have 60 whose prime representation is 22 ⋅ 31 ⋅ 51 . Thus, we have 2 twos, 1 three and 1 five with us. (You could imagine that we have 2 apples, 1 orange and 1 banana). To form a factor of 60, what we have to do is to make a selection of prime factors from amongst the available prime factors. For 60, we have 2 twos (apples), 1 three (orange) and 1 five(banana). In how many ways can we make our selection? The last part tells is that we can do it in (m + 1)(n + 1)( p + 1) ways or (2 + 1)(1 + 1)(1 + 1) = 12 in this particular case. (We also allow no selection of any prime factor - this corresponds to the factor 1 of 60) Do you feel the elegance of this solution? For the general case of a natural number N whose prime representation is of the α α α form p1 1 p2 2........ pn n , the number of factors (including 1 and N) is (α1 + 1)(α 2 + 1)........(α n + 1) For example, 144000 can be represented as

144000 = 27 ⋅ 32 ⋅ 53 The required number of factors is (7 + 1)(2 + 1)(3 + 1) = 96 Mathematics / P & C, Binomial Theorem

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Example – 13

12 person are sitting in a row. In how many ways can we select 4 persons out of this group such that no two are sitting adjacent in the row? Solution: To make the question more clear, here’s a valid and a non-valid selection:

Valid

Non-valid Fig - 9 We have to find some (mathematical) way of specifying a valid-selection. An obvious method that strikes is this: In our row, we represent every unselected person by U and every selected person by S. Thus, the valid selection in the figure above becomes:

U S U U S UU S U S UU To count the numbers of valid selections, we count the number of permutation of this string above, consisting of 8 “U” symbols and 4 “S” symbols, subject to the constraint that no two “S” symbols are adjacent. To count such permutation, we first fix the 8 “U” symbols. There will then be 9 blank spaces generated for the “S” symbols as shown below:

__U__U__U__U__U__U__U__U__ From these 9 blank spaces, any 4 spaces can be chosen and the “S” symbols can be put there and it’ll be guaranteed that no two “S” symbols will be adjacent. Thus, our task now reduces to just selecting 4 blank spaces out of the 9 available to us, which can be done simply in 9 C4 ways. These are the required number of selections.

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Example – 14

DIVISION INTO GROUPS

We will use a standard pack of 52 cards to illustrate the concepts involved. (a) (i) In how many ways can a pack of 52 cards be divided equally into 4 sets? (ii) Equally among 4 players? (b) (i) In how many ways can this pack be divided into 4 sets of 9, 13, 14 and 16 cards each? (ii) Distributed among 4 players with 9, 13, 14 and 16 cards? (c) (i) In how many ways can this pack be divided into 4 sets of 10, 14, 14 and 14 cards each? (ii) Distributed among 4 players with 10, 14, 14 and 14 cards? (iii) Distributed among 4 players with 12, 12, 14 and 14 cards? Solution: For this question, you must understand that the division of a set into a number of subsets is different from the division of a set into a number of subsets and then distribution of those subsets to different persons. For example, suppose you are alone by yourself and you are dividing your deck of cards into 4 sets. What you’ll do is do the division of the deck and keep the 4 sets in front of you. Which set lies where in front of you doesn’t matter. Your only objective was to divide the pack into 4 sets which you did. Precisely speaking, the order of the groups doesn’t matter. On the other hand, suppose you were playing a game of Bluff with 3 other people. Before starting the game, you divide the deck into 4 sets of 13 cards each. But now you have to give each set of 13 cards to a different person. Where a set of cards goes matters. In other words, the order of the groups matters. As another example, suppose you had to divide a class into 3 sub-groups. You do your job-the order of the groups doesn’t matter, you just have to divide them. But suppose you had to divide the class into 3 sub-groups and place each sub-group in a different bogey of a roller coaster. In this case the order of the groups matters because which sub-group sits in which bogey matters. Once again, make sure that you understand very clearly when the order of the groups matters and when it doesn’t (a) (i) We want to divide the pack into 4 equal sets of 13 cards each. We proceed as follows; First select a group of 13 cards from the 52 cards; call this group A. This can be done in 52 C13 ways. Group B of another 13 cards can now be selected from the remaining 39 cards in 39 C13 ways. Group C of another 13 cards from the remaining 26 can now be selected in 26 C13 ways. This automatically leaves group D of the last 13 cards. Thus, the division can be carried out in 52

C13 × 39C13 × 26C13 ways.

However, notice that the order of the groups doesn’t matter, as discussed earlier. This means that even if our order of group selection was BACD or ADBC (or for that matter any other permutation of ABCD), such a selection would essentially be the same as ABCD. The number of permutations of ABCD is 4! = 24.

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Thus, the actual number of ways of division is 1 52 ( C13 × 39C13 × 26C13 ) 24

=

1 52! 39! 26! × × 24 13! 39! 13! 26! 13! 13!

=

52! (13!)4 ⋅ 4!

(ii) If we had to distribute the cards equally among 4 players, the order of group selection would have mattered. For example, ABCD would be different from BACD. Thus, we 52! don’t divide by 4! in this case. The number of possible distributions is ( )4 13!

(b) (i) For this part, observe that the group sizes are all unequal. This is a somewhat different situation than the previous part where the group sizes were equal. We’ll soon see why. We follow a similar sequence of steps as described earlier. (i) Select group A of 9 cards from the deck of 52 : 52 C9 ways (ii) Select group B of 13 cards from the remaining 43 : 43C13 ways (iii) Select group C of 14 cards from the remaining 30 : 30 C14 ways. (iv) This automatically leaves group D of the remaining 16 cards. The number of ways for this division is 52

C9 × 43C13 × 30C14

=

52! 43! 30! × × 9! 42! 13! 30! 14! 16!

52! = 9! 13! 14! 16!

From now on, note and remember that m+ n+ p + q Cm × n+ p + qCn × p + qC p × qCq = (m + n + p + q) m! n! p ! q !

Why don’t we divide by 4! here as we did earlier, even when the order of groups doesn’t matter? This is because the group sizes are unequal. A particular selection in a particular order, say ABCD, will never be repeated in any other order of selection. Our order of selection (in terms of group size) is 9, 13, 14 and 16 cards. Carefully think about it; a particular selection like ABCD will be done only once (and not 4! times as in the previous part due to equal group sizes) Mathematics / P & C, Binomial Theorem

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(ii) If we had to distribute the 4 sets to 4 players, the situation changes because which subset of cards goes to which player matters. A particular group of sets of cards, say ABCD, can be distributed among the 4 players in 4! ways. Thus, now the number of distributions will be no. of ways of division of the no. of ways of Distribution × deck as specified into the 4 sets of those sets to the 4 players 52! ( ) = × 4! 9! 13! 14! 16! (c) (i) In this part, the situation is a hybrid of the previous two parts. 3 of the 4 groups are equal in size while the 4th is different. We first select 10 cards out of 52. This can be done in 52C10 ways. The remaining 42 cards can be divided into 3 equal groups (by the logic of part -(a)) in 42! 1 × (14!)3 3! (Division by 3! is required since the order of groups doesn’t matter. Thus, a particular order of 15-card groups, say ABC, is, for example, the same as ACB) Thus, the required number of ways is 42! 1 × (14!) 3 3!

52

C10 ×

=

52! 1 × 3 10!(14!) 3!

...(1)

(ii) For the second question where we have 4 players and we want to give them 10, 14, 14 and 14 cards, we already have obtained the number of possible division in (1). To distribute the subsets, each division of 4 sets can be given to the 4 players in 4! ways. Thus, the number of distributions of the 4 subsets is =

52! 1 4 × 52! × × 4! = 3 3 10!(14!) 3! 10!(14!)

We can also look at this current problem in the following way. We want to distribute the 52 cards to the 4 players in sets of 10, 14, 14 and 14. We first decide which guy to give the set of 10 cards. This can be done in 4 ways. Now we choose 10 cards for him (in possibly 52 C10 ways). The other three players will now get the remaining 42 cards equally. We can select 12 cards for one of them in 42 C14 ways. The third one can get another 14 cards in 28 C14 ways and finally the fourth one gets the remaining 14 cards. Thus, the number of possible ways of distribution is 4 × 52C10 × 42C14 × 28C14 = Mathematics / P & C, Binomial Theorem

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(iii) Finally, we now have two sets of one size and two of another size. We first just carry out the division into subsets, without assigning any subset to any player. This can be done in C12 ×

52

! Group A of 12 cards

40

×

C12

Group B of 12 cards

=

×

28

C14

! Group C of 14 cards

14

C14

! Group D of 14 cards

1 2!

×

↓ Groups A and B are of the same sizer

×

1 2!

↓ Groups C and D are of the same sizer

52! 1 × 2 2 (12!) (14!) ( 2!)2

Once the division of the deck into the 4 subsets has been accomplished, we assign a subset to each of the 4 players. This can be done in 4! ways. Therefore, the possible number of ways of distribution of the cards to the 4 players is 52! 1 × × 4! 2 2 (12!) (14!) ( 2!)2 Example – 15

CIRCULAR PERMUTATIONS

In this example, we talk about circular permutations, i.e., arrangements in a circular fashion (a) In how many ways can n people be seated around a circular table? (b) We have a group of 5 men and 5 women. In how many ways can we seat this group around a circular table such that: (i) all the 5 women sit together (ii) no two women sit together. (c) In how many ways can a necklace be formed from n different beads? Solution: Circular permutations are somewhat different from linear permutations. Lets see why. Consider the letters A, B, C and D. The 4 linear permutations ABCD, BCDA, CDAB and DABC correspond to a single circular permutation as shown in the figure below: A

B

In a circular permutation, only the relative ordering of the symbols matters. The 4 linear permutations ABCD, BCDA, CDAB and DABC all have the same relative ordering so that they correspond to a single circular permutation.

D

C Fig - 10

This means that the number of circular permutations of ABCD is only Mathematics / P & C, Binomial Theorem

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(a) We have n people, say P1, P2............Pn. As just described, n linear permutations of these people will correspond to a single circular permutation as depicted in the figure below:

PP 1 2...............Pn P2 P3...............Pn P1 P3P4...............PP 1 2 : : : : Pn P1...............Pn −2 Pn −1 n linear permutations correspond to →

PP 1 2...........Pn

a single circular permutation because of the same relative ordering.

Thus, the number of circular permutations is

n! = (n − 1)! n

We can also arrive at this number in another way. We take a particular person, say P1 , and seat him anywhere on the table. Once P1 ' s seat becomes fixed, the rest of the (n – 1) seats bear a fixed relation to P1 ' s . In other words, once P1 ' s seat becomes fixed, we can treat the (n – 1) seats left as a linear row of (n –1) seats. Thus, the remaining (n –1) people can be seated in (n –1)! ways. (b) Let the 5 women be represented by W1, W2 ,W3,W4 , and W5 and the 5 men by M1, M 2 , M 3, M 4 , and M 5 . Since we need all the women to sit together, we first treat all the 5 women as a single entity W. Now, the 5 men and the entity W can be seated around the table in ((5 + 1) − 1)! ways, i.e., 120 ways. (i) Once the 5 men and the entity W have been seated, we now permute the 5 women inside the entity W. This can be done in 5! = 120 ways. The total number of ways is thus 120 × 120 = 14400 W3 W 4

1

W

W2

W

The entity W

5

M2

(i) First seat the 5 men and the entity W

M4

(ii) Then permute the women inside the entity W. M1

M3 M5 Fig - 11

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(ii) Since we want no two women to sit together, we first seat all the 5 men, which can be done in (5 –1)! = 24 ways. Seating the 5 men first creates 5 non-adjacent seats where the women can then be seated in 5! = 120 ways.

M1

M4

(i) First seat the 5 men, 4! ways are possible

M3

M2

(ii) Then seat the 5 women in the 5 spaces created as shown in the figure 5! ways are possible

M5

Fig - 12

The total number of ways is this 5! × 4!. (c) The situation of a necklace is slightly different than that of seating people around a circular table. The reason is as follows: Suppose we have 5 beads A, B, C, D and E. Consider two circular necklaces of these 5 beads A B

A E

C

E

D

B

D

Necklace-1

C

Necklace-2 Fig - 13

Are these two necklaces different? No, because a-necklace can be worn from both ways. Necklace-2 is the same as necklace-1 if I look into it from the other side of the page. In other words, for a necklace, a clockwise permutation and its corresponding anti-clockwise permutation are identical. Thus, the number of circular permutations would reduce by a factor 1 of two, i.e., the number of different necklaces possible is (n − 1)!. 2

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TRY YOURSELF - I

Q. 1

How many words can be made with the letters of the word “TECHNOLOGY” which do not begin with T but end with Y ?

Q. 2

In how many ways can the letters of the word CINEMA be arranged so that the order of vowels do not change?

Q. 3

How many sides are there in a polygon which has 35 diagonals?

Q. 4

In how many ways can 6 boys and 4 girls sit in a row so that no boy is between two girls?

Q. 5

How many words can be made with the letters of the word BHARATI so that all the vowels are consecutive?

Q. 6

Let n ∈ N and 300 < n < 3000. If n is made of distinct digits by taking from 0, 1, 2, 3, 4, 5 then find the greatest possible number of values of n.

Q. 7

How many words can be made with the letters of the word INSTITUTION such that vowels and consonants alternate?

Q. 8

How many different committees of 5 members can be formed from 6 men and 4 ladies if each committee is to contain at least one lady?

Q. 9

Six Xs are to be placed in the squares of the given figure, such that each row contains at least one X. In how many different ways can this be done?

Q. 10 In how many ways can 16 identical mangoes be distributed among 4 persons if none gets less than 3 mangoes?

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Section -1 MORE APPLICATIONS

Section - 5

By now, you should have a pretty good idea about the basics of permutations and combinations. In this section, we will encounter more advanced problems. All the questions discussed in the following pages are directly or indirectly based on the concepts already discussed in the previous sections. In case you find anything confusing, refer to the relevant parts again.

Example – 16

On a standard 8 × 8 chessboard, find the (a) number of rectangles (b) number of squares Solution: (a) Visualise any arbitrary rectangle on the chessboard, say the one depicted on the left in the figure below:

P

Q

An arbitrary rectangle on the chessboard. How to specify this rectangle is explained in the figure on the right

X Y To mathematically characterise the rectangle that we selected, we specify the pair of vertical edges X and Y and the pair of horizontal edges P and Q. Doing so uniquely determines the rectangle

As explained in the figure above, any rectangle that we select can be uniquely determined by specifying the pair of lines X and Y that make up the vertical edges of the rectangle and the pair of lines P and Q that make up the horizontal edges of the rectangle. On the chessboard, there are 9 vertical lines available to us from which we have to select 2. This can be done in 9C2 ways. Similarly, 2 horizontal lines can be selected in 9C2 ways. Thus, the total number of rectangles that can be formed is 9C2 × 9C2 = 1296 . (b) To select a square, observe that the pair of lines X and Y must have the same spacing within X and Y as the pair of lines P and Q. Only then can the horizontal and vertical edges of the selected rectangle be of equal length (and thus, the selected rectangle is actually a square). Mathematics / P & C, Binomial Theorem

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In how many ways can we select a pair (X, Y) of lines which are spaced a unit distance apart ? Its obviously 8. Corresponding to each of these 8 pairs, we can select a pair (P, Q) of lines in 8 ways such that P and Q are a unit distance apart. Thus, the total number of unit squares is 8 × 8 = 64 (This is obvious otherwise also). Now we count the number of 2 × 2 squares. In how many ways can we select a pair (X, Y) of lines which are 2 units apart ? A little thought shows that it will be 7. Corresponding to each of these 7 pairs, we can select a pair (P, Q) of lines in 7 ways such that P and Q are 2 units apart. Thus, the total number of 2 × 2 squares is 7 × 7 = 49. Reasoning this way, we find that the total number of 3 × 3 squares will be 6 × 6 = 36, the total number of 4 × 4 squares will be 5 × 5 = 25 and so on. Thus, the total number of all possible squares is 64 + 49 + 36 + ........... + 4 + 1 = 204

↑ 1×1 squares

↑ 2× 2 squares

↑ 7× 7 squares

↑ 3×3 squares

↑ 8×8 squares

Example – 17

Consider an n-sided convex polygon. (a) In how many ways can a quadrilateral be formed by joining the vertices of the polygon ? (b) How many diagonals can be formed in the polygon? Solution: (a) Observe that a selection of any 4 points out of the n vertices of the quadrilateral will give rise to a unique quadrilateral (since the polygon is convex, the problem of our selection containing all or 3 collinear points does not exist). 4 points out of n can be selected in nC4 ways. Thus, we can have nC4 different quadrilaterals. (b) To form a diagonal, we need 2 non-adjacent vertices ( because 2 adjacent vertices will form a side of the polygon and not a diagonal). The total number of ways of selecting 2 vertices out if n is nC2 . This number also contains the selections where the 2 vertices are adjacent. Those selections are simply n in number because the polygon has n sides. Thus, the total number of diagonals is n

C2 − n

=

n( n − 1) −n 2

=

n(n − 3) 2

Example – 18

Give a combinatorial (logical) justification for this assertion: n

C0 + n +1C1 + n + 2C2 + ... + n + rCr =

Mathematics / P & C, Binomial Theorem

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Solution: The right hand side tells us that we have to select r persons out of a group of n + r + 1 persons. To do so, we consider any particular group of r persons from these n + r + 1 persons. Specify these r persons by the symbols A1 , A2 ... Ar . Now, to count all the possible r-groups from this group of n + r + 1, we consider the following mutually exclusive cases: (1)

The r -group does not contain A1 Such r-groups can be formed in

(2)

n+ r

Cr ways since we have to select r people out of n + r.

The r -group contains A1 but not A2 We have to select (r – 1) people out of ( n + r − 1) because we already have selected A1 so we need only r –1 more people and since we are not taking A2 , we have ( n + r − 1) people to choose from. This can be done

(3)

(r)

n + r −1

Cr −1 ways.

The r -group contains A1, A2 but not A3 We now have to select (r – 2) people out of (n + r − 2) . This can happen in ways. Proceeding in this way, we arrive at the last two possible cases.

n+ r −2

Cr − 2

' The r -group contains A1, A2 ... Ar – 1 but not Ar . We need to select only 1 person out of (n + 1) available for selection. This can be done in n +1

C1 ways.

(r + 1)The r-group contains A1, A2 ... Ar In this case, our r-group is already complete. We need not select any more person. This can be done in nC0 or equivalently 1 way. Convince yourself that these (r + 1) cases cover all the possible cases that can arise in the formation of the r - groups. Also, all these cases are mutually exclusive. Thus, adding the number of possibilities of each case will give us the total number of r-groups possible, i.e. n

C0 + n +1C1 + n + 2C2 + ... + n + r −1Cr −1 + n + r Cr =

n + r +1

Cr

Example – 19

How many distinct throws are possible with a throw of n dice which are identical to each other, i.e. indistinguishable among themselves ? Solution:

The important point to be realised here is that the dice are totally identical. Suppose we had just 2 dice, say die A and die B. Suppose that, upon throwing these dice, we get a “two” on A and a “three” on B. This case would be the same as the one where we get a “three” on A and a “two” on B because we cannot distinguish between A and B. What we are concerned with is only what numbers show up on the top of the dice. We are not concerned with which die shows what number. This means that if we have n dice and we throw them, we are only concerned with how many “ones”, “twos”, “threes” etc show on the top faces of the dice; we are not at all interested in which die throws up what number.

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If we denote the number of “ones” we get by x1 , number of “twos” we get by x2 and so on, we will have x1 + x2 + ... + x6 = n Thus, the total number of distinct throws will be simply the number of non-negative solutions to this integral equation. As discussed earlier, this number will be

n + 6 −1

C6 −1 =

n+5

C5 .

What would be the number of distinct throws if the n dice were not identical? Example – 20

Give combinatorial arguments to prove that n

∑r⋅

n

r =1

Cr = n ⋅ 2n −1

Solution: Let us first interpret what the left hand side of this assertion says. Suppose we have a group of n people. We select a sub-group of size r from the group of n people. This can be done in nCr ways. Once the sub-group has been formed, we select a leader of that sub-group, and send that sub-group on an excursion. The leader can be selected in r ways. Thus, the total number of different ways in which an r - group can be formed with a unique leader is r × nCr . Now r can take any integer value from 1 to n, i.e. 1 ≤ r ≤ n . Thus, the total number of all possible n

sub-groups, each sub-group being assigned a unique leader, will be

∑r⋅ r =0

hand side of our assertion.

n

Cr which is the left

To prove this equal to the right hand side, we count the sub-groups from a different angle. We count all those sub-groups in which a particular person, say A, is the leader. Since A is the leader, A is fixed in our sub-group. For each of the remaining (n – 1) people, we have two options. We either put the person in the group led by A or we don’t. Thus, the total number of sub-groups in which A is the leader will be 2"# × 2# ×$## 2 × ... ×%2 = 2n −1 ( n −1) times

Since any of the n persons can be the leader, and under each person’s leadership, 2n−1 groups can be formed, the total number of sub-groups, each sub-groups under some unique person’s leadership is n ⋅ 2n −1 . This proves our assertion that n

∑r⋅ r =1

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Example – 21

A composition of a natural number N is a sequence of non-zero integers {a1, a2 .......... ak} which add up to N. How many compositions of N exist? Solution: Let us make the question more clear by taking a particular example for N, say N = 4. As described in the question, the compositions of N = 4 will be

{4}, {1, 3},{2, 2}, {3,1}, {1,1, 2}, {1, 2,1}, {2,1,1} and {1,1,1,1} which are 8 in number. Observe carefully the compositions listed out. How can we characterize each of these compositions mathematically? Recall the problem of finding the number of non-negative solutions of the integral equation x1 + x2 + ....... + xn = r where each solution corresponded to a unique permutation of r “ 1 “symbols and n - 1 " " symbols. Can we do something like that here? In other words, can we represent each composition in another form whose permutations are easier to count? It turns out that we can, as follows: {4}

=

1 1

1

1

{1, 3}

=

1+1

1

1

{3, 1}

=

1

1

1+ 1

{2, 2}

=

1

1 + 1

1

{1, 1, 2}

=

1+1 + 1

1

{1, 2, 1}

=

1+1

{2, 1, 1}

=

1 1 + 1+ 1

{1, 1, 1, 1} =

1+1 + 1+ 1

1+ 1

On the right hand side, we have 4 “1”s and thus 3 blank spaces between the 4 “1” s. We can insert “+” signs in these blank spaces; each different arrangement of “+” signs in these blank spaces will correspond to a different composition. To count the number of these arrangements, we proceed as follows: For each blank space, we have 2 options, we can either insert or not insert a “+” sign into that space. There are 3 blank spaces; so the total number of all arrangements of “+” signs in the 3 blank spaces is 2 × 2 × 2 = 8 (which is the number of compositions we already listed out). In the general case, we will have (N – 1) blank spaces and 2 options for each such blank space. Thus, the total number of ways in which we can arrange “+” signs in these blank spaces, and 2 × 2 × 2 × ........2 = 2 N −1 therefore, the total number of compositions, will be "##$##% ( N −1) times

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Example – 22

Prove logically the following assertion: n

∑2

k

k =0

⋅ n Ck = 3n

Solution: Suppose that we have to form a string of length n, consisting of only letters from the set {A, B, C}. Thus, we have 3 options to fill any particular place in the string: We fill that place with either A, B or C. Thus the total number of different strings of length n would be 3"##$##% × 3 × 3 × ........ × 3 = 3n n times

We now approach the task of formation of these strings from a different perspective. Suppose that our string contains a total of r “A” s. How many such strings will exist? We first select r places out of n which we will fill with “A”. This can be done in nCr ways. For each of the remaining (n – r) places, we have two options; we either fill it with “B” or “C”. Thus, the number of strings containing r “A” s will be nCr .2n–r. Now we vary r from 0 to n and thus get the total number of strings as n

∑ r =0

n

n

Cr ⋅ 2 n − r = ∑ n Cn − k ⋅ 2 k

( where k = n − r )

k =0

n

(since

= ∑ n Ck ⋅ 2 k k =0

n

∑

Thus,

k =0

n

n

Cn − k = n Ck )

Ck ⋅ 2 k = 3n

Example – 23

Prove logically the following assertion: n

Cr + n −1Cr + n − 2Cr + ...... + r Cr = n +1Cr +1

Solution: The right hand side says that we have to select (r + 1) people out of a group of (n +1). To do so, we list down the following (mutually exclusive) cases which exhaust all the possible cases: (1)

The group contains A1

Mathematics / P & C, Binomial Theorem

:

n

Cr ways (since we have to select r people more apart from A1 from n that are available for selection)

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(2)

n −1

:

Cr ways (since we have to select r people more apart from A2 from (n – 1) that are available for selection)

:

n−2

Cr ways (since we have to select r people more apart from A3 from (n – 2) that are available for selection)

(n – r) The group does not contain A1, A2 ....... An–r–1 but contains An–r

:

r +1

Cr ways (since we have to select r people more apart from An–r from the (r + 1) that are available for selection

(n – r + 1) The group does not contain

:

r

(3)

The group does not contain A1 but contains A2

The group does not contain A1 and A2 but contains A3

'

A1 , A2 ... An − r but contains An − r +1

Cr ways (since we have to select r

more people apart from rn − r +1 from the remaining r that are available for selection)

Convince yourself that all these cases are mutually exclusive and exhaust all the possibilities. Thus, the total number of (r + 1) - groups from (n + 1) people is n +1

Cr +1 = nCr + n −1Cr + n− 2Cr + ... + r Cr

Example – 24

Consider n points in a plane such that no three of them are collinear. These n points are joined in all possible ways by straight lines. Of these straight lines so formed, no two are parallel and no three are concurrent. Find the number of the points of intersection of these lines exclusive of the original n points. Solution: To gain more insight into the problem, let us consider the case when n = 4.

A B

1

3 C D 2 Fig - 15 Mathematics / P & C, Binomial Theorem

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We originally have 4 points in the plane labelled as A, B, C and D. When we draw all the straight lines possible (by joining every possible pair of points), we see that 3 new intersection points are generated, labelled as 1, 2 and 3. As we now discuss the general case, refer to this figure for help. Since we have n points, the number of straight lines that can be generated is equal to the number of pairs of points that can be selected from n points, which is equal to nC2 . Every pair of straight lines so generated will intersect at some point. Thus, the total number of intersection points if we count this way should be equal to the number of pairs of straight lines possible from the nC2 lines, which is (

n

C2 )

C2 . However, observe that in this number, the original intersection points have also been counted, and that too multiple times. Lets determine how many times a particular point, say A has been counted. (n – 1) lines pass through A. Thus n −1C2 pairs of straight lines are possible from these (n – 1) lines. Thus, A has been counted n −1C2 times. This means that the n original points have been counted n × n−1C2 times. To calculate the new intersection points, we subtract this number from the (supposed) number of intersection points we obtained earlier. Thus, the actual number of new intersection points is C2 − n × n −1C2

( n C2 )

n( n − 1) n( n − 1) − 1 2 2 − n × (n − 1)(n − 2) = 2 2 n(n − 1)(n 2 − n − 2) n( n − 1)(n − 2) − 8 2 2 n(n − 1){(n − n − 2) − 4(n − 2)} = 8 =

n(n − 1)(n 2 − 5n + 6) 8 n( n − 1)( n − 2)(n − 3) = 8 Verify that this formula works for n = 4 =

Example – 25

How many 4 letters strings can be formed from the letters of the work INEFFECTIVE ? Solution: Observe that some letters repeat more than once: Letter E F I C T V N Frequency 3 2 2 1 1 1 1 This means that our string of 4 letters could contain repeated letters. (It’s thus obvious that we cannot straightway use n Pr to evaluate the number of strings. Mathematics / P & C, Binomial Theorem

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In such a case, we divide the types of strings that we can form into different mutually exclusive cases: Case 1:

All 4 letters are different In this case, we have 7 letters to choose from and we have to arrange them in 4 places. The number of such strings will be 7 P4 = 840

Case 2:

2 letters are same, the other 2 are different We have 3 types of letters (E, F and I) that can be repeated twice. Thus, the twicerepeated letter can be selected in 3 ways. Once that is done, the rest of the 2 letters can be selected from amongst the remaining 6 options in 6C2 = 15 ways. Once we’ 4! ve formed the combination of the 4 letters we can permute them in = 12 ways. The 2! “2!” occurs in the denominator because of the twice-repeated letter. Thus, the total number of such strings will be 3 ×15 ×12 = 540 .

Case 3:

2 letters are the same, the other 2 are also the same For example, the string “EFEF” will be such a string. There are 2 letters now that we want to be twice-repeated; observe that there are only 3 types of letters (E, F and I) that can be twice repeated. Thus, the letters that will occur in the string can be selected in 3C2 = 3 ways. Once that is done, we can 4! = 6 ways. permute the 4 letters in 2!2! The total number of such strings is therefore 3 × 6 = 18.

Case 4:

3 letters are the same, the 4th is different Only “E” can be repeated thrice so that E must occur in this string. The 4th letter can 4! be selected in 6 ways. Then, the combination so formed can be permuted in = 4 3! ways. The number of such strings is 6 × 4 = 24. Verify that these four cases are mutually exclusive and they exhaust all the possibilities. The total number of strings is 840 + 540 + 18 + 24 = 1422

Example – 26

(a) (b)

In how many ways can 3 girls and 9 boys be seated in 2 vans, each having numbered seats with 3 seats in the front and 4 seats at the back? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats?

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Solution: (a) We have 12 people to seat and 14 available seats. We first select the 12 seats on which to seat these people. This can be done in 14C12 ways. Once that is done, we can permute the 12 people amongst these 12 seats in 12! ways. Thus, the total number of seating arrangements is 14C12 × 12! . (b) Since the 3 girls need to be seated together in one back row, we first select the back row. This can be done in 2 ways since there are 2 vans. In this row, 3 adjacent seats can be selected in 2 ways. "#$#%

"#$#% selection of 3 adjacent seats can be done in 2 ways.

In the 3 adjacent seats, the 3 girls can be permuted in 3! ways. Thus, the girls can be seated in 2×2×3! = 24 ways. We now have 11 remaining seats. The 9 boys can be seated in these 11 seats in 11P9 ways. Thus, the total number of ways to seat the entire group is 24 × 11P9 = 24 ×

11! = 12 × 11! = 12! . 2!

Example – 27

We have 3n objects, of which n are identical and the rest are all different. In how many ways can we select n objects from this group ? Solution: To select n objects, we select k objects from the identical ones and the remaining (n – k) from the different ones. ( k will vary from 0 to n) As discussed somewhere earlier, from the group containing identical objects, there will always be only 1 way of selection. From the group of 2n nonidentical objects, (n – k) objects can be selected in 2nCn −k ways. Thus, the group of n objects containing k identical objects and the remaining (n – k) as nonidentical objects can be formed in 1× 2 nCn −k = 2 nCn − k ways. The total number of ways is S where n

S = ∑ 2 nCn − k = 2 nCn + 2 nCn −1 + ... 2 nC0 k =0

A closed-form expression for S can be obtained by the following manipulation : 2 S = 2( 2 nCn + 2 nCn −1 + ... + 2 nC0 ) 2 nCn + 2 nCn −1 + ... + 2 nC0 = + 2 nCn + 2 nCn −1 + ... + 2 nC0 Mathematics / P & C, Binomial Theorem

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2 nCn + 2 nCn −1 + ... + 2 nC0 = + 2 nCn + 2 nCn +1 + ... + 2 nC2 n

For all the terms in the lower row, we can use 2n 2n Cr = C2 n− r

= 2 nCn + ( 2 nC0 + 2 nC1 + 2 nC2 + ... + 2 nC2 n ) = 2 n Cn + 2 2 n

⇒

S=

(This step uses the result of example - 7)

1 2n ⋅ Cn + 22 n −1 2

Example – 28

5 balls are to be placed in 3 boxes. Each box can hold all the 5 balls. In how many ways can we place the balls into the boxes so that no box remains empty, if (a) balls and boxes are all different (b) balls are identical but boxes are different (c) balls are different but boxes are identical (d) balls as well as boxes are identical (e) balls as well as boxes are identical, but the boxes are kept in a row. Solution: One of the constraints that should always be satisfied is that no box should remain empty. Thus, each box should get at least one ball. This means that the distribution of balls can have the configuration ( 1, 1, 3) or (1, 2, 2). Only in these two configurations does no box remain empty. Readers might observe some similarity between this problem and Example - 14 where we discussed division of a deck of cards into groups. Observe carefully which part in this problem corresponds to which case in Example - 14

(a) In this case, the balls and the boxes are all different. If you observe carefully, you will note that this case is equivalent to dealing cards to players. Here, we are dealing balls (all different) into boxes (all different). In the card game, we were dealing cards (all different) to players (all different). Suppose we distribute the balls in the configuration (3, 1, 1). We first divide the group of balls into this configuration. This can be done in 5 C3 ways since we just need to select a group of 3 balls and our division will be accomplished. Once the division of the group of balls into 3subgroups in the configuration (3, 1, 1) has been done, we can permute the 3 sub-groups among the 3 different boxes in 3! ways. Thus, the number of ways to achieve the (3, 1, 1) configuration is 5 C3 × 3! = 60 We now find the number of ways to achieve the (1, 2, 2) configuration . Mathematics / P & C, Binomial Theorem

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We first select 2 balls out of the 5 which can be done in 5 C2 ways. We then select 2 balls from the remaining 3, which can be done in 3 C2 ways. Thus simple division into the configuration C 2 × 3C 2 ways. Division by 2! is required since two subgroups 2! are of the same size and right now we are just dividing into sub groups so the order to the subgroups doesn’t matter. Once division has been accomplished, we permute the 3 subgroups so formed amongst the 3 different boxes is 3! ways. 5

(2, 2, 1) can be achieved in

5

Thus, the number of ways to achieve the (2, 2, 1) configuration is

C2 × 3C2 × 3! = 90 2!

The total number of ways is 60 + 90 = 150 To make things more clear, let us list down in detail the various configurations possible for the 3 different boxes, A, B & C: Box - A Box - B Box - C Number of ways 1 1 3 20 1 3 1 20 3 1 1 20 1 2 2 30 2 1 2 30 2 2 1 30 Total:150 (b) The balls are now identical so it doesn’t matter which ball goes into which box. What matters is only the configuration of the distribution. By simple enumeration, only 6 configurations exist for this case. (The notation [a, b, c] implies that Box - A has a balls, Box - B has b balls and Box - C has c balls.)

[1,1,3] [1, 2, 2]

[1,3,1] [2,1, 2]

[3,1,1] [2, 2,1]

Thus, 6 possible ways exist for this case (c) The boxes are identical. This means that it does not matter which sub-group of balls you put in which box. What matters is only the division of the group of balls. This case is akin to the one where you have to divide a deck of cards into sub-groups. (you aren’t required to distribute those sub-groups to players). For the configuration (1, 1, 3), the number of ways of division is 5C3 = 10 (we just choose 3 balls out the 5 and the division is automatically accomplished. For the configuration (1, 2, 2), 5 C 2 × 3C 2 = 15 (division by 2! is required since two the number of ways of division is 2! sub-groups are of the same size, and here the order of the group doesn’t matter) Thus, the total number of ways is 10 + 15 = 25 Mathematics / P & C, Binomial Theorem

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(d) This case is quite straightforward. The balls are identical. The boxes are identical too. The only 2 possible configuration are (1, 1, 3) and (1, 2, 2). There can be no permutation of these configurations since the boxes are indistinguishable. Thus, only 2 ways of division exist. (e) If we keep the boxes in a row, we have inherently ordered them and made them non-identical, since the boxes can be numbered now. Therefore, in this case, the balls are identical but the boxes are different so this question becomes the same as the one in part - (b)

Example – 29

PIRATES OF THE CARRIBEAN

In the Caribbean Sea, 13 pirates, while plundering an English ship, come upon a chest full of gold. Since the pirates have found the chest simultaneously, no one can claim the chest as his own. To protect the chest from the avarices of the pirates, the pirate leader Captain Jack Sparrow suggests a scheme. “We must put a certain number of locks on this chest and distribute their keys amongst ourselves in such a way that it can be opened only when 7 or more then 7 pirates decide to open it. In other words, only when a majority of the 13 pirates agree should the chest be able to be opened.” How can this scheme be implemented? What would be the minimum number of locks required and how must their keys be distributed? Solution: This interesting problem can be easily solved by the application of the elementary concepts developed in this chapter. Let us rephrase our problem slightly. We want that only a group of 7 or more pirates should be able to open the chest. This means that whenever such a majority group decides to open the chest, they should have amongst themselves the keys to all the locks on the chest. Suppose on the other hand, that only 6 pirates decide to open the chest. Then there should be at least one lock on the chest whose key(s) are not with anyone amongst that group of 6. Thus, that single lock will prevent the minority group of 6 pirates from opening the chest. This is the approach we now follow. For every possible group of 6 pirates, we put a lock on the chest and distribute 7 keys of the lock amongst the remaining 7 pirates. That lock will prevent our group of 6 pirates from opening the chest. Such a lock will exist for every group of 6 pirates. Thus, whenever any group of 6 pirates decides to open the chest, they will be prevented by one lock whose keys are with the other 7 pirates (the case when even less than 6 pirates decide to open the chest is automatically solved because then there will be more than one lock to prevent that group from opening the chest) Also, whenever any group of 7 pirates decides to open the chest, there is no lock whose key is not amongst one of the 7 members of that group. Thus, any group of 7 pirates (or more) will be able to open the chest. 13 13 Thus, what we need to do is put C6 ( = C7 ) locks on the chest and for each lock, we select a

group of 7 pirates, make 7 keys of that lock and give one key to each member of this group. Mathematics / P & C, Binomial Theorem

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Example – 30

Prove that (n!)! is divisible by (n!)(n–1)! Solution: We can equivalently show that

( n !)! n −1 ! ( n !)( )

is an integer.

To keep things easier, let us take n = 6. Thus, we need to show that

(6!)! 5! (6!)

is an integer. (You must

appreciate the magnitude of the numbers we are dealing with here!) Consider the following sequence of symbols:

a 1a 1a 1a 1a 1a 1a 2 a 2 a 2 a 2 a 2 a 2 ....................a120 a120 a120 a120 a120 a120 This is a sequence of symbols whose length is 720, because each symbol ai occurs 6 times in this sequence and there are in total 120 symbols, and thus the total number of symbols is 6 × 120 = 720. Let us find the number of permutations of this string. Since there are 120 ‘types’ of symbols, each symbol being repeated 6 times, we will have to divide by 6! a total of 120 times, i.e, the number of permutations of this string will be (720 )! 6! × 6! × 6! × .......6! "## #$### % 120 times

=

(6!)! 5! (6!)

Since the number of permutations of any string must obviously be an integer, the term = must be an integer! We can now easily generalise this result.

(6!)! 5! (6!)

Example – 31

n persons are seated around a circular table. In how many ways can we select 3 persons such that no 2 of them are adjacent to each other on the table? Solution: We will solve this problem by first considering the total number of selections possible and then subtracting from this, the number of cases where 2 persons or all the 3 persons are adjacent. Observe the following figure for reference. Pn

P1

P2 P3

Pn-1

P4

Fig - 16 Mathematics / P & C, Binomial Theorem

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The total number of ways of selecting 3 persons out of these n is nC3 . Let us count the cases where only 2 of the 3 persons are adjacent. 2 adjacent people can be selected as follows:

{P1P2 } or {P2 P3} or {P3 P4 }...............or {Pn P1} Thus, there are n ways in which 2 adjacent persons can be selected. Once these 2 have been selected (for example, say we select {P2 P3}), we can select the third non-adjacent person in (n – 4) ways (since for example, if we selected {P2 P3}, we cannot select P1 or P4). Thus the cases where only 2 of the 3 people are adjacent are n (n – 4) in number. We now count the cases where all 3 persons are adjacent. 3 persons can be adjacent in the following manner:

{P1P2 P3} or {P2 P3 P4 } or ..................or {Pn PP 1 2} Thus, observe that there are n ways of selecting 3 persons who are adjacent. Finally, the number of ways to select 3 persons such that no 2 are adjacent is n

C3 − n ( n − 4 ) − n

=

n ( n − 1)( n − 2 ) − n (n − 4) − n 6

=

1 n ( n − 4 )( n − 5 ) 6

Example – 32

We have 21 identical balls available with us which we need to be distributed amongst 3 boys A, B and C such that A always gets an even number of balls. How many ways of doing so are possible? Solution: The only constraint is that A should get an even number of balls. There’s no constraint on the minimum number of balls a boy should get. This means that a boy can also not be given any ball. We can represent the number of balls given to A by 2x since A must get an even number of balls. If we represent the number of balls given to B and C by y and z, we should have 2 x + y + z = 21

... (1)

This means that to find the number of distributions possible, we find the number of non-negative integral solutions to the equation (1). Note that x can take a maximum value of 10 and a minimum value of 0. We rearrange (1) so that we get an integral equation with y and z as variables, treating x as a constant y + z = 21 − 2 x Mathematics / P & C, Binomial Theorem

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The number of non-negative integer solutions of (2) is 21− 2 x + 2 −1

C1 =

22 − 2 x

C1 = 22 − 2 x

We now add the number of solutions so obtained for all the possible values of x. The total number 10

of solutions is therefore ∑ ( 22 − 2 x ) = 132. x=0

Note: An alert reader must have noticed that we can form arbitrarily complex integer equations. For example, what do we do if we intend to find out the number of non-negative integer solutions to the equation α1 x1 + α 2 x2 + ...........α n xn = r where the αi ' s are integers that are not necessarily equal to unity. In addition, what if we impose constraints on the xi' s themselves say, we define an upper and lower bound for xi , i.e li ≤ xi ≤ ui For example, how do we distribute 20 apples among 4 boys such that each boy gets more than 2 apples but less that 8 apples, in addition to the constraint that one of the boys, say A, must always get an even number of apples? Think about it. We will revisit the problem of such general integral equations in Example - 39 and in the chapter on Binomial Theorem. There we’ll learn to solve such problems using the Multinomial theorem. Example – 33

Find the sum of the divisors of 120. Generalise the result for an arbitrary natural number N. Solution: The divisors of 120 are listed out below: {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120} The sum of these divisors is 360. We have to determine an elegant way to deduce this sum because we cannot repeat everytime the procedure of listing down all the factors and summing them. For this purpose, we again resort to the use of the prime factorization form. 120 = 23 ⋅ 3' ⋅ 51

The sum of the divisors will be S=

∑

2i ⋅ 3 j ⋅ 5k

0≤ i ≤3 0 ≤ j ≤1 0 ≤ k ≤1

This notation is simply a shorthand which implies that we vary the integral indices i, j and k (in their respective allowed ranges) and this way we will have listed down all the factors and hence evaluated the required sum. Mathematics / P & C, Binomial Theorem

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To generate the expression for the sum S, we can alternatively use the following method: S = (1 + 21 + 22 + 23 )(1 + 31 )(1 + 51 )

Did you realize the trick? Writing S this way also gives rise to all the factors. You are urged to convince yourself about this by expanding this expression and observing that all possible factors will be generated. Thus, S can now be simply evaluated as follows: S = (1 + 21 + 22 + 23 ) (1 + 31 ) (1 + 51 )

= 15 × 4 × 6 = 360 This is the same result that we got earlier! To do the general case, assume that the prime factorization form of N is N = p1α1 p2α 2 ........ pnα n

where the α i' are all positive integers and the pi' s are all primes. The sum Sf of all the factors will be

(

S f = (1 + p1 + p12 + ..... p1α1 ) (1 + p2 + p22 + ..... + p2α 2 )......... 1 + pn + pn2 + ..... pnα n

)

Example – 34

Find the sum of all the five-digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 if no digit is repeated. Solution: This problem can be solved very easily if we view it from an individual digit’s perspective. Suppose that we only consider the digit “4”. How many numbers will there be with “4” in the units place?

' '

4 4 4

There are 24 numbers with 4 in the units place because the remaining four digits can be permuted among the remaining 4 places in 4!=24 ways

From these 24 numbers, what is the total contribution of the digit “4” to the sum we are required to calculate? Since “4” is at the units place and it occurs 24 times, its contribution will be 4 × 24 = 96 Mathematics / P & C, Binomial Theorem

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Similarly, there will be 24 numbers where “4” is at the tens place. The total contribution of “4” from these 24 numbers will be 4 × 240 = 960 Proceeding in this way, we see that the total contribution of the digit “4” from all the 120 numbers that can be possibly formed is:

4 ( 24 + 240 + 2400 + 24000 + 240000 ) = 4 × 24 × (1 + 10 + 100 + 1000 + 10000 ) = 4 × 24 × 11111

This is the contribution to the sum from only the digit “4”. To calculate the entire sum S, we calculate the contributions from all the five digits. Thus, the sum is

S = (1 + 2 + 3 + 4 + 5 ) × 24 ×11111 = 3999960

Example – 35

DEARRANGEMENTS

Find the number of ways in which 5 different letters can be taken out of their 5 addressed envelopes and put back into the envelopes in such a way that all letters are in the wrong envelope. Solution:

The problem of rearranging objects so that each object is assigned to a place different from its original place is referred to as the problem of dearrangments. Here, we need to find out the number of dearrangements possible with 5 letters and 5 envelopes. Let us denote by Dn the number of dearrangements possible with n things. We will attempts a general solution, that is for an arbitrary n, and then substitute n = 5 Denote by Li the ith letter and by Ei, the original envelope of the ith letter. Consider the envelope E1. It can be assigned a wrong letter in (n – 1) ways. Suppose that we assign the letter L2 to E1. E1 E2 ............ En ( L1 L2 ............ Ln The dearrangements that can now arise can be divided into two mutually exclusive classes: (i) Those in which L1 is assigned to E2 (ii) Those in which L1 is not assigned to E2 If L1 is assigned to E2, we have the following configuration: E1 E2 ............ En ( ! L1 L2 ............ Ln In this case, we have a remaining of (n – 2) letters which can be dearranged in Dn–2 ways.

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Suppose the other case now, where we do not assign L1 to E2. In this case, we have (n – 1) letters to dearrange (L1 will also be counted as a letter to be dearranged since it is being assigned to an envelope other than E2). This can be done in Dn–1 ways. Thus, if we give L2 to E1, the total dearrangements possible are Dn −1 + Dn − 2 . Since E1 can assigned a wrong letter in (n –1) ways, the overall total number of dearrangements Dn is

Dn = ( n − 1)( Dn −1 + Dn− 2 ) We have thus related the nth order 'dearrangements–coefficient' Dn to lower order coefficients. We now have to somehow use this relation to obtain Dn only in terms of n. This is how we do it:

Dn = ( n − 1)( Dn −1 + Dn− 2 )

(

Dn − nDn−1 = ( −1) Dn−1 − (

⇒

n −1)

Dn− 2

)

... (1)

But if we substitute n → ( n − 1) we get

Dn−1 − ( n − 1) Dn− 2 = ( −1) ( Dn− 2 − ( n − 2 ) Dn−3 )

... (2)

We substitute (2) back into (1) to get Dn − nDn −1 = ( −1) ( Dn − 2 − ( n − 2 ) Dn −3 ) 2

= ( −1) ( Dn −3 − ( n − 3) Dn − 4 ) {By the same process} 3

' = ( −1)

n−2

( D2 − 2 D1 )

It is obvious that D1 = 0 since one letter cannot be dearranged while D2 = 1 because two letters can be dearranged in only one way : by exchanging them. Thus, Dn − nDn −1 = ( −1)

n−2

... (3)

We still have not obtained a relation involving only Dn. We do it using (3) repeatedly Dn − nDn −1 = ( −1)

= ( −1)

n−2

n

( −1) Dn Dn −1 − = n ! ( n − 1)! n!

n

⇒

Mathematics / P & C, Binomial Theorem

(Division by n!)

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Now we substitute n → ( n − 1) , n → ( n − 2 ) .... successively and add the corresponding sides of all the relations so obtained to get

( −1) + ...... + ( −1) Dn D1 ( −1) ( −1) − = + + n ! 1! n! 2! ( n − 1)! ( n − 2 )! n

n −1

n−2

2

Since D1 = 0, we finally get n 1 1 1 −1) ( Dn = n ! − + − ........ n! 2! 3! 4!

n 1 1 1 −1) ( = n !1 − + − + ... 1! 2! 3! n!

The first two terms "1"and " 1 " 1! are just added to make the series look more sequenced

This is the number of de arrangements possible with n things For n = 5, we have 1 1 1 1 1 D5 = 5!1 − + − + − 1! 2! 3! 4! 5!

= 44 Thus, there are 44 ways to rearrange the letters back into their envelopes so that each letter goes to a wrong envelope. Since n = 5 is a small number, we could have worked out an alternative solution as follows: Total No. of arrangements No. of ways in which all letters go to correct envelopes - No. of ways in No. of dearrangemets = which one letter goes to incorrect envelope No. of ways in which two letters go to incorrect envelopes -.............................and so on

You are urged to work out the solution by this way yourself. Example – 36

Prove that for n ≥ 4, the sum 1! + 2! + 3! + ... + n ! cannot be the square of a positive integer. Solution: The approach lies in realizing the general properties of a perfect square. Note that a square will have, at its unit place, a digit from amongst the set {1, 4, 5, 6, 9}. It cannot have a ‘3’ or an ‘8’ as its unit digit. Work it out yourself. Mathematics / P & C, Binomial Theorem

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Now, for the given sum, the first four terms add to 33:

1!+ 2!+ 3!+ 4! = 1 + 2 + 6 + 24 = 33 For any integer greater than 4, its factorial will always have a ‘0’ at its unit place. Can you see why? This means that the given sum will have a ‘3’ at its unit place, so it cannot be a perfect square! Example – 37

A set P contains x elements while Q is another set which contains y elements. How many functions f : P → Q exist which are (a) one - one (b) onto? Solution: Obviously, the total number of functions possible is y x (a) If the function is to be one-one, y, the number of elements in the co-domain of f, must be greater than or equal to x : y≥x In that case, the number of one-one functions will simply be y Px , since you are creating a mapping from x elements in P to x elements in Q, out of a total y elements in Q. (b) For the function to be onto, the only constraint that needs to be satisfied is that each element in the co-domain Q must have a pre-image, which means that x must be greater than or equal to y. The remaining part is left to the reader as an exercise. Example – 38

Four non-identical dice are rolled simultaneously. How many rolls are possible in which 6 shows up on at least one dice? Solution: Its easiest to proceed by considering the complementary case: Total number of rolls possible : 64

(Why ?)

Rolls in which no 6 shows up : 54 (Why ?) Our answer is therefore 64 – 54. Now, a very important issue which we leave to you to resolve: what if the four dice were identical? Example – 39

In an examination, the maximum marks for each of three papers is n and that for the fourth paper is 2n. Find the number of ways in which a candidate can get 3n marks. Mathematics / P & C, Binomial Theorem

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Solution: We need to find the number of solutions to the non-negative integral solution x1 + x2 + x3 + x4 = 3n,

... (1)

where x1 , x2 , x3 all lie in [0, n], while x4 can vary upto 2n, i.e., x4 ∈ [0, 2n], where all the xi' s are integers. Let us construct the polynomial P ( x ) given by P( x) = (1 + x + x 2 + ... + x n )3 (1 + x + x + ... + x 2 n ) Now, think hard ! The first bracket in P ( x ) is actually three identical brackets (1 + x + x 2 + ... + x n ). When P ( x ) is expanded , each of these three brackets can contribute at the most an nth power in x, whereas the fourth bracket can contribute upto 2n in powers of x. When P ( x ) is expanded and the terms written out, note that x 3n will also be formed. But what will be the coefficient of x3n ? It will be precisely the number of solutions to the equation (1) under consideration ! Why ? Because, each time x 3n is formed is a unique combination of power contribution from each of the four brackets. For example, P( x) = (1 + x + ... xn )(1 + x + ... xn )(1 + x + ... + xn )(1 + x + ... + x2n ) ↓ Take 1

↓ Take 1

↓ Take x n

↓ Take x2 n

"##########$##########% Product formed = x3 n

In this example, we take the term 1( or x0) from the first two brackets, x n from the third and x 2n from the fourth. This is a unique combination in which x 3n can be generated, but this is only one such. There will be many more ways in which x 3n can be generated. How many ? Precisely, the coefficient of x 3n when P ( x ) is expanded ! To find out the coefficient, you’ll have to wait until you are familiar with the Binomial / Multinomial expansion.

Example – 40

In how many ways can one fill an m × n table with ±1 such that the product of the entries in each row and each column equals –1? Solution: Denote by aij the entries in the table. For 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1, we let aij = ±1 in an arbitrary way. This can be done in 2( m −1)( n −1) ways. The values for amj with 1 ≤ j ≤ n − 1, and for ain with 1 ≤ i ≤ m − 1, are uniquely determined by the condition that the product of the entries in each row and each column equals –1. The value of amn is also uniquely determined, but it is necessary that n −1

m−1

j =1

i =1

∏ amj = ∏ ain ⋅ Mathematics / P & C, Binomial Theorem

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If we denote by m −1 n −1

P = ∏∏ aij i =1

j =1

we observe that n −1

P∏ amj = (−1) n −1 j =1

and m −1

P∏ ain = ( −1) m −1 i =1

Hence (*) holds if and only if m and n have the same parity.

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CONCLUSION : A MAGIC TRICK

For those of you interested in cards, here is a very interesting card trick that has its basis in an elementary application of this chapter’s concepts. A magician sends out his assistant into the audience with a well-shuffled standard deck of cards. The assistant asks 5 different people chosen at random from this audience to pick a card each from the deck. The magician is on-stage and cannot see the cards drawn from the deck. The magician then asks the assistant to reveal 4 of the 5 drawn cards while keeping the fifth card secret. The magician then thinks for a while and tells the audience the number and suit on the fifth secret card ! We have to figure exactly how the assistant conveys the information about the fifth card to the magician. You can assume that no one is cheating!

The trick explained Somehow, the assistant has been able to convey information about the fifth card to the magician, even though he reveals only 4 of the 5 cards. The information must be embedded in the order in which the assistant reveals the 4 cards. Thus, the magician and assistant must have pre-decided on a scheme whereby the magician can deduce the secret card from the order of the 4 revealed cards. Here is how this can be done. Since 5 cards are drawn, at least 2 cards must be of the same suit. If there are more than 2 cards of the same suit, the assistant simply chooses any 2 of them. Now, one of these 2 cards will be revealed first by the assistant while the other card will become the secret card. So, the magician knows from the suit of the first revealed card what the suit of the secret card is. The first issue is that, of the 2 cards of the same suit, how does the assistant decide which one to reveal and which one to keep secret? For that, the assistant refers to the following diagram (mentally of course!): A

K

2 3

Q 4 J 5 10 6 9 8

7

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In this circular arrangement, observe that the difference between any two members is at the most 6. For example, the difference between “7” and “Q” is 6 while that between “J” and “2” is 4

A

K

4

2

ste

ps

A

K

2

3

3

Q

Q 4

4 J

5

5 10

10

6

6

6

steps

J

9

9 7

8

7

8 Fig - 18

Thus, of the 2 cards of the same suit, the assistant reveals that card from which it is possible to reach the other card in at most 6 steps. For example, if the assistant finds a “J” of Hearts and a “2” of Hearts in the 5 drawn cards, he will reveal the “J” of Hearts first and keep the “2” of Hearts secret since in our circular arrangement it is possible to reach from “J” to “2” in 4 steps. Thus, till now the scheme is that the secret card is of the same suit as the first revealed card and lies within at the most 6 steps from the first card, that is, the offset of the secret card from the first card is at the most 6. We now require a scheme whereby the assistant can communicate the offset to the magician. For this purpose, they pre-decide upon an order of all the 52 cards of the deck. For example, suppose they order the deck like follows:

1

2

3

4 ... 14

↓ ↓ ↓ ↓ A of 2 of 3 of 4 of clubs clubs clubs clubs

15

16 ... 27 ...

↓ ↓ ↓ A of 2 of 3 of spades spades spades

↓ A of hearts

40

...

52

↓ ↓ A of K of diamonds diamonds

→ Increasing order

Fig 19 Now, after the assistant has revealed the first card, he has 3 more cards to reveal. Those 3 cards can be ordered according to the order of Fig 19. Thus, one of those 3 cards will be the ‘smallest’, one will be the ‘middle’ card and one will be the ‘largest’. Label the 3 cards as S, M and L.

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To communicate the offset to the magician, the assistant reveals the 3 cards in a particular order as shown in the table below Offset Order in which the assistant reveals the 3 cards 1 SML 2 SLM 3 MSL 4 MLS 5 6

LSM LMS

Since the maximum offset can be 6, the 6 permutations of S, M and L cover all the possible offsets! Once the offset is communicated, the magician simply adds it to the first card to deduce the fifth, secret card! Let us try this out with a particular example. Suppose that the following 5 cards are drawn: “A of spades” “2 of Hearts” “3 of clubs” “7 of Hearts” “8 of clubs”. The assistant picks 2 cards of the same suit. Here we have 2 options for the 2 cards. We can choose either. Suppose the assistant chooses “2 of Hearts” and “7 of Hearts”. The offset of “7” from “2” is 5 which is less than 6. Therefore, “2 of Hearts” is the first card that the assistant reveals. To communicate the offset, the assistant orders the three other cards (to be revealed) according to Fig. 19 as follows: "3 of clubs" < "8 of clubs" < "A of spades" S M L

Since the offset required is 5, the assistant reveals these three cards (according to the Table on the previous page) in the order L S M. Thus, this is what the assistant says: “2 of Hearts” “A of spades” “3 of clubs” “8 of clubs” The magician thinks for a while and says, “The fifth card is the 7 of Hearts !” And the audience gasps!

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P & C [ ASSIGNMENT ]

[ OBJECTIVE ] LEVEL - I Q. 1

The number of ways in which 4 balls can be selected from a bag containing 4 identical and 4 different balls, is (a) 120 (b) 80 (c) 14 (d) 16

Q. 2

The maximum number of intersection points of n straight lines, none of which are parallel to each other, is n

C2 (d) n C2 − n 2 There are 4 bulbs in a room. The number of ways of illuminating the room, is (a) 16 (b) 15 (c) 24 (d) 4 n

n

(b) P2

(a) C2 Q. 3 Q. 4

There are n different books and each book has p copies. The number of selection of books from these is (a) ( p + 1) − 1 n

Q. 5

(c)

(b) ( p + 1) − 2 n

(c) n ( p + 1) − 1

(d) none of these

(c) 19 C16

(d) none of these

The value of 18 C2 +17 C2 +16 C2 +15 C2 +14 C3 is (a) 17 C3

(b) 18 C3

Q. 6

The number of ways in which 6 men can be made to sit round a table in 6 numbered seats is (a) 5! (b) 6! (c) 6 × 6! (d) none of these

Q. 7

The number of diagonals that can be formed by joining the vertices of an n-sided polygon, is (a) n C2 − n

Q. 8

(b) n P3

(c) n ( n − 1)

(d) n C2 − ( n + 1)

The maximum number of intersection points of n circles, is n

C2 (d) n C2 − n 2 In an examination there are 5 multiple choice questions. Each question has four choices of which only one is correct. The number of ways in which an examinee can get at least one answer wrong, is n

Q. 9

n

(a) C2

(b) P2

(c)

(a) 45 − 1

(b) 54 − 1

(c) 5!4!− 1

(d) 5 C4 − 1

(c) 47

(d) 65

Q. 10 The exponent of 4 in 200!, is (a) 86 (b) 101

Q. 11 The number of different messages by signals with three dots and two dashes is (a) 120 (b) 12 (c) 10 (d) 20 Q. 12 If n Cr −1 = 36, n Cr = 84 , n Cr +1 = 126 then the value of r is (a) 1

(b) 2

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(c) 3

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Q. 13 Five persons including one lady are to deliver lectures to an audience. The organiser can arrange the presentation of their lectures, so that the lady is always in the middle, in (a) 5 P5 ways

(b) 4.4 P4 ways

(d) 5 C4 ways

(c) 4! ways

Q. 14 The number of ways in which a set A where n(A) = 12 can be partitioned in three subsets P, Q, R each of 4 elements if P ∪ Q ∪ R = A , P ∩ Q = φ , Q ∩ R = φ , R ∩ P = φ , is 12!

12!

(a) 4! ( )

(b) 3! 4 ( )

3

(c)

12! 1 . 3 ( 4!) 3!

(d) none of these

Q. 15 In a test of 10 multiple-choice questions of one correct answer, each having 4 alternative answers, the number of ways to put ticks at random for the answers to all the questions is (a) 410

(c) 410 − 4

(b) 10 4

(d) 10 4 − 10

Q. 16 The number of positive integral solutions of the equation x + y + z = 50, is (a) 51 C2

(b)

50

(c)

C3

Q. 17 The number of zeros at the end of 50!, is (a) 17 (b) 5

49

(d)

C2

(c) 12

48

C3

(d) 47

Q. 18 The number of ways in which 9 different objects can be distributed to three different people giving 2, 2 and 5 objects to each of them, is (a) 328 (b) 984 (c) 756 (d) 2268 Q. 19 If m parallel straight lines intersect n parallel straight lines, then the number of parallelograms thus formed, is (a) m n

(b)

m+n

C2

(d) m C2 × n C2

(c) mn

Q. 20 There are 2 points on a line, 3 points on another line and 4 points on yet another line. The total number of triangles that can be formed by joining these points, is (a) 30 (b) 205 (c) 79 (d) 85 [ LEVEL - II ] Q. 1 Q. 2

The number of divisors of 1008 of the form 4n + 2( n ≥ 0, n ∈ N) , is (a) 5 (b) 6 (c) 11

Numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 all at a time. The number of such numbers in which the odd digits occupy even places, is (a) 162

Q. 3

(d) 7

(b) 175

(c) 180

(d) 160

The number of ways of distributing 10 different toys among 4 children C1, C2, C3, C4 such that C1 and C2 get 2 toys each and C3 and C4 get 3 toys each, is 10! (a) 2! 2 3! 2 ( )( )

10! 4! (b) 2! 2 3! 2 × 2! 2 (c) ( )( ) ( )

Mathematics / P & C, Binomial Theorem

(

10

C2 )

2

(

10

C3 )

2

(d) none of these

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There are n different colours and p balls of each colours. The number of ways in which one or more than one ball can be chosen, is (a) ( n + 1) − 1

(b) ( p + 1) − 1

p

Q. 5

Q. 6

n

(c) p Cn − 1

(d)

p +1

Cn − 1

A four digit number is formed using the digits 1, 2, 3, 4, 5. The total number of numbers which have at least one digit repeated, is (a) 625 (b) 500 (c) 120 (d) 505 The number of integral solutions of the equation x1 + x2 + x3 = 0 ( x1 , x2 , x3 ≥ −3) , is 9! (b) 3! 3 (c) 11 C3 (d) 11 C2 ( ) The number of natural numbers less than 2000 that can be formed using the digits 0, 1, 2, 3, 4 without repeating any digit, is

(a) (3!)

3

Q. 7

(a) 78

(b) 102 n

Q. 8

In the identify ∑ k =0

(a) n Ck Q. 9

(c) 88

(d) 92

Ak n! = the value of Ak is x + k x ( x + 1)( x + 2 )... ( x + n ) (b) n Ck +1

(c) ( −1) .n Ck k

(d) ( −1)

k −1 n

. Ck −1

The number of times the digit 1 will occur in natural numbers less than 1000, is (a) 250

(b) 300

(c) 350

(d) 450

Q. 10 The number of ways in which m prizes can be distributed among n players if each player can receive any number of prizes, is (a) m n

(b) n m

(c) m Cn

(d) n cm

Q. 11 There are n identical seats (unmarked) around a circular table. The number of ways in which m ( m < n ) people can be seated, is ( n − 1)! (a) ( m − 1)! (b) n − m ! (c) ( n − m + 1)! (d) n Cm ( m − 1)! ( ) Q. 12 The number of non-negative integral solutions of the equation x + y + 3 z = 10 , is (a) 26

(b) 10 C2 × 9C3

(c) 160

(d)

10! 2!3!

Q. 13 There are p points in space, no four of which are in the same plane with the exception of q points which are all in the same plane. The number of different planes determined by the points is (a)

p ( p − 1)( p − 2 ) q ( q − 1)( q − 2 ) − 6 6

(c)

p ( p − 1)( p − 2 )( p − 3) q ( q − 1)( q − 2 )( q − 3) − +1 24 24

Mathematics / P & C, Binomial Theorem

(b)

p ( p − 1)( p − 2 ) q ( q − 1)( q − 2 ) − +1 6 6 (d) none of these

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Q. 14 The number of ways in which 8 candidates A1 , A2 , A3 ,....., A8 can be ranked such that A3 is always comes before A4 and A4 always comes before A5 , is (a) 2 × 8 P6

(b)

8! 6

(c) 8!

(d)

8! 3

Q. 15 The number of ways in which 8 candidates A1 , A2 ,....., A8 can be ranked such that A4 is always above A5 , is (a) 2 × 8 P6

(b) 8P6

(c) 8!

(d)

8! 3

Q. 16 The number of triangles that can be formed from n points of which m are collinear, is (a) n C3 − ( m + n ) (b) n C3 (c) n C3 − m C3 (d) n − m C3 Q. 17 The number of ways in which a sum of 10 can be obtained by throwing a dice thrice, is (a) 21 (b) 27 (c) 6 (d) 15 Q. 18 There are 6 identical blue balls and 6 identical red balls. The number of ways in which 6 balls out of the given 10 balls can be arranged in a row such that the number of blue balls is equal to the number of red balls, is (a) 6 (b) 64 (c) 20 (d) 9 Q. 19 The sum of the digits in the units place of all numbers formed with the digits 1, 1, 2, 3 when taken all at a time, is (a) 21 (b) 17 (c) 32 (d) 28 Q. 20 If n objects are arranged in a row, then the number of ways of selecting three of these objects so that no two adjacent object is selected, is n! (a) n C3 − n C2 (b) 3! 3 (c) n C3 (d) n − 2 C3 ( )

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[ SUBJECTIVE ] Q. 1

A letter lock consists of 4 rings each marked with 15 different letters. In how many ways is it possible to make an unsuccessful attempt to open the lock?

Q. 2

Show that the greatest value of

Q. 3

There are an unlimited number of identical balls of four different colors. How many arrangements of at the most 8 balls in a row can be made by using them?

Q. 4

How many different numbers greater than 5000 can be formed with the digits 1, 2, 5, 9, 0 when repetition of digits is not allowed?

Q. 5

A polygon has 44 diagonals. How may sides does it have?

Q. 6

There are m points on one straight line AB and n points on another straight line AC, none of them being A. How many triangles can be formed with these points as vertices (excluding A)?

Q. 7

In how many ways can five different rings be worn in four fingers with at least one ring in each finger?

Q. 8

m men and n women are to be seated in a row so that no two women sit together. If m > n, show that

2n

Cr for 0 ≤ r ≤ 2n is

the number of ways in which they can be seated is Q. 9

2n

Cn

m !( m + 1)! . ( m − n + 1)!

How many four digit numbers with distinct digits can be formed using the digits 0, 1, 2, 3, 4, 5 which are (a) divisible by 3? (b) divisible by 6?

Q. 10 How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS? Q. 11 Find the total number of ways of selecting 5 letters from the letters of the word INDEPENDENT. Q. 12 Two numbers a and b are chosen from the set {1, 2, 3, ....., 3n}. In how many ways can these integers be selected such that (a) a2 – b2 is divisible by 3?

(b) a3 + b3 is divisible by 3?

Q. 13 What is the number of non-negative integral solutions of the equation x1 + x2 + x3 + 4 x4 = 20? Q. 14 An exam consists of four papers. Each paper has a maximum of m marks. Show that the number of m +1 ways in which a student can get 2m marks in the exam is ( 2m 2 + 4m + 3 ) 3

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Q. 15 Find the total number of positive unequal integral solutions of the equation x + y + z + w = 20 Q. 16 In how many ways can 10 persons take seats in a row of 24 fixed seats so that no two persons take consecutive seats? Q. 17 There are 15 seats in a row numbered 1 to 15. In how many ways can 4 persons sit in such a way that seat number 6 is always occupied and no two persons sit in adjacent seats? Q. 18 In how many ways can 2n + 1 identical balls be placed in distinct boxes so that any two boxes together will contain more balls than the third? Q. 19 12 seats are to be occupied by 4 people. Find the number of possible arrangements if (i) no two persons sit side by side. (ii) there should be at least two empty seats between any two persons. (iii) each person has exactly one neighbour. Q. 20 The sides of a triangle are a, b, c inches where a, b, c are integers and a ≤ b ≤ c . If c is given, show (c + 1) 2 c (c + 2) if c that the number of different triangles that can be formed is if c is even and 4 4 is odd.

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P & C / ANSWER

[ TRY YOURSELF - I ] 1. 161280 6. 180

2. 120 7. 1200

3. 10 8. 246

4. 120960 9. 26

5. 360 10. 35

[ ASSIGNMENT ] OBJECTIVE LEVEL - I

LEVEL - II

1. d

2. a

1. b

2. c

3. b

4. a

3. a

4. b

5. c

6. b

5. d

6. d

7. a

8. b

7. d

8. c

9. a

10. d

9. b

10. b

11. c

12. c

11. b

12. a

13. c

14. c

13. b

14. b

15. a

16. c

15. d

16. c

17. c

18. d

17. b

18. c

19. d

20. c

19. a

20. d

SUBJECTIVE 1. 154 – 1

3. 87380

4. 48

5. 11

6.

mn (m + n − 2) 2

7. 480

9. 96, 52

10. 2454

11. 72

12.

5n 2 − 3n 3n 2 − n , 2 2

13. 536

15. 552

16.

17. 3072

18.

n ( n + 1) 2

19.

9

15

P10

P4 , 6P4 , 4! × 9C2

Mathematics / P & C, Binomial Theorem

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APPENDIX : BINOMIAL THEOREM

Binomial theorem is something that has been known to mathematicians since many centuries ago. In this introduction, we’ll trace the origins of this theorem to the coefficients we obtain when we expand any binomial term raised to an integral power. Consider the following expansions, which can be verified by direct multiplication:

(x + y)

=

1

(x + y)

=

x+ y

(x + y)

=

x 2 + 2 xy + y 2

(x + y)

3

=

x 3 + 3 x 2 y + 3 xy 2 + y 3

(x + y)

4

=

x 4 + 4 x 3 y + 6 x 2 y 2 + 4 xy 3 + y 4

(x + y)

5

=

x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 xy 4 + y 5

(x + y)

6

=

x 6 + 6 x 5 y + 15 x 4 y 2 + 20 x 3 y 3 + 15 x 2 y 4 + 6 xy 5 + y 6

0

1

2

' and so on Do you notice anything special about these expansions, in particular, any general rule or trend these expansions follow that might enable us to expand ( x + y ) directly for a general n ? First of all, notice that the number of n

terms in each expansion is one more than the power of the binomial term. For example, ( x + y ) has 6 terms. 5

However, mathematicians long back also realized another important fact, namely, the relation between the coefficients obtained upon expansion. To see what this relation is, let us write the coefficients in the following ‘triangular’ pattern: n=0

1 1

1

1 3

1

5

1

6

1 3

4

1 1

2

n=1

6 10

4

20

n=3

1

10

15

n=2

5 15

n=4

1

n=5

1 6

1

n=6

Fig - 20 Mathematics / P & C, Binomial Theorem

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61

Do you observe any relation between the various coefficients. If not, consider this same arrangement in a slightly modified form: 1 1

1

1 1 1 1

(1+1) (1+2)

(1+3) (1+4)

1

(2+1)

(3+3)

(4+6)

1

(3+1)

(6+4)

1

(4+1)

1

Fig - 21 The ‘rule’ for constructing this triangular pattern should be pretty obvious now. All edge-numbers are 1. Any other is obtained by adding the number directly above and to the left with the number directly above and to the right, as in Fig-34 Extending this process gives us all the ‘binomial coefficients.’ This geometrical arrangement of the binomial coefficients in a triangle is called Pascal’s triangle. The figure below shows a Pascal triangle containing the coefficients upto n = 15. 1 2

1 3

1 5

1 6

1 7

1 8

1 9

1

26

10 20

35 56

84

4 5 15 35

70 126

1 6 21

56 126

1 1 7 28

84

1 8

26

1 9

1

120 210 252 210 120 45 10 1 165 330 462 462 330 165 55 11 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 1

10

15

28

1

6 10

21

1 3

4

1

Pascal's triangle, upto n = 15

1

1

45

55

1

14

91

364

1001 2002 3003 3432 3003 2002 1001

364

91

1

14

1

Fig - 22

If we denote the (i + 1)th number at the nth level by Tn, i , then we have Tn , i +1 = Tn −1, i + Tn −1, i +1 ... (1) Later on, when we actually write Tn , i +1 in terms of combinational notation (in fact, we’ll see that Tn , i +1 corresponds to n Ci ), we will immediately understand that (1) is equivalent to n Ci = n −1Ci −1 + n −1Ci which, as we already know from the last chapter on P & C, is true. Mathematics / P & C, Binomial Theorem

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Section -1 BINOMIAL THEOREM, POSITIVE INTEGRAL INDEX

Section - 1

Let us now consider more formally the binomial theorem. We need to expand ( x + y ) , where x, y are two arbitrary quantities, but n is a positive integer. n

In particular, if we write

(x + y)

n

= ( x + y )( x + y )( x + y ) ... ( x + y )

( n times )

... (1)

we need to find out the coefficient of xi y j . Note that i + j must always equal n, so that we can write a general term of the expansion (without the coefficient) as x r y n − r so that 0 ≤ r ≤ n . Now, to find the coefficient of x r y n − r , note that we need the quantity x, r times, while y is needed n – r times. Thus, in (1), x r y n − r will be formed whenever x is ‘contributed’ by r of the binomial terms, while y is ‘contributed’ by the remaining n - r of the binomial terms. For example, in the expansion of ( x + y ) , to form x 2 y 3 , we 5

need x from 2 terms and y from 3: take x 5

( x + y ) = (x + y )(x + y )(x + y )(x + y )(x + y ) take y 2

One of the ways to form x y

3

Fig - 23

How many ways are there to form x 2 y 3 ? In other words, how many times will x 2 y 3 be formed? The number of times x 2 y 3 is formed is what is the coefficient of x 2 y 3 . That number, which would be immediately obvious to the alert reader, is simply 5 C2 . Why? Because this is the number of ways in which we can select any 2 binomial terms from 5. These 2 terms will contribute x. The remaining will automatically contribute y. In the general case of ( x + y ) , we see that the coefficient of x r y n − r would be n Cr . (which is infact the same n

as n Cn − r ). Thus, the general binomial expansion is n

= nC0 x n + nC1 x n- 1 y + nC2 x n- 2 y 2 + ... + nCn y n

(x + y )

⇒

(x + y)

n

Mathematics / P & C, Binomial Theorem

n

= ∑ n Cr x n − r y r r =0

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63

The coefficients n Ci are called the binomial coefficients, for a reason that should now be obvious. Note that the (i + 1) coefficient in this expansion is n Ci , which now explains the relation th

Tn , i +1 = Tn −1,i + Tn −1, i +1

we observed in the Pascal triangle; this relation simply corresponds to n

Ci = n −1Ci −1 + n −1Ci

Also, the binomial coefficients of terms equidistant from the beginning and the end are equal, because we have n Cr = n Cn − r . The general term of expansion, n Cr x n − r y r , is the (r + 1)th term from the beginning of the expansion and is conventionally denoted by Tr +1 , i.e. Tr +1 = nCr x n − r y r Since we have (n + 1) terms in the general expansion, we see that if n is even, there will be an odd number of terms, and thus there will be only one middle term, which would be n Cn / 2 x n / 2 y n / 2 . For example,

(x + y)

4

= x 4 + 4 x3 y + 6 x 2 y 2 + 4 xy 3 + y 4 only one middle term

On the other hand, if n is odd , then there will be an even number of terms in the expansion, and thus there will be two middle terms, namely n C n −1 x

n +1 2

y

n −1 2

n

and C n −1 x

2

n −1 2

y

n +1 2

2

For example;

(x + y)

5

= x5 + 5 x 4 y + 10 x3 y 2 + 10 x 2 y 3 + 5 xy 4 + y 5 Two middle terms

Example – 1

1 Find the middle term(s) in the expansion of x − x

9

Solution: Since n = 9, there will be 10 terms in the expansion, which means that there will be 2 middle terms in the expansion, the 5th and the 6th: T5 = C4 ( x )

−1 = 126 x x

T6 = 9C5 ( x )

−1 −126 = x x 5

9

9−4

9 −5

Mathematics / P & C, Binomial Theorem

4

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64

Example – 2 10

2x2 1 Is there any term in the expansion of − that will be independent of x? x 5

Solution: The general term in the expansion is 10 − r

2x2 Tr +1 = Cr 5 10

10 − r

2 = 10Cr 5

10 − r

2 = 10Cr 5

−1 x

r

( −1) x

20 − 2 r −

( −1)

20 −

r

r

x

r 2

5r 2

Thus, for the term that is independent of x, we have 20 −

⇒

5r =0 2

r =8

Thus, the term free of x is the 9th term given by 2

36 8 2 T9 = C8 ( −1) = 5 5 10

Example – 3

Evaluate the sum n C0 + n C1 + n C2 + ... + n Cn . Solution: We have already evaluated this sum in the chapter on P & C. That approach was as follows: this sum basically counts the number of all sub-groups of a set of size n; this can also be counted by focusing on each element of the set, which has two corresponding choices - you either include it into your sub-group or you don’t, which means that the total number of ways to form sub-groups is 2 × 2 × 2 .... n times = 2n. The sum of the binomial coefficients therefore equals 2n. Here, we evaluate the same sum using a binomial approach. Consider the following expansion:

(1 + x )

n

= nC0 + nC1 x + n C2 x 2 + nC3 x 3 + ... + nCn x n

If we put x = 1, we simply obtain 2n = n C0 + n C1 + n C2 + ... n Cn Mathematics / P & C, Binomial Theorem

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65

Thus, the same result is obtainable from both a combinatorial and a binomial approach. We can also derive another useful result by putting x = –1 in the above relation, so that we obtain 0 = nC0 − nC1 + nC2 − nC3 + ... + ( −1) nCn n

⇒

n

C0 + n C2 + n C4 + ... = n C1 + n C3 + n C5 + ...

This states the sum of the even-numbered coefficients is equal to the sum of the odd-numbered coefficients. Can you prove this using a combinatorial approach? As an exercise, prove the following relations: n

C0 2n + n C1 2n −1 + n C2 2n − 2 + ... n Cn = 3n

n

C0 − nC1 x + nC2 x 2 − ... + ( −1) nCn x n = (1 − x ) n

n ( −1) nCn = 1 C nC C C0 − 1 + 2 2 − 3 3 + ... + 2 2 2 2n 2n n

n

n

n

Example – 4

(a) What is the greatest coefficient in the expansion of ( x + y ) ? n

(b) What is the greatest term in the expression of ( x + y ) ? n

n Solution: (a) For this part, we basically need to only determine max ( Cr ) for 0 ≤ r ≤ n ; x and y have no role to play in this part. To find the greatest coefficient, consider the following ratio:

n! ( r + 1)!( n − r − 1)! C q = n r +1 = n! Cr r !( n − r )! n

=

n−r r +1

Thus, q >1

Mathematics / P & C, Binomial Theorem

⇒

n−r >1 r +1

⇒

n − r > r +1

⇒

r<

n −1 2 www.locuseducation.org

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66

Similarly, q nCr

whenever

n

Cr +1 < nCr

whenever

n −1 2 n −1 r> 2

r<

... (1)

If n is odd, we have n

C n −1 > nC n −3 2

n

and

2

C n +3 < nC n +1 2

2

Also, since n

C n −1 = n C n +1 2

2

we see that for odd n, the two middle coefficients are the greatest. This can be verified by considering the following expansion:

(x + y)

5

= x5 + 5 x 4 y + 10 x3 y 2 + 10 x 2 y3 + 5 xy 4 + y5 ! ( The two middle coefficients are the greatest for odd n

If n is even, (1) gives n

C n > nC n 2

n

and

2

−1

C n < nC n 2

+1

2

In this case therefore, the greatest coefficient is the single middle coefficient n C n . Lets verify 2

this again:

(x + y)

6

Mathematics / P & C, Binomial Theorem

= x6 + 6 x5 y + 15 x 4 y 2 + 20 x3 y3 + 15 x 2 y 4 + 6 xy 5 + y 6 ! The single middle coefficient is the greatest for even n

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(b) To find the greatest term, we must also consider x and y. We again follow the approach of part (a): n Tr +1 Cr x n − r y r q= = n Tr Cr −1 x n − r +1 y r −1

=

( n − r + 1) . y r

x

Observe that q >1

If

⇒

( n − r + 1) ⋅ y > 1

⇒

( n + 1) y x + y − r⋅ >0 x + y r x

r

x

( n + 1) y

is an integer m , which must lie in (0, n ], we see that there are two greatest terms x+ y Tm and Tm + 1. (Why). Here’s the explanation: We have q >1

Now, if

q m

1 ≤ r < m and Tr > Tr +1 for m + 1 ≤ r ≤ n

⇒

Tr < Tr +1 for

⇒

Tm −1 < Tm , Tm +1 > Tm + 2 , Tm = Tm +1

⇒

Tm and Tm +1 are the two greatest terms

( n + 1) y x+ y

for 1 ≤ r < m and

( n + 1) y = m. is a non-integer, assume x+ y

We now have q > 1 for 1 ≤ r ≤ m and q < 1 for r > m

⇒

Tr < Tr +1 for 1 ≤ r ≤ m and Tr > Tr +1 for r > m

⇒

Tm < Tm +1 , Tm +1 > Tm + 2

⇒

Tm +1 is the greatest term.

Mathematics / P & C, Binomial Theorem

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Example – 5

How will you expand the multinomial expression ( x1 + x2 + ... + xm ) ? n

Solution: We will approach this problem using combinatorics. Note that a general term of the expansion would be of the form (without the coefficient) x1n1 x2n2 x3n 3 ....xmnm

... (1)

where the various powers must always sum to n (why?). i.e., n1 + n2 + n3 + ... + nm = n Now, to evaluate the coefficient of the term in (1), we consider the multinomial expression in expanded form: x1 + x2 + ... + xm )( x1 + x2 + ... + xm ).............. ( x1 + x2 + ...xm ) ("########## #$########### % n times

To generate the term in (1), we must get x1 from n1 terms, x2 from n2 terms and so on. Let us find the number of ways in which this can be done. First select those n1 multinomials that will contribute x1 : this can be done in n Cn1 ways. Now, from the remaining ( n − n1 ) multinomials, select those n2 multinomials that will contribute x2 : this

can be done in (

n − n1 )

Cn2 ways. Continuing this process, we see that the number of ways to get

x1 from n1 , x2 from n2 ... and so on, that is, the number of times the term in (1) will be generated in the expansion is n

Cn1 × (

n − n1 )

Cn2 × (

n − n1 − n2 )

Cn3 × ...

=

( n − n1 )! × ( n − n1 − n2 )! × ... n! × n1 !( n − n1 )! n2 !( n − n1 − n2 )! n3 !( n − n1 − n2 − n3 )!

=

n! n1 ! n2 !...nm !

This is what is known as the general multinomial coefficient. The multinomial expansion can now be written compactly as

( x1 + x2 + ... + xm )

n

Mathematics / P & C, Binomial Theorem

=∑

n! x1n1 x2n2 ... xmnm n1 ! n2 !...nm !

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where the summation is carried out over all possible combinations of the ni 's such that ∑ ni = n . For example, in ( x1 + x2 + x3 ) , let us consider some terms in the expansion: 4

Multinomial

Term

Coefficient 4! = 12 2!1!1!

x12 x2 x 3

4! =4 3!1!

x13 x2 ( x1+x2 + x3)

4

4! 4!

x24

=1

4! = 12 1!1!2!

x1 x2 x 23 etc

Example – 6 10

Find the coefficient of x4 in the expansion of 1 + x − 22 . x Solution: From the previous example, the general term in the expansion will be n −2 10! (1) n1 ( x ) 2 2 n1 ! n2 ! n3 ! x

n3

10 x n2 −2 n3 (−2) n3 n1 ! n2 ! n3 !

=

where n1 + n2 + n3 must be 10. Now, x4 is generated whenever n2 − 2n3 = 4. The possible values of the triplet (n1 , n2 , n3 ) can now simply be listed out: (n1 , n2 , n3 ) ≡ (6, 4, 0), (3, 6, 1), (0, 8, 2) Thus, the (total) coefficient of x4 is 10! 10! 10! ( −2) 0 + ( −2)1 + ( −2) 2 6! 4! 0! 3! 6!1! 0!8! 2!

= – 1290 (verify) Mathematics / P & C, Binomial Theorem

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Section - 2

70

-1 DIFFERENTIATIONSection & INTEGRATION TECHNIQUES

The techniques of calculus enable us to sum a lot of series involving binomial coefficients. This is the subject of this section. Suppose that we have to evaluate the sum S given by S = n C1 + 2 n C2 + 3 n C3 + ...... + n n Cn

From now on, to avoid clutter, we’ll write n Cr as simply Cr, where the upper index n should be understood to be present. Thus, S = C1 + 2C2 + ..... + n Cn

=∑ rCr This series can be generated using a manipulation involving differentiation, as follows: Consider the binomial expansion (1 + x ) n = C0 + C1 x + C2 x 2 + ...... + Cn x n

If we differentiate both sides with respect to x, look at what we’ll obtain: n(1 + x ) n −1 = C1 + 2C2 x + 3C3 x 2 + ..... + nCn x n −1

Now, all that remains is to substitute x = 1, upon which we obtain: n ⋅ 2n −1 = C1 + 2C2 + 3C3 + ..... + n Cn

This is what we were looking for. Thus, S = n ⋅ 2n −1 Had we substituted x = −1 , we would’ve obtained 0 = C1 − 2C2 + 3C3 ....... + ( −1) n −1 n Cn

Thus, we have evaluated another interesting sum. Suppose that we now wish to evaluate S1 given by S1 = C0 +

C C1 C2 + + .... + n 2 3 n +1

The alert reader would immediately realize that integration needs to be applied here. How exactly to do so is now described. Consider again the expansion. (1 + x ) n = C0 + C1 x + C2 x 2 + ..... + Cn x n

If we integrate this with respect to x, between some limits say a to b, we obtain b

b

b

b

(1 + x)n+1 x2 x3 x n+1 b = C0 x a + C1 + C2 + .... + Cn n +1 a 2 a 3 a n +1 a Mathematics / P & C, Binomial Theorem

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To generate the sum S1 , a little thought will show that we need to use a = 0, b = 1, so that we obtain C C C 2n +1 − 1 = C0 + 1 + 2 + ..... + n n +1 n +1 2 3 2n +1 − 1 Thus, S1 equals n +1 Try some other values for a and b and hence generate other series on your own. Be as varied as you can in choosing these limits.

Example – 7

Find the sum S given by S = 12 ⋅ C1 + 22 ⋅ C2 + 32 ⋅ C3 + .... + n 2 ⋅ Cn

Solution: We have to plan an approach wherein we are able to generate r2 with Cr. We can generate one r with every Cr, as we did earlier, and which is now repeated here: (1 + x)n = C0 + C1x + C2 x2 + ...... + Cn xn

Differentiating both sides with respect to x, we have n(1 + x)n−1 = C1 + 2C2 x + 3C3 x 2 + ...... + nCn x n−1

Now we have reached the stage where we have an r with every Cr. We need to think how to get the other r. If we differentiate once again, we’ll have r(r – 1) with every Cr instead of r2(understand this point carefully). To ‘make-up’ for the power that falls one short of the required value, we simply multiply by x on both sides of the relation above to obtain: nx (1 + x ) n −1 = C1 x + 2C2 x 2 + 3C3 x 3 + .... + nCn x n

It should be evident now that the next step is differentiation: n ( n − 1) x (1 + x ) n − 2 + n (1 + x ) n −1 = C1 + 2 2 ⋅ C2 x + 32 ⋅ C3 x 2 + .... + n 2 ⋅ Cn x n −1

Now we simply substitute x = 1 to obtain n ( n − 1) ⋅ 2 n − 2 + n ⋅ 2 n −1 = C 1 + 2 2 ⋅ C 2 + 3 2 ⋅ C 3 + ..... + n 2 ⋅ C n

The required sum S is thus S = n(n − 1) ⋅ 2n − 2 + n ⋅ 2n −1 = n ⋅ 2n −2 {(n − 1) + 2}

= n(n + 1) ⋅ 2n− 2 Mathematics / P & C, Binomial Theorem

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Example – 8

Evaluate the following sums: (a) S1 =

C0 C2 C4 + + + ....... 1 3 5

(b) S 2 =

C1 C3 C5 + + + ....... 2 4 6

Solution: The first sum contains only the even-numbered binomial coefficients, while the second contains only odd-numbered ones. Recall that we have already evaluated the sum S given by C C1 C2 2n +1 − 1 + + ...... + n = n +1 n +1 2 3 Note that S is the sum of S1 and S2, i.e., S = C0 +

2n +1 − 1 S1 + S2 = n +1 Thus, if we determine S1, S2 is automatically determined, and vice-versa. Let us try to determine S1 first. (a) Consider again the general expansion (1 + x ) n = C0 + C1 x + C2 x 2 + .... + Cn x n

Integrating with respect to x, we have (we have not yet decided the limits) b

b

b

b

(1 + x)n+1 x2 x3 xn+1 b = C0 x a + C1 + C2 + ..... + Cn n +1 a 2 a 3 a n +1 a Since we are trying to determine S1 which contains only the even-numbered terms, we have to choose the limits of integration such that the odd-numbered terms vanish. This is easily achievable by setting a = – 1 and b = 1 (understand this carefully). Thus, we have

2n+1 C C = 2 C0 + 2 + 4 + .... n +1 3 5 which implies that 2n S1 = n +1 (b) S2 is now simply given by S 2 = S − S1 =

2 n +1 − 1 2n − n +1 n +1

=

2n − 1 n +1

Mathematics / P & C, Binomial Theorem

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Section - TECHNIQUES 1 MISCELLANEOUS

Section - 3

Not all questions can be subjected to the method(s) described earlier. For example, consider the sum S given by S = C0C1 + C1C2 + C2C3 + ...... + Cn −1Cn Let us first go through a combinatorial approach, using the observation that C0 = Cn , C1 = Cn−1 and so on, so that S can be rewritten as S = CnC1 + Cn −1C2 + Cn − 2C3 + ...... + C1Cn Consider a general term of this sum, which is of the form Cn − r Cr +1 . We can think of this as the number of ways of selecting (n – r) boys from a group of n boys and (r + 1) girls from a group of n girls. The total number of people we are thus selecting is (n − r ) + (r + 1) = (n + 1) . Therefore, S represents the total number of ways of selecting (n + 1) people out of a group of 2n, so that S is simply

2n

Cn +1 .

Now to a binomial approach. This will involve generating the general term Cr Cr +1 somehow, which is the same as Cn − r Cr +1 . Consider the general expansion of (1 + x) n . (1 + x ) n = C0 + C1 x + C2 x 2 + ..... + Cn x n

...(1)

We have to have the terms CnC1 , Cn−1C2 and so on, which suggests that we write (1) twice, but in the second expansion we reverse the terms, multiply, and see what terms contain the (combinations of) coefficients we require. n

(1 + x) = C0 n

2

+ C1x n

n

+ C2x n–1

+ ..........+ Cnx n–2

(1 + x) = Cnx + Cn – 1x + Cn – 2x

+ .....+ C1x + C0

Multiplying, we find on the left hand side we have (1 + x) 2 n , while on the right hand side, the terms containing the (combinations of) coefficients we want will always be of the form (

) x n+1 , that is, the power of x will be

(n + 1). No other terms will contain x n +1 , verify this for yourself. Thus, the sum Cn C1 + Cn−1C2 + ........ + C1Cn is actually the total coefficient of x n +1 on the right hand side, and from the left hand side we know that the coefficient of x n +1 would be simply 2 n Cn +1 . Thus, S = 2 n Cn +1 A very similar approach could have been: n

2

n

(1 + x) = C0 + C 1x + C2x + ........ + Cnx 1 1+ x

n

Mathematics / P & C, Binomial Theorem

1 1 = C0 + C1 1 + C2 2 + ........ + Cn n x x x www.locuseducation.org

LOCUS

Thus,

74

S

n

=

1 n Coefficient of x in (1 + x ) 1 + x

=

Coefficient of x in

=

Coefficient of x n +1 in (1 + x ) 2 n = 2 nCn +1

(1 + x) 2 n xn

Example – 9

Find the sum S given by S = C02 + C12 + C22 + ...... + Cn2

Solution: Note that S can be rewritten as S = C0Cn + C1Cn−1 + C2Cn− 2 + ...... + Cn C0 Using a combinatorial approach, the sum should be immediately obvious to the alert reader as 2n Cn . In brief, this is because the right hand side represents, as an example, the total number of ways of selecting n people from a group of n boys and n girls, etc. Now, we discuss the binomial expansion approach:

(1 + x ) n = C 0 *

+ C1 x

+ C 2 x 2 + C 3 x 3 + ..... + C n x n

*

*

)

(1 + x ) n = C n x n + C n −1 x n −1 + C n − 2 x n − 2 + ..... + C 0

Thus, we observe that the required sum is the coefficient of xn in (1 + x) 2 n , which is simply

2n

Cn .

Example – 10

Find the sum S given by S = n C0 +

n +1

C1 +

n +2

C2 + ........ +

n+r

Cr

Solution: We have already evaluated this sum in P & C; here we’ll use a binomial approach. Note that n+ r

⇒

∑

n+r

Cr = Coeff. of x n in (1 + x) n+ r Cr = ∑ (Coeff. of x n in (1 + x) n + r )

= Coeff. of x n in ∑ (1 + x)n + r

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Thus, S = Coeff. of x n in (1 + x )n + (1 + x )n +1 + ...... + (1 + x )n + r

{

= Coeff. of x n in (1 + x) n 1 + (1 + x ) + (1 + x) 2 + ...... + (1 + x ) r = Coeff. of x in n

(1 + x)n {(1 + x) r +1 − 1} x

= Coeff. of x n +1 in {(1 + x ) n + r +1 − (1 + x) n } =

⇒

S=

n + r +1

Cr

n + r +1

Cr , which is the same result we obtained in P & C.

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TRY YOURSELF - I

Q 1.

Find the sum of all the rational terms in the expansion of (31/ 4 + 41/ 3 ) .

Q. 2

Prove that three consecutive terms in a binomial expansion can never be in G.P.

12

( ) is even (b) Show that the integral part of (8 + 3 7 ) is odd 67

Q. 3* (a) Show that the integral part of 5 5 + 11

n

Q. 4

Use the binomial theorem to show that 7103 when divided by 5 leaves a remainder 3.

Q. 5

Find the coefficient of x301 in the expansion of (1 + x )500 + x (1 + x)499 + x 2 (1 + x )498 + ...... + x 500 2n

Q. 6* Suppose (1 + x + x ) is written in expanded form, i.e., in the form 2 n

∑ax r =0

r

r

. Show that

ar = a2n−r 32

Q. 7* Use the binomial theorem to show that 3232 when divided by 7 leaves the remainder 4. Q. 8

Prove the following relations: (a) n C0 ⋅ 2 nCn − n C1 ⋅ 2 n −1Cn + n C2 ⋅ 2 n −2Cn − ....... + n Cn (−1)n ⋅ n Cn = 1 (b) n C0 ⋅ 2 nCn − nC1 ⋅ 2 n − 2Cn + nC2 ⋅ 2 n − 4Cn − ..... = 2n (c) C0Cr + C1Cr +1 + C2Cr + 2 + ...... + Cn − r Cn = 2 nCn + r (d)* m C1 n Cm − mC2 2 n Cm + mC3

3n

Cm − ..... + (−1) m −1 ⋅ mCm

(e) (C 0 + C1 )(C1 + C 2 )..... (C n −1 + C n ) = Q. 9

nm

Cm = ( −1) m +1 ⋅ n m

( n + 1) n C1C 2 .....C n n!

Find the sums of the following series. (a) C0 − 2C1 + 3C2 − 4C3 + .....

(b)*

C0 C1 C2 C3 − + − + ..... 3 4 5 6

Q. 10 Using the binomial theorem, show that r

n

Cr = ∑ k C j ⋅ n − k Cr − j j =0

Q. 11 Find the sum of the series n

∑ r =0

1 3r 7r 15r (−1)r n Cr r + r + r + r + ....m terms 2 4 8 16

Q. 12* Show that n

Cm + 2 ⋅ n −1Cm + ....... + (n − m + 1) ⋅ mCm =

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Section - RATIONAL 1 BINOMIAL THEOREM, INDEX

Section - 4

In the previous section, we discussed the expansion of ( x + y ) n , where n is a natural number. We’ll extend that discussion to a more general scenario now. In particular, we’ll consider the expansion of (1 + x) n , where n is a rational number and | x | < 1. Note that any binomial of the form (a + b )n can be reduced to this form.:

b ( a + b) = a 1 + a

n

n

b = a n 1 + a

(we are assuming | a | > | b |)

n

= a n (1 + x) n where | x | < 1 The general binomial theorem states that (1 + x )n = 1 + nx + ........ +

n(n − 1) 2 n(n − 1)(n − 2) 3 x + x + ........ 2! 3! n( n − 1)( n − 2)......(n − r + 1) r x + .......∞ r!

That is, there are an infinite number of terms in the expansion with the general term given by Tr +1 =

n( n − 1)(n − 2).......(n − r + 1) r x r!

For an approximate proof of this expansion, we proceed as follows: assuming that the expansion contains an infinite number of terms, we have:

(1 + x )

= a0 + a1 x + a2 x 2 + a3 x 3 + ... + an x n + ...∞

n

Putting x = 0 gives a0 = 1. Now differentiating once gives n (1 + x )

n −1

= a1 + 2a2 x + 3a3 x 2 + .....∞

Putting x = 0 gives a1 = n. Proceeding in this way, we find that the rth coefficient is given by an =

n ( n − 1)( n − 2 )... ( n − r + 1) r!

Note that if n is a natural number, then this expansion reduces to the expansion obtained earlier, because Tr +1 becomes n Cr x r , and the expansion terminates for r > n . For the general Tr +1 , we obviously cannot use n Cr since that is defined only for natural n. One very important point that we are emphasizing again is that the general expansion holds only for | x | < 1 . Mathematics / P & C, Binomial Theorem

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Let us denote the genral binomial coefficient by Vr. Thus, we have Vr =

n(n − 1)(n − 2)......(n − r + 1) r! ∞

and

(1 + x) n = ∑ Vr x r r =0

Let us discuss some particularly interesting expansions. In all cases, | x | < 1: (1) (1 + x) −1 :

Since

n = −1, we see that

Vr =

( −1)(( −1) − 1)((−1) − 2)......(( −1) − r + 1) r!

= (−1)r so that the expansion is (1 + x )−1 = 1 − x + x 2 − x 3 + ......∞ (2) (1 − x) −1 :

Again, Vr = ( −1) r and thus (1 − x) −1 = 1 + x + x 2 + x 3 + ......∞

(3) (1 + x) −2 :

We have n = – 2; Vr =

( −2)((−2) − 1)((−2) − 2)....((−2) − r + 1) r!

(−1) r (r + 1)! = r! = (−1)r ⋅ (r + 1) Thus, (1 + x)−2 = 1 − 2x + 3x2 − 4x3 + ......∞ (4) (1 − x) −2 :

Again, Vr = ( −1) r ⋅ ( r + 1) so that (1 − x) −2 = 1 + 2 x + 3 x 2 + 4 x 3 + ........∞

Example – 11

(a) For | x | < 1 , expand (1 − x) −3 (b) Find the coefficient of x n in the expansion of (1+ 3x + 6x2 +10x3 + .......∞)−n Mathematics / P & C, Binomial Theorem

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Solution: (a) We have Vr =

( −3) (( −3) − 1)((−3) − 2))......(( −3) − r + 1) r!

=

(−1) r (r + 2)! 2(r !)

=

(−1) r (r + 1) ( r + 2) 2

Thus, V0 = 1, V1 = −3, V2 = 6, V3 = −10........ so that (1 − x) −3 = 1 + 3 x + 6 x 2 + 10 x3 + .......∞ (b) We use the result of part –(a) in this: (1 + 3 x + 6 x 2 + 10 x 3 .......∞ ) − n = ( (1 − x ) −3 )

−n

= (1 − x)3n The coefficient of x n in this binomial expansion (note: the power is now a positive integer) would be ( −1) n ⋅ 3 nCn . Example – 12

Find the magnitude of the greatest term in the expansion of (1 − 5y )

−2 / 7

1 for y = . 8

Solution: Let us first do the general case: what is the greatest term in the expansion of (1 + x) n , where n is an arbitrary rational number. We have, Tr +1 = Vr x r

and

Tr = Vr −1 x r −1 Tr +1 Vr = ⋅x Tr Vr −1

so that

=

n − r +1 ⋅x r

Now, let us find the conditions for which this ratio exceeds 1. We have

Tr +1 ≥ Tr ⇒ Mathematics / P & C, Binomial Theorem

1 n +1 −1 ≥ r |x|

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For this particular problem, (1) becomes −2 +1 1 7 −1 ≥ −5 r 8 5 8 −1 ≥ 7r 5

⇒ ⇒

5 13 ≥ 7r 5

⇒

r≤

⇒

r=0

25 91

Thus, Tr +1 = T1 is the greatest term, with magnitude 1. Example – 13

Find the sum of the series 1+

2 2.5 2 ⋅5 ⋅8 + + + ......∞ 6 6.12 6 ⋅12 ⋅18

if you are told that this corresponds to an expansion of a binomial, of the form (1 + x) n . Solution: We need to determine n and x. For that, we can compare the terms of this series with the corresponding terms in the following general expansion. (1 + x ) n = 1 + nx +

n( n − 1) 2 n( n − 1) ( n − 2) 3 x + x + ......∞ 2! 3!

Thus, nx =

2 6

n( n − 1) 2 2.5 x = 2! 6.12

Solving for n and x from these two equations, we get n = −

2 1 and x = − . Thus, the sum of the 3 2

series is

S = (1 + x )

n

1 = 1 − 2

−

2 3

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Example – 14

Find the sum of the series 1 1 1 + + + ...... to n terms a + b a + 2b a + 3b

for b

View more...
01.

Fundamental Principle of Counting

01 - 04

02.

Introduction To Permutations

05 - 07

03.

Introduction To Combinations

08 - 09

04.

Applications of Basics

10 - 26

05.

More Applications

27 - 59

06.

Appendix : Binomial Theorem

60 - 86

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P & C, Binomial Theorem This chapter is one of the most interesting chapters that we’ll study at this level. The beauty and challenge of this branch of mathematics lies in the innumerous tricks and mathematical artifices that abound in this subject. Clarity of thought, more than any thing else, is what is required to understand the subject properly. Also, you’d do best if you refrain from memorising any formulae or particular cases here; concentrate on building a logical approach, solving everything from first principles. The main objective of this chapter is to count. Given a set U of things or objects or persons ( or whatever), we need to arrange a subset S of U (according to some constraints) or select a subset S of U (again, according to certain criteria). In fact, we are actually interested in counting the number of such arrangements or selections. Read the following questions: “From a team of 15 cricket players, how can we select a playing team of 11players? “There are 20 people whom we need to seat in 2 rows of 10 seats each. How many ways exist of doing so? “From a deck of 52 playing cards, in how many ways can we select two red cards and three black cards?” “How many rectangles exist on a standard 8×8 chessboard?” “How many factors does 144000 have? In general, how many factors does a natural number N have?” These are some of the many types of questions which we’ll learn to solve in this chapter. We’ll build a systematic approach to deal with such counting issues. To really appreciate the beauty of the solving techniques that we’ll develop, you are urged to try out each and every question that we solve here, on your own first, and only then look at the solution. Only this approach will help you solve counting questions elegantly. Section - 1

Section - 1 OF COUNTING FUNDAMENTAL PRINCIPLE

The fundamental principle of counting is so fundamental that you already must have used it practically a lot many times without realising it. In other words, this principle is already programmed into our minds. A logical, step-by-step application of this principle gives rise to the entire subject of permutations and combinations. Suppose you have 4 boys and 3 girls. From this group of 7, you want to select a couple ( a boy and a girl). How many ways exist of forming this couple? Let us label the boys as B1 , B2 , B3 and B4 and the girls as G1 , G2 and G3 . Mathematics / P & C, Binomial Theorem

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There are 4 ways to choose a boy. Once you’ve chosen a boy, say B2 , there are now 3 ways to choose a girl. In other words, the boy B2 can form the couple ( B2 G1 ), ( B2 G2 ) or ( B2 G3 ) . Similarly, for every other boy, there exist 3 girls with whom that boy can be paired. Thus, the total number of ways in which a pair can be formed is:

N = ( No. of ways of selecting a boy ) × ( No. of ways of selecting a girl ) = 4×3 = 12

The most crucial aspect in this calculation is that you must realise that the task of selecting a boy is independent of the task of selecting a girl. This means that which boy you select has no effect what so ever on which girl you select; the selection of a boy and that of a girl are independent of each other. Another subtle point must be made here. There are 4 ways to select a boy. These 4 ways are mutually exclusive. This means that a selection of any particular boy, once made, rules out the selection of the other 3 boys. Similarly, there are 3 mutually exclusive ways to select a girl. Let us consider another example now. We need to travel from New Delhi, India to Fiji Islands (in the South Pacific Ocean). We must change flights, first at Singapore and then at Sydney, Australia. There are 6 different flights available from New Delhi to Singapore, 5 from Singapore to Sydney and 3 from Sydney to Fiji.

Singapore

New Delhi

6 flights

Sydney

5 flights

Fiji Islands

3 flights

Fig - 1

How may ways exist of making a flight plan from New Delhi to Fiji? Assume that we are in Singapore. From Singapore, we have 5 available flights to Sydney (5 mutually exclusive ways). For each of these 5 flights, we have 3 further ways (flights) from Sydney to Fiji. Thus, we have a total of 5 × 3 = 15 ways of travelling from Singapore to Fiji. Now, how many ways do we have to reach Singapore from New Delhi in the first place? 6 flights. For each of these 6 flights, we have 15 further ways of reaching Fiji From Singapore (as we just calculated in the preceding paragraph). Thus, the total number of ways N of travelling from New Delhi to Fiji is 6 × 15 = 90. In other words,

(

)(

)(

N = No. of flights from New × No. of flights from × No. of flights from Delhi to Singapore Singapore to Sydney Sydney to Fiji

)

Intuitively easy, isn’t it? Let us now generalise the results of these two examples into our fundamental principle of counting. Consider the set of tasks {T1 , T2 , T3 ... Tn } which are all independent of each other. This means that task Ti has “no relation ” to task T j if i ≠ j; the choice of how to accomplish task Ti has thus “no effect” on the choice of how to accomplish talk T j for i ≠ j , (you’ll realise the meaning of “no relation” and “no effect” more specifically later). Task Ti can be accomplished in ki mutually exclusive ways. Mathematics / P & C, Binomial Theorem

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The fundamental principle of counting says that the set of tasks {T1 , T2 , T3 ... Tn } can be accomplished in k1 × k2 × k3 × ... × kn ways. Here are more examples that illustrate this simple yet extremely powerful principle. A proper understanding of this principle is absolutely essential for the subject of counting to be fully comprehensible. • Think of a standard six-faced die, with the markings 1, 2, 3, 4, 5 and 6. Suppose that you have two such dice. When these two dice are thrown call the number that shows up on the first die x and the one on the second die y. How many pairs ( x, y ) are possible? There are 6 mutually exclusive ways in which a number can show up on the first die. Similarly, 6 such ways exist for the second die. The fundamental principle of counting says that the total number of possible pairs is 6 × 6 = 36. •

Suppose that we have a (unlimited) supply of the letters A, B, C , D, E , F available with us. How many 5-letter chains can we form using these letters? Imagine 5 blank spaces for the 5-letter chain that we are supposed to form (numbered 1 to 5):

1

2

3 Fig - 2

4

5

•

We now have 5 tasks at hand; each task correspond to filling a space in Fig. 1. Realise that these 5 tasks are independent of each other. For example, what you fill in place 2 has no effect on what you fill in place 5 since you’ ve been assured an unlimited supply of letters. Each task can be accomplished in 6 possible ways. For example, you can fill place -1 in 6 possible ways : with A, B , C , D , E , or F . Thus, by the fundamental principle of counting, the total number of ways to form the 5-letter chain would be 6 × 6 × 6 × 6 × 6 = 65 .

•

Now consider the scenario when you don’t have an unlimited supply of letters available. Suppose that you have only one of each of the 5 letters available. How many 5-letter chains can be formed now? You must realise that this limited-supply-of-letters situation is very different from the previous one. In this situation, once you fill a particular place with a particular letter, you are left with one letter less to choose from, from the remaining places. Let us start with filling the places from left to right. To fill place -1, w e have 6 letters to choose from, but once we’ve filled place-1 , we now have only 5 letters to choose from, to fill place-2. Continuing in this way, place-3, place - 4 and place -5 can then be filled in 4, 3 and 2 ways respectively. 1

2

3

! 6 ways to fill this place

! 5 ways to fill this place

! 4 ways to fill this place

4 ! 3 ways to fill this place

5 ! 2 ways to fill this place

Fig - 3 The fundamental principle of counting tells us that the total number of chains in this case will be 6 × 5 × 4 × 3 × 2 = 720 . You might think that in this case filling up place-i and place-j are not independent tasks. In other words, how we fill place-i has a certain relation to how we Mathematics / P & C, Binomial Theorem

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fill place-j . For example, if we fill place-1 with A, we cannot fill any other place with A and thus, how we fill place -1 has a definite effect on how we are able to fill the other places. This is definitely true. But when we talk about the independence of two events, it is in a different sense: Let event X be filling place-1 and event Y be filling place-2 after event X has been accomplished, i.e. after place-1 has been filled. There are 6 ways of accomplishing X and 5 ways of accomplishing Y. Which 5 letters contribute to event Y (i.e which of the 5 remaining letters can we choose for place -2) definitely depends on how event X was accomplished, but the event Y is itself the choice of one of the 5 symbols contributing to event Y. The number of ways in which this choice can be made is still independent of how event X was accomplished. Whatever selection was made for event X, the number of ways in which Y can be accomplished still remains 5. This is the sense that should be attached to the phrase “independent events”. ____________________________________________________________________________________ These three examples should have given you a pretty good idea about three concepts : mutually exclusive events, independent events and the fundamental principle of counting. Ponder over these examples for a bit longer and think up examples of you own till your feel very comfortable with these new concepts.

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Section 1 INTRODUCTION TO -PERMUTATIONS

Section - 2

We now formally study the concept of permutations, by generalising the last example in the preceding section. The fundamental issue in Permutations is the arrangement of things. In the last example, we had 6 letters and 5 places where we could arrange (5 of) those 6 letters, We calculated that there are 720 ways of arranging those 6 letters, taken 5 at a time. In mathematical terminology, we calculated as 720 the number of permutations (arrangements) of 6 letters taken 5 at a time. Let us generalise this: Suppose we have n people. It we had n seats available to seat these people, the total number of ways to do so would be (by the logic discussed in the preceding section) n × ( n − 1) × ( n − 2) × ... × 1. This quantity is denoted by n!. n ! = n × ( n − 1) × ( n − 2) × ... × 1

Suppose now that we have only r seats, where r < n. The total number of ways now would be n × ( n − 1) × (n − 2) × ... × ( n − r + 1) . This is the number of ways of permuting n things, taken r at a time, and the notation used for this number is n Pr . Thus : n

Pr = n × (n − 1) × (n − 2) × ... × (n − r + 1)

=

n × (n − 1) × ( n − 2) × ... × (n − r + 1) × (n − r ) × ...1 (n − r ) × ... × 1

=

n! (n − r )!

Thus, using the fundamental principle of counting, you see that we’ve been able to calculate the value of n Pr . Finally, note that n Pn = n ! Lets apply this discussion to an example: Example – 1

Find the number of permutations of the word EDUCATION which (a) consist of all the letters of this word

(b) start with E and end with N

(c) contain the string “CAT”

(d) have the letters C, A and T occurring together

(e) start and end with vowels

(f) have no two vowels occurring together.

Solution: (a) We have 9 letters and we want to permute all of them. The required number of arrangements would be 9 P9 = 9! Mathematics / P & C, Binomial Theorem

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(b) Fix E at the start and N at the end of the word

E

N Fig - 4

We now have 7 places which we need to fill using the remaining 7 letters. The number of such permutations will be 7 P7 = 7!. (c) We want all those permutations in which the string “CAT” occurs. Let us treat “CAT” as a single letter/object since this is what we want - we want “CAT” to appear as a single entity. We now have the following objects (letters) which we need to permute: E

D

U

“CAT ”

I

O

N

These are in total 7 objects. Thus, they can be permuted in 7 P7 = 7! ways. This is the number of permutations that contain the string “CAT ”. (d) We now need the permutations which contain the letters C, A and T occurring together. This means that they can occur together in any order. There are 3! = 6 ways in which C, A and T can be arranged among themselves, namely CAT, CTA, ACT, ATC, TAC, and TCA. Now consider from the previous part all those permutations in which the string “CAT” occurs. These are 7! in number. Corresponding to each such permutation, we can have 6 permutations in which the constraint is that C, A and T occur together. This is because the string “CAT” can itself be permuted in 6 ways as described above. For example, consider the permutation “EDUCATION”; to this permutation will correspond 6 permutations in which C, A and T occur together: EDU EDU EDU CAT ION → EDU EDU EDU EDU

CAT ION CTA ION ACT ION ATC ION TAC ION TCA ION

The required number of permutations is therefore 7!×3! . (e) In the word EDUCATION, there are a total of 5 vowels. We want to have the starting and ending letters of our permutations as vowels. Let us first arrange the starting and ending letters and then fill in the rest of the 7 places. Out of the 5 vowels available to us, the first and the last place can be filled in 5 P2 places. Once we’ve fixed the first and the last letters, the remaining 7 places can be filled in 7 P7 = 7! ways. Thus, the total number of required permutations is 5 P2 × 7! .

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(f) There are 5 vowels and 4 consonants in the word EDUCATION. We require permutations in which no two vowels occur together. We can ensure such permutations by first fixing the 4 consonants and then arranging the 5 vowels in the 5 possible places that arise as depicted in the following figure. D

C

T

N

Fig - 5 Convince yourself that any arrangement of the 5 vowels in the 5 blank spaces above will correspond to a permutation with no two vowels together. The number of ways of arranging the 4 consonants is 4 P4 = 4!. After the consonants have been arranged, the number of ways of arranging the 5 vowels in the 5 blank spaces as depicted in Fig.5 is 5!. Thus, the required number of permutations is 4! × 5!

____________________________________________________________________________________ It would be a good exercise for you to construct more of such examples on your own and solve them. You can take an arbitrary word and find the number of permutations of that word, that satisfy any particular condition that you can think of.

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Section 1 INTRODUCTION TO -COMBINATIONS

Section - 3

In the previous section, we encountered permutations, which correspond to arrangement of objects /things/ entities. In this section, we will encounter combinations which correspond to selection of things (and not their arrangement). We do no intend to arrange things. We intend to select them. For example, suppose we have a team of 15 cricket players. We intend to select a playing team of 11 out of these 15 players. Thus, we want the number of ways in which we can select 11 players out of 15 players. (We are not interested in arranging those 11 players in a row - only the group/ combination of those 11 players matters). Let us make this concept more specific. Suppose we have a set of 6 letters { A, B, C , D, E , F } . In how many ways can we select a group of 3 letters from this set? Suppose we had to find the number of arrangements of 3 letters possible from those 6 letters. That number would be 6 P3 . Consider the permutations that contain the letters A, B and C. These are 3! = 6 in number, namely ABC, ACB, BAC, BCA, CAB and CBA. Now, what we want is the number of combinations and not the number of arrangements. In other words, the 6 permutations listed above would correspond to a single combination. Differently put, the order of things is not important; only the group/combination matters. This means that the total number of combinations of 3 letters from the set of 6 letters available to us would be 6 P3 / 3! since each combination is counted 3! times in the list of permutations. Thus, if we denote the number of combinations of 6 things taken 3 at a time by 6C3 , we have C3 =

6

6

P3 3!

In general, suppose we have n things available to us, and we want to find the number of ways in which we can select r things out of these n things. We first find the number of all the permutations of these n things taken r at a time. That number would be n Pr . Now, in this list of n Pr permutations, each combination will be counted r! times since r things can be permuted amongst themselves in r! ways. Thus, the total number of combinations of these n things, taken r at a time, denoted by nCr , will be

Cr =

n

n

Pr n! = r ! r !(n − r )!

You should now be able to appreciate the utility of the fundamental principle of counting. Using only a step-bystep application of this principle, we have been able to obtain an expression for n Cr . As we progress through the chapter, you’ll slowly realise that each and every concept that we discuss and each and every expression that we obtain follows logically as a consequence of this simple principle.

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Example – 2

Consider the word EDUCATION. (a) In how many ways can we select a pair of letters consisting of a vowel and a consonant ? Both vowels? Both consonants? (b) In how many ways can a 5-letter selection be made with more than 2 vowels? Solution: (a) We have 5 vowels and 4 consonants available in the word EDUCATION. A vowel can be selected in 5 ways while a consonant can be selected in 4 ways. Thus, a pair consisting of a vowel and a consonant can be selected in 5 × 4 = 20 ways. Two vowels can be selected in 5 C2 ways while two consonants can be selected in 4 C2 ways. (b) Since we want more than 2 vowels, we could have 3, 4 or 5 vowels in our 5-letter selection. Suppose we select 3 vowels for our 5-letter group. This can be done in 5 C3 ways. The two remaining letters (consonants) can be selected in 4 C2 ways. Thus, 5-letter groups containing 3 vowels can be formed in 5C3 × 4C2 ways. Similarly, if we had 4 vowels, the possible number of groups would be 5C4 × 4C1 . There’s only 1 group possible containing (all) the 5 vowels. Thus, the number of required selections is 5

C3 × 4C2 +5 C4 × 4C1 + 1

____________________________________________________________________________________ Now that we’re done with the introductions, lets move on and see some really interesting and diverse applications of the basic concepts covered till now.

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Section - OF 1 BASICS APPLICATIONS

Section - 4

Our approach in the applications of the basic concepts of counting will be to keep everything really basic - and try to do everything from a first principles approach. Memorization of formulae will do no good! A good deal of thinking is required. Lets start with proving certain combinatorial assertions.

Example – 3

Prove that n Cr =n Cn −r Solution: We can easily prove this relation using the expression we obtained for n Cr : n

Cr =

n! n! = = n Cn − r r !(n − r )! (n − r )!(n − (n − r ))!

However, what we would really like is not an analytical justification like the one above but a logical justification, that involves no mathematical manipulations and is instead purely based on the interpretation of n Cr . For this example, let us discuss such a logical justification in a good amount of detail. Suppose you have a group of 6 letters, say {A, B, C, D, E, F}. Out of this group, you want to count the number of subsets containing 4 letters. Consider a particular selection of 4 letters, say {B, D, E, F}. This selection can equivalently be obtained if we say that we exclude the group {A, C} from our original group of 6 letters. This means that each selected group of 4 letters corresponds to an excluded group of 2 letters. The number of (selected) 4-letter groups will therefore equal the number of (excluded) 2-letter groups, or in other words, to count the number of 4 letter groups, we can equivalently count the number of 2-letter groups. Thus, 6

C4 = 6 C 2

Generalising this logic, we obtain n Cr =n C n−r

Example – 4

Prove that n Cr =n−1 Cr + n−1 Cr −1 Solution:

You can prove this assertion yourself very easily analytically. Let us discuss a logical justification. The left hand side of this assertion says that we have a group of n people out of which we want to select a subset of r people; more precisely, we want to count the number of such r-subsets. Fix a particular person in this group of n people, say person X. Now, all the r-groups that we form will either contain X or not contain X. These are the only two options possible.

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To count the number of groups that contain X, we proceed as follows: we already have X; we need (r – 1) more people from amongst (n – 1) people still available for selection. Thus, such groups will be n−1Cr −1 in number. For the number of groups that do not contain X we need to select r people from amongst (n –1) options available. Therefore, such groups are n −1Cr in number. The total number of r-groups are hence

n −1

Cr −1 + n−1 Cr in number, which is the same as n Cr . With this example, you should begin

to realise the beauty of the logic and skill required in this subject.

Example – 5

Prove that n Pr = n−1 Pr + r n−1Pr −1 Solution: The analytical justification is again very straightforward and is left to you as an exercise. The left hand side of this assertion says that we need to count the number of arrangements of n people, taken r at a time. We again fix a particular person, say person X. All the possible r-arrangements will either contain X or not contain X. These are the only two (mutually exclusive) cases possible. If we do not keep X in our permutation, we have r people to select from a potential group of (n – 1) people. The number of arrangements not containing X will therefore be n −1 Pr . To count the number of permutations containing X, we first seat X in one of the r seats available. This can be done in r ways. The remaining (r – 1) seats can be filled by (n –1) people in n −1 Pr −1 ways. Thus, the number of arrangements containing X is r ⋅ n−1Pr −1 . These arguments prove that n Pr = n−1 Pr + r ⋅n −1 Pr −1

Example – 6 n (a) Prove that Cr =

n r

n −1

Cr −1

n (b) Prove that Cr =

n − r +1 n Cr −1 r

Solution: (a) Let us consider this assertion in a particular example, say with n = 6 and r = 4. This will make things easier to understand. Our purpose is to select 4 people out of 6 people, say the set {A, B, C, D, E, F}. To select a group of 4, we can first select a single person : this can be done in 6 ways. The rest of the 3 people can now be selected in 5C3 ways. The total number of groups possible would thus be

6 × 5C3 . But some careful thought will show that we have ‘overcounted’, doing the calculation this way. Mathematics / P & C, Binomial Theorem

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Suppose that in our first step, we select A. While selecting the remaining 3 persons, we then select B, C, D thus forming the group {A, B, C, D}. But this same group would have been formed had we selected B in our first step and A, C, D in the second step, or C in the first step and A, B, D in the second step, or D in the first step and A, B, C in the second step. We have thus counted the group {A, B, C, D} 4 times in the figure 6 × 5C3. The actual number of groups will hence be

6 × 5C3 . 4

We now generalise this: to select r people out of a group of n, we first select one person; this can be done in n ways. The remaining (r – 1) persons can be selected in n−1Cr −1 ways. The total number of r-groups thus becomes n × n –1Cr −1 . However, as described earlier, in this figure each group has been counted r times. The actual number of r-groups is therefore n

n × n −1Cr −1 Cr = r

(b) The logic for this part is similar to that of part - (a) To select r people out of n, we first select (r – 1) people out of n. This can be done in n Cr −1 ways. We now have n − (r − 1) = (n − r + 1) persons remaining for selection out of which we have to choose 1 more person. This can therefore be done in (n – r + 1)ways. The total number of r-groups thus becomes (n − r + 1) × nCr −1. However, each r-group has again been counted r times in this figure (convince yourself about this by thinking of a particular example). The actual number of r-group is thus n

Cr =

(n − r + 1) × nCr −1 r

Example – 7

Prove that n C0 + nC1 + nC2 + .......... + nCn = 2n Solution: This looks like a tough one! Lets first interpret what the left side means. n C0 is the number of ways in which we can select ‘nothing’ out of n things (there will obviously be only one such way: that we do nothing!). n C1 is the number of ways in which we can select 1 thing out of n. n Cr is the n

n number of ways of selecting r things out of n. We want the value of the sum ∑ Cr , which is the r =0

number of all groups possible of any size what so ever. Thus, our selection could be any size from 0 to n (both inclusive); what we want is the total number of selections possible. For example, consider the set {A, B, C}. The set of all possible selections that can be made from this set is {φ , {A}, {B}, {C}, {AB}, {AC} , {BC}, {ABC}}. Thus, 8 total different selections are possible (note that 8 = 23 ) Mathematics / P & C, Binomial Theorem

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To count the total number of selections possible if we have n persons, we adopt an individual’s perspective. An individual can either be or not be in our selection. Thus, we have two choices with respect to any individual; we either put him in our group or do not put him in our group. These two choices apply to every individual. Also, choosing or not choosing any individual is independent of choosing or not choosing another. Thus, the total number of ways in which an × 2 ×# × ........ ×%2 = 2n 2$### arbitrary number of individuals can be selected from n people is 2"## n times

This proves that n

∑ nCi = 2n i =0

Example – 8

Prove that

n+ m

Cr = nC0 mCr + nC1 mCr −1 + nC2 nCr −2 + ......... + nCr mC0

Solution: We interpret the left hand side as the number of ways of select r people out of a group of (n + m) people. Let this group of (n + m) people consist of n boys and m girls. A group of r people can be made in the following ways: Sl.

Group contains

1 2

0 boys, r girls 1 boy, (r − 1) girls 2 boys, (r − 2) girls

3 . . . r. (r − 1) boys, 1 girl (r + 1) r boys, 0 girsl

No. of ways possible n

C0 × mCr

n

C1 × mCr −1

n

C2 × mCr −2

n

Cr −1 × mC1

n

Cr × mC0

This table is self-explanatory. The ( r + 1) types of groups that have been listed are mutually exclusive. Thus, the total number of r-groups is n+ m

Cr = nC0 mCr + nC1mCr −1 + ........ + nCr mC0

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Example – 9

(a) Consider the set of letters {a, a, a, b, c, d }. How many permutations of this set exist? (b) Consider the set of letters {a, a, a, b, b, c, d , e}. How many permutations of this set exist? (c) Generalise the results of the previous two parts. If we have a set of ( m + n + p + .....) things where a particular thing, say X, is repeated m times, Y is repeated n times, Z is repeated p times, and so on, find the number of permutations of this set. Solution: The main issue now is that we have a repetition of things. If the things we have were all different, the number of permutations would have been easy to calculate. But now, with repetition of things, the number of permutations will change (it will actually decrease, if you think about it carefully). Let us calculate the permutations in this case from first principles (a) We have 3 (repeated) “a” letters, a “b”, a “c” and a “d”. In all, 6 letters. Consider, for a moment, that the 3 “a” s are all different. Let us denote the 3 different “a”s by a1, a2 and a3. Our set of letters is now {a1, a2 , a3, b, c, d }. The number of permutations of this set is simply 6

P6 = 6!

If we list down all these 6! permutations, we will see that 6 permutations from this list will correspond to only one permutation, had the “a”s been all the same. Why? Consider any particular permutation with the “a” s all different, say {b, a1, a2 , c, a3, d }. If we fix the letters “b”, “c” and “d”, the 3 different “a” s can be permuted amongst themselves in 3! = 6 ways. We now list down all the 6 permutations so generated on the left hand side in the figure below, and see that these 6 permutations correspond to a single permutation if the “a” s were all the same: b a1 a2 c a3 d b a1 a3 c a2 d b a2 a1 c a3 d baacad b a2 a3 c a1 d b a3 a1 c a2 d b a3 a2 c a1 d The 6 permutations on the left with different “a”s as correspond to a single permutation with the “a” s all same. Fig - 6

Thus, the actual number of permutation with the “a”s all same will be 6! 6! = = 120 3! 6

(b) We now have 3 repeated “a” s and 2 repeated “b” s, and a total of 8 letters. If we for a moment take the “a” s and “b” s as all different, the total number of permutations of this set of 8 letters would be 8 P8 = 8! Mathematics / P & C, Binomial Theorem

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However, once we list down these 8! permutations, we will see that (as in the previous part) 3! = 6 permutations in this list will correspond to a single permutation if the “a”s were all the same. Similarly, 2! = 2 permutations in this list will correspond to a single permutation if both the “b”s were the same. Thus, the actual number of permutations if “a”s and “b”s were the 8! same would be 3!2! (c) These results can now easily be generalised for this general set and the number of permutations will be (m + n + p + .....)! m !n ! p !...... ____________________________________________________________________________________ From this example once again, the power of the fundamental principle of counting should be quite evident. Using a logical development/extension of this principle, we see that we’ve been able to solve non-trivial questions like the one above. Example – 10

Consider the integral equation x1 + x2 = 4 where x1, x2 ∈ & . The non-negative solutions to this equation

can be listed down as {0, 4} , {1,3} , {2, 2} , {3,1} and {4, 0} . Thus, 5 non-negative integral solutions exist for this equation. We would like to solve the general case. How many non-negative, integral solutions exist for the equation x1 + x2 + ..... + xn = r

Solution: You might be surprised to know that this question can be solved using the general result obtained in the previous example. Can you think how? Let us consider an arbitrary integral equation, say x1 + x2 + x3 = 8. Consider any particular nonnegative integral solution to this equation, say {2,3,3} . We some how need to “tag” this solution in a new form; a form which is easily countable. This is how we do it. We break up the solution 2 + 3 + 3 = 8 as shown below: 11

111 = 8

...(1)

1=8

...(2)

1111111 = 8

...(3)

111

Similarly, 1 + 6 + 1=8 would be written as 1

111111

and 0 + 1 + 7 = 8 would be written as 1

and 0 + 0 + 8 = 8 would be written as 11111111 = 8 Mathematics / P & C, Binomial Theorem

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An alert reader must have realised the ‘trick’ by now. In each of (1), (2), (3) and (4), we have on the left hand side 8 “1” symbols and 2 “

” symbols, in different orders. Any non-negative integral ” symbols.

solution can thus be represented by a unique permutation of 8 “1” symbols and 2 “ Conversely, every permutation of 8 “1” symbols and 2 “ negative integral solution to the equation .

” symbols represents a unique non-

Thus, the set of non-negative integral solutions to the equation and the set of permutations of 8 “1” symbols and 2 “

” symbols are in one-to-one-correspondence. To count the required number

of solutions, we simply count the permutations of 8 “1” symbols and 2 “ described in the last example would be

” symbols, which as

(8 + 2)! 10! = = 45 8!2! 8!2!

This beautiful artifice described about should make it clear to you the significance of (and the challenge of producing!) elegant proofs/solutions. We now generalise this result. Any non-negative integral solutions to the equation x1 + x2 + ..... + xn = r can be represented using r “1” symbols and n – 1 “ number of permutation of these symbols will be

” symbols. The total

(n + r − 1)! n + r −1 = Cr r !(n − 1)! and hence, this is the required number of solutions.

Example – 11

Consider a rectangular integral grid of size m × n. For example, a 4 × 5 integral grid is drawn alongside: A person has to travel from point A(0, 0) to the diagonally opposite point C(m, n). He moves one step at a time, towards the east or towards the north (that is, never moves towards the west or south at any time). How many distinct paths exist from the point A to the point C?

D

C

A

B Fig - 7

Solution: Let us draw a random path on our 4 × 5 grid in Fig. 6 and think of some way to mathematically specify/describe this path D

C

A

B

A random path across the 4 × 5 grid that our travelling person can follow. The question now is: How do we mathematically characterise this path?

Fig - 8 Mathematics / P & C, Binomial Theorem

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Suppose you had to describe this path to a blind person. If you use E for a step towards the east and N for a step towards the north, you’d tell the blind person that the travelling person took the following path. “EENENEENN” This string that we just formed should immediately make you realise how to calculate the number all the possible paths. We have 5 “E” steps and 4 “N” steps. Any permutation of these 9 steps gives rise to a different unique path. For example, the string “ E E E E E N N N N ” is the path that goes straight east from A to B and then straight north from B to C. Thus, any path can be uniquely characterised by a permutation of these 9 steps. The number of permutations of these 9 9! . This is therefore the number of different paths 5!4! that the travelling person can take from A to C. For an m × n grid we will have (m + n) total steps, m of them being “E” s and the remaining n being “N” s Thus, the number of possible paths

letters, 5 of which are “E”s and 4 are “N”s, is

is

(m + n)! . m !× n !

Example – 12

(a) We have m apples, n oranges and p bananas. In how many ways can we make a non-zero selection of fruit from this assortment? (b) How many factors does 144000 have? In general, how many factors does a natural number N have? Solution: These two seemingly unrelated questions have exactly the same approach to their solutions! Before reading the solution, can you imagine how? (a) The most important point to realise in this question is the nature of objects to be selected. We have m apples. These m apples are exactly identical to each other. You cannot make out one apple from another. This means that if you have to choose r apples out of n, there’s only one way of doing it: you just pick (any of the) r apples. It doesn’t matter which apples you choose, because all the apples are identical. Thus, only 1 r-selection is possible. So how many total selections are possible? We either select 0 apples, 1 apple, 2 apples, .... r apples, ... or m apples. Thus (m + 1) ways exist to select apples. Make sure you properly understand the essence of this discussion and why n choosing r apples out of n can be done in only 1 and not Cr ways.

We therefore have (m + 1) ways to select apples, (n +1) ways to select oranges and (p + 1) ways to select bananas. The selection of a particular fruit is independent of the selection of another fruit. Hence, we have (m +1) (n +1) (p +1) ways to select a group of fruit.

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But wait! In the (m +1) ways of selecting apples, there’s also one way in which we select no apple. Similarly, in the (n + 1) ways of selecting oranges, there’s one in which we select no orange, and in the (p + 1) ways of selecting bananas, there’s one in which we select no banana. Thus, in the product (m +1) (n +1) (p +1), there’ll be one case involving no fruit of any type. We have to exclude this case if we want a non-zero selection of fruit. Therefore, the number of ways of making a non-zero selection is (m +1)(n +1)(p +1) —1. (b) You might be wondering how the factors problem is related to the first part! Read on. Lets consider a smaller number first. Take 60, for example, and list down all its factors (including 1 and 60):

{1, 2,3, 4,5, 6,10,12,15, 20,30, 60} From elementary mathematics, you know that any number can be factorized into a product of primes. For example, 60 can be written in its prime factorization from as:

60 = 2 × 2 × 3 × 5 = 22 ⋅ 31 ⋅ 51

Such a representation exists for every natural number N. Can we somehow use this representation to find the number of factors? Consider any factor of 60, say 12, The prime factorization from of 12 is 22 ⋅ 31 . Similarly, this representation for 15, for example, is 31 ⋅ 51 and for 30 would be 21 ⋅ 31 ⋅ 51 . With discussion in our mind, we rephrase our problem: We have 60 whose prime representation is 22 ⋅ 31 ⋅ 51 . Thus, we have 2 twos, 1 three and 1 five with us. (You could imagine that we have 2 apples, 1 orange and 1 banana). To form a factor of 60, what we have to do is to make a selection of prime factors from amongst the available prime factors. For 60, we have 2 twos (apples), 1 three (orange) and 1 five(banana). In how many ways can we make our selection? The last part tells is that we can do it in (m + 1)(n + 1)( p + 1) ways or (2 + 1)(1 + 1)(1 + 1) = 12 in this particular case. (We also allow no selection of any prime factor - this corresponds to the factor 1 of 60) Do you feel the elegance of this solution? For the general case of a natural number N whose prime representation is of the α α α form p1 1 p2 2........ pn n , the number of factors (including 1 and N) is (α1 + 1)(α 2 + 1)........(α n + 1) For example, 144000 can be represented as

144000 = 27 ⋅ 32 ⋅ 53 The required number of factors is (7 + 1)(2 + 1)(3 + 1) = 96 Mathematics / P & C, Binomial Theorem

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Example – 13

12 person are sitting in a row. In how many ways can we select 4 persons out of this group such that no two are sitting adjacent in the row? Solution: To make the question more clear, here’s a valid and a non-valid selection:

Valid

Non-valid Fig - 9 We have to find some (mathematical) way of specifying a valid-selection. An obvious method that strikes is this: In our row, we represent every unselected person by U and every selected person by S. Thus, the valid selection in the figure above becomes:

U S U U S UU S U S UU To count the numbers of valid selections, we count the number of permutation of this string above, consisting of 8 “U” symbols and 4 “S” symbols, subject to the constraint that no two “S” symbols are adjacent. To count such permutation, we first fix the 8 “U” symbols. There will then be 9 blank spaces generated for the “S” symbols as shown below:

__U__U__U__U__U__U__U__U__ From these 9 blank spaces, any 4 spaces can be chosen and the “S” symbols can be put there and it’ll be guaranteed that no two “S” symbols will be adjacent. Thus, our task now reduces to just selecting 4 blank spaces out of the 9 available to us, which can be done simply in 9 C4 ways. These are the required number of selections.

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Example – 14

DIVISION INTO GROUPS

We will use a standard pack of 52 cards to illustrate the concepts involved. (a) (i) In how many ways can a pack of 52 cards be divided equally into 4 sets? (ii) Equally among 4 players? (b) (i) In how many ways can this pack be divided into 4 sets of 9, 13, 14 and 16 cards each? (ii) Distributed among 4 players with 9, 13, 14 and 16 cards? (c) (i) In how many ways can this pack be divided into 4 sets of 10, 14, 14 and 14 cards each? (ii) Distributed among 4 players with 10, 14, 14 and 14 cards? (iii) Distributed among 4 players with 12, 12, 14 and 14 cards? Solution: For this question, you must understand that the division of a set into a number of subsets is different from the division of a set into a number of subsets and then distribution of those subsets to different persons. For example, suppose you are alone by yourself and you are dividing your deck of cards into 4 sets. What you’ll do is do the division of the deck and keep the 4 sets in front of you. Which set lies where in front of you doesn’t matter. Your only objective was to divide the pack into 4 sets which you did. Precisely speaking, the order of the groups doesn’t matter. On the other hand, suppose you were playing a game of Bluff with 3 other people. Before starting the game, you divide the deck into 4 sets of 13 cards each. But now you have to give each set of 13 cards to a different person. Where a set of cards goes matters. In other words, the order of the groups matters. As another example, suppose you had to divide a class into 3 sub-groups. You do your job-the order of the groups doesn’t matter, you just have to divide them. But suppose you had to divide the class into 3 sub-groups and place each sub-group in a different bogey of a roller coaster. In this case the order of the groups matters because which sub-group sits in which bogey matters. Once again, make sure that you understand very clearly when the order of the groups matters and when it doesn’t (a) (i) We want to divide the pack into 4 equal sets of 13 cards each. We proceed as follows; First select a group of 13 cards from the 52 cards; call this group A. This can be done in 52 C13 ways. Group B of another 13 cards can now be selected from the remaining 39 cards in 39 C13 ways. Group C of another 13 cards from the remaining 26 can now be selected in 26 C13 ways. This automatically leaves group D of the last 13 cards. Thus, the division can be carried out in 52

C13 × 39C13 × 26C13 ways.

However, notice that the order of the groups doesn’t matter, as discussed earlier. This means that even if our order of group selection was BACD or ADBC (or for that matter any other permutation of ABCD), such a selection would essentially be the same as ABCD. The number of permutations of ABCD is 4! = 24.

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Thus, the actual number of ways of division is 1 52 ( C13 × 39C13 × 26C13 ) 24

=

1 52! 39! 26! × × 24 13! 39! 13! 26! 13! 13!

=

52! (13!)4 ⋅ 4!

(ii) If we had to distribute the cards equally among 4 players, the order of group selection would have mattered. For example, ABCD would be different from BACD. Thus, we 52! don’t divide by 4! in this case. The number of possible distributions is ( )4 13!

(b) (i) For this part, observe that the group sizes are all unequal. This is a somewhat different situation than the previous part where the group sizes were equal. We’ll soon see why. We follow a similar sequence of steps as described earlier. (i) Select group A of 9 cards from the deck of 52 : 52 C9 ways (ii) Select group B of 13 cards from the remaining 43 : 43C13 ways (iii) Select group C of 14 cards from the remaining 30 : 30 C14 ways. (iv) This automatically leaves group D of the remaining 16 cards. The number of ways for this division is 52

C9 × 43C13 × 30C14

=

52! 43! 30! × × 9! 42! 13! 30! 14! 16!

52! = 9! 13! 14! 16!

From now on, note and remember that m+ n+ p + q Cm × n+ p + qCn × p + qC p × qCq = (m + n + p + q) m! n! p ! q !

Why don’t we divide by 4! here as we did earlier, even when the order of groups doesn’t matter? This is because the group sizes are unequal. A particular selection in a particular order, say ABCD, will never be repeated in any other order of selection. Our order of selection (in terms of group size) is 9, 13, 14 and 16 cards. Carefully think about it; a particular selection like ABCD will be done only once (and not 4! times as in the previous part due to equal group sizes) Mathematics / P & C, Binomial Theorem

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(ii) If we had to distribute the 4 sets to 4 players, the situation changes because which subset of cards goes to which player matters. A particular group of sets of cards, say ABCD, can be distributed among the 4 players in 4! ways. Thus, now the number of distributions will be no. of ways of division of the no. of ways of Distribution × deck as specified into the 4 sets of those sets to the 4 players 52! ( ) = × 4! 9! 13! 14! 16! (c) (i) In this part, the situation is a hybrid of the previous two parts. 3 of the 4 groups are equal in size while the 4th is different. We first select 10 cards out of 52. This can be done in 52C10 ways. The remaining 42 cards can be divided into 3 equal groups (by the logic of part -(a)) in 42! 1 × (14!)3 3! (Division by 3! is required since the order of groups doesn’t matter. Thus, a particular order of 15-card groups, say ABC, is, for example, the same as ACB) Thus, the required number of ways is 42! 1 × (14!) 3 3!

52

C10 ×

=

52! 1 × 3 10!(14!) 3!

...(1)

(ii) For the second question where we have 4 players and we want to give them 10, 14, 14 and 14 cards, we already have obtained the number of possible division in (1). To distribute the subsets, each division of 4 sets can be given to the 4 players in 4! ways. Thus, the number of distributions of the 4 subsets is =

52! 1 4 × 52! × × 4! = 3 3 10!(14!) 3! 10!(14!)

We can also look at this current problem in the following way. We want to distribute the 52 cards to the 4 players in sets of 10, 14, 14 and 14. We first decide which guy to give the set of 10 cards. This can be done in 4 ways. Now we choose 10 cards for him (in possibly 52 C10 ways). The other three players will now get the remaining 42 cards equally. We can select 12 cards for one of them in 42 C14 ways. The third one can get another 14 cards in 28 C14 ways and finally the fourth one gets the remaining 14 cards. Thus, the number of possible ways of distribution is 4 × 52C10 × 42C14 × 28C14 = Mathematics / P & C, Binomial Theorem

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(iii) Finally, we now have two sets of one size and two of another size. We first just carry out the division into subsets, without assigning any subset to any player. This can be done in C12 ×

52

! Group A of 12 cards

40

×

C12

Group B of 12 cards

=

×

28

C14

! Group C of 14 cards

14

C14

! Group D of 14 cards

1 2!

×

↓ Groups A and B are of the same sizer

×

1 2!

↓ Groups C and D are of the same sizer

52! 1 × 2 2 (12!) (14!) ( 2!)2

Once the division of the deck into the 4 subsets has been accomplished, we assign a subset to each of the 4 players. This can be done in 4! ways. Therefore, the possible number of ways of distribution of the cards to the 4 players is 52! 1 × × 4! 2 2 (12!) (14!) ( 2!)2 Example – 15

CIRCULAR PERMUTATIONS

In this example, we talk about circular permutations, i.e., arrangements in a circular fashion (a) In how many ways can n people be seated around a circular table? (b) We have a group of 5 men and 5 women. In how many ways can we seat this group around a circular table such that: (i) all the 5 women sit together (ii) no two women sit together. (c) In how many ways can a necklace be formed from n different beads? Solution: Circular permutations are somewhat different from linear permutations. Lets see why. Consider the letters A, B, C and D. The 4 linear permutations ABCD, BCDA, CDAB and DABC correspond to a single circular permutation as shown in the figure below: A

B

In a circular permutation, only the relative ordering of the symbols matters. The 4 linear permutations ABCD, BCDA, CDAB and DABC all have the same relative ordering so that they correspond to a single circular permutation.

D

C Fig - 10

This means that the number of circular permutations of ABCD is only Mathematics / P & C, Binomial Theorem

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(a) We have n people, say P1, P2............Pn. As just described, n linear permutations of these people will correspond to a single circular permutation as depicted in the figure below:

PP 1 2...............Pn P2 P3...............Pn P1 P3P4...............PP 1 2 : : : : Pn P1...............Pn −2 Pn −1 n linear permutations correspond to →

PP 1 2...........Pn

a single circular permutation because of the same relative ordering.

Thus, the number of circular permutations is

n! = (n − 1)! n

We can also arrive at this number in another way. We take a particular person, say P1 , and seat him anywhere on the table. Once P1 ' s seat becomes fixed, the rest of the (n – 1) seats bear a fixed relation to P1 ' s . In other words, once P1 ' s seat becomes fixed, we can treat the (n – 1) seats left as a linear row of (n –1) seats. Thus, the remaining (n –1) people can be seated in (n –1)! ways. (b) Let the 5 women be represented by W1, W2 ,W3,W4 , and W5 and the 5 men by M1, M 2 , M 3, M 4 , and M 5 . Since we need all the women to sit together, we first treat all the 5 women as a single entity W. Now, the 5 men and the entity W can be seated around the table in ((5 + 1) − 1)! ways, i.e., 120 ways. (i) Once the 5 men and the entity W have been seated, we now permute the 5 women inside the entity W. This can be done in 5! = 120 ways. The total number of ways is thus 120 × 120 = 14400 W3 W 4

1

W

W2

W

The entity W

5

M2

(i) First seat the 5 men and the entity W

M4

(ii) Then permute the women inside the entity W. M1

M3 M5 Fig - 11

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(ii) Since we want no two women to sit together, we first seat all the 5 men, which can be done in (5 –1)! = 24 ways. Seating the 5 men first creates 5 non-adjacent seats where the women can then be seated in 5! = 120 ways.

M1

M4

(i) First seat the 5 men, 4! ways are possible

M3

M2

(ii) Then seat the 5 women in the 5 spaces created as shown in the figure 5! ways are possible

M5

Fig - 12

The total number of ways is this 5! × 4!. (c) The situation of a necklace is slightly different than that of seating people around a circular table. The reason is as follows: Suppose we have 5 beads A, B, C, D and E. Consider two circular necklaces of these 5 beads A B

A E

C

E

D

B

D

Necklace-1

C

Necklace-2 Fig - 13

Are these two necklaces different? No, because a-necklace can be worn from both ways. Necklace-2 is the same as necklace-1 if I look into it from the other side of the page. In other words, for a necklace, a clockwise permutation and its corresponding anti-clockwise permutation are identical. Thus, the number of circular permutations would reduce by a factor 1 of two, i.e., the number of different necklaces possible is (n − 1)!. 2

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TRY YOURSELF - I

Q. 1

How many words can be made with the letters of the word “TECHNOLOGY” which do not begin with T but end with Y ?

Q. 2

In how many ways can the letters of the word CINEMA be arranged so that the order of vowels do not change?

Q. 3

How many sides are there in a polygon which has 35 diagonals?

Q. 4

In how many ways can 6 boys and 4 girls sit in a row so that no boy is between two girls?

Q. 5

How many words can be made with the letters of the word BHARATI so that all the vowels are consecutive?

Q. 6

Let n ∈ N and 300 < n < 3000. If n is made of distinct digits by taking from 0, 1, 2, 3, 4, 5 then find the greatest possible number of values of n.

Q. 7

How many words can be made with the letters of the word INSTITUTION such that vowels and consonants alternate?

Q. 8

How many different committees of 5 members can be formed from 6 men and 4 ladies if each committee is to contain at least one lady?

Q. 9

Six Xs are to be placed in the squares of the given figure, such that each row contains at least one X. In how many different ways can this be done?

Q. 10 In how many ways can 16 identical mangoes be distributed among 4 persons if none gets less than 3 mangoes?

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Section -1 MORE APPLICATIONS

Section - 5

By now, you should have a pretty good idea about the basics of permutations and combinations. In this section, we will encounter more advanced problems. All the questions discussed in the following pages are directly or indirectly based on the concepts already discussed in the previous sections. In case you find anything confusing, refer to the relevant parts again.

Example – 16

On a standard 8 × 8 chessboard, find the (a) number of rectangles (b) number of squares Solution: (a) Visualise any arbitrary rectangle on the chessboard, say the one depicted on the left in the figure below:

P

Q

An arbitrary rectangle on the chessboard. How to specify this rectangle is explained in the figure on the right

X Y To mathematically characterise the rectangle that we selected, we specify the pair of vertical edges X and Y and the pair of horizontal edges P and Q. Doing so uniquely determines the rectangle

As explained in the figure above, any rectangle that we select can be uniquely determined by specifying the pair of lines X and Y that make up the vertical edges of the rectangle and the pair of lines P and Q that make up the horizontal edges of the rectangle. On the chessboard, there are 9 vertical lines available to us from which we have to select 2. This can be done in 9C2 ways. Similarly, 2 horizontal lines can be selected in 9C2 ways. Thus, the total number of rectangles that can be formed is 9C2 × 9C2 = 1296 . (b) To select a square, observe that the pair of lines X and Y must have the same spacing within X and Y as the pair of lines P and Q. Only then can the horizontal and vertical edges of the selected rectangle be of equal length (and thus, the selected rectangle is actually a square). Mathematics / P & C, Binomial Theorem

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In how many ways can we select a pair (X, Y) of lines which are spaced a unit distance apart ? Its obviously 8. Corresponding to each of these 8 pairs, we can select a pair (P, Q) of lines in 8 ways such that P and Q are a unit distance apart. Thus, the total number of unit squares is 8 × 8 = 64 (This is obvious otherwise also). Now we count the number of 2 × 2 squares. In how many ways can we select a pair (X, Y) of lines which are 2 units apart ? A little thought shows that it will be 7. Corresponding to each of these 7 pairs, we can select a pair (P, Q) of lines in 7 ways such that P and Q are 2 units apart. Thus, the total number of 2 × 2 squares is 7 × 7 = 49. Reasoning this way, we find that the total number of 3 × 3 squares will be 6 × 6 = 36, the total number of 4 × 4 squares will be 5 × 5 = 25 and so on. Thus, the total number of all possible squares is 64 + 49 + 36 + ........... + 4 + 1 = 204

↑ 1×1 squares

↑ 2× 2 squares

↑ 7× 7 squares

↑ 3×3 squares

↑ 8×8 squares

Example – 17

Consider an n-sided convex polygon. (a) In how many ways can a quadrilateral be formed by joining the vertices of the polygon ? (b) How many diagonals can be formed in the polygon? Solution: (a) Observe that a selection of any 4 points out of the n vertices of the quadrilateral will give rise to a unique quadrilateral (since the polygon is convex, the problem of our selection containing all or 3 collinear points does not exist). 4 points out of n can be selected in nC4 ways. Thus, we can have nC4 different quadrilaterals. (b) To form a diagonal, we need 2 non-adjacent vertices ( because 2 adjacent vertices will form a side of the polygon and not a diagonal). The total number of ways of selecting 2 vertices out if n is nC2 . This number also contains the selections where the 2 vertices are adjacent. Those selections are simply n in number because the polygon has n sides. Thus, the total number of diagonals is n

C2 − n

=

n( n − 1) −n 2

=

n(n − 3) 2

Example – 18

Give a combinatorial (logical) justification for this assertion: n

C0 + n +1C1 + n + 2C2 + ... + n + rCr =

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n + r +1

Cr www.locuseducation.org

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Solution: The right hand side tells us that we have to select r persons out of a group of n + r + 1 persons. To do so, we consider any particular group of r persons from these n + r + 1 persons. Specify these r persons by the symbols A1 , A2 ... Ar . Now, to count all the possible r-groups from this group of n + r + 1, we consider the following mutually exclusive cases: (1)

The r -group does not contain A1 Such r-groups can be formed in

(2)

n+ r

Cr ways since we have to select r people out of n + r.

The r -group contains A1 but not A2 We have to select (r – 1) people out of ( n + r − 1) because we already have selected A1 so we need only r –1 more people and since we are not taking A2 , we have ( n + r − 1) people to choose from. This can be done

(3)

(r)

n + r −1

Cr −1 ways.

The r -group contains A1, A2 but not A3 We now have to select (r – 2) people out of (n + r − 2) . This can happen in ways. Proceeding in this way, we arrive at the last two possible cases.

n+ r −2

Cr − 2

' The r -group contains A1, A2 ... Ar – 1 but not Ar . We need to select only 1 person out of (n + 1) available for selection. This can be done in n +1

C1 ways.

(r + 1)The r-group contains A1, A2 ... Ar In this case, our r-group is already complete. We need not select any more person. This can be done in nC0 or equivalently 1 way. Convince yourself that these (r + 1) cases cover all the possible cases that can arise in the formation of the r - groups. Also, all these cases are mutually exclusive. Thus, adding the number of possibilities of each case will give us the total number of r-groups possible, i.e. n

C0 + n +1C1 + n + 2C2 + ... + n + r −1Cr −1 + n + r Cr =

n + r +1

Cr

Example – 19

How many distinct throws are possible with a throw of n dice which are identical to each other, i.e. indistinguishable among themselves ? Solution:

The important point to be realised here is that the dice are totally identical. Suppose we had just 2 dice, say die A and die B. Suppose that, upon throwing these dice, we get a “two” on A and a “three” on B. This case would be the same as the one where we get a “three” on A and a “two” on B because we cannot distinguish between A and B. What we are concerned with is only what numbers show up on the top of the dice. We are not concerned with which die shows what number. This means that if we have n dice and we throw them, we are only concerned with how many “ones”, “twos”, “threes” etc show on the top faces of the dice; we are not at all interested in which die throws up what number.

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If we denote the number of “ones” we get by x1 , number of “twos” we get by x2 and so on, we will have x1 + x2 + ... + x6 = n Thus, the total number of distinct throws will be simply the number of non-negative solutions to this integral equation. As discussed earlier, this number will be

n + 6 −1

C6 −1 =

n+5

C5 .

What would be the number of distinct throws if the n dice were not identical? Example – 20

Give combinatorial arguments to prove that n

∑r⋅

n

r =1

Cr = n ⋅ 2n −1

Solution: Let us first interpret what the left hand side of this assertion says. Suppose we have a group of n people. We select a sub-group of size r from the group of n people. This can be done in nCr ways. Once the sub-group has been formed, we select a leader of that sub-group, and send that sub-group on an excursion. The leader can be selected in r ways. Thus, the total number of different ways in which an r - group can be formed with a unique leader is r × nCr . Now r can take any integer value from 1 to n, i.e. 1 ≤ r ≤ n . Thus, the total number of all possible n

sub-groups, each sub-group being assigned a unique leader, will be

∑r⋅ r =0

hand side of our assertion.

n

Cr which is the left

To prove this equal to the right hand side, we count the sub-groups from a different angle. We count all those sub-groups in which a particular person, say A, is the leader. Since A is the leader, A is fixed in our sub-group. For each of the remaining (n – 1) people, we have two options. We either put the person in the group led by A or we don’t. Thus, the total number of sub-groups in which A is the leader will be 2"# × 2# ×$## 2 × ... ×%2 = 2n −1 ( n −1) times

Since any of the n persons can be the leader, and under each person’s leadership, 2n−1 groups can be formed, the total number of sub-groups, each sub-groups under some unique person’s leadership is n ⋅ 2n −1 . This proves our assertion that n

∑r⋅ r =1

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Example – 21

A composition of a natural number N is a sequence of non-zero integers {a1, a2 .......... ak} which add up to N. How many compositions of N exist? Solution: Let us make the question more clear by taking a particular example for N, say N = 4. As described in the question, the compositions of N = 4 will be

{4}, {1, 3},{2, 2}, {3,1}, {1,1, 2}, {1, 2,1}, {2,1,1} and {1,1,1,1} which are 8 in number. Observe carefully the compositions listed out. How can we characterize each of these compositions mathematically? Recall the problem of finding the number of non-negative solutions of the integral equation x1 + x2 + ....... + xn = r where each solution corresponded to a unique permutation of r “ 1 “symbols and n - 1 " " symbols. Can we do something like that here? In other words, can we represent each composition in another form whose permutations are easier to count? It turns out that we can, as follows: {4}

=

1 1

1

1

{1, 3}

=

1+1

1

1

{3, 1}

=

1

1

1+ 1

{2, 2}

=

1

1 + 1

1

{1, 1, 2}

=

1+1 + 1

1

{1, 2, 1}

=

1+1

{2, 1, 1}

=

1 1 + 1+ 1

{1, 1, 1, 1} =

1+1 + 1+ 1

1+ 1

On the right hand side, we have 4 “1”s and thus 3 blank spaces between the 4 “1” s. We can insert “+” signs in these blank spaces; each different arrangement of “+” signs in these blank spaces will correspond to a different composition. To count the number of these arrangements, we proceed as follows: For each blank space, we have 2 options, we can either insert or not insert a “+” sign into that space. There are 3 blank spaces; so the total number of all arrangements of “+” signs in the 3 blank spaces is 2 × 2 × 2 = 8 (which is the number of compositions we already listed out). In the general case, we will have (N – 1) blank spaces and 2 options for each such blank space. Thus, the total number of ways in which we can arrange “+” signs in these blank spaces, and 2 × 2 × 2 × ........2 = 2 N −1 therefore, the total number of compositions, will be "##$##% ( N −1) times

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Example – 22

Prove logically the following assertion: n

∑2

k

k =0

⋅ n Ck = 3n

Solution: Suppose that we have to form a string of length n, consisting of only letters from the set {A, B, C}. Thus, we have 3 options to fill any particular place in the string: We fill that place with either A, B or C. Thus the total number of different strings of length n would be 3"##$##% × 3 × 3 × ........ × 3 = 3n n times

We now approach the task of formation of these strings from a different perspective. Suppose that our string contains a total of r “A” s. How many such strings will exist? We first select r places out of n which we will fill with “A”. This can be done in nCr ways. For each of the remaining (n – r) places, we have two options; we either fill it with “B” or “C”. Thus, the number of strings containing r “A” s will be nCr .2n–r. Now we vary r from 0 to n and thus get the total number of strings as n

∑ r =0

n

n

Cr ⋅ 2 n − r = ∑ n Cn − k ⋅ 2 k

( where k = n − r )

k =0

n

(since

= ∑ n Ck ⋅ 2 k k =0

n

∑

Thus,

k =0

n

n

Cn − k = n Ck )

Ck ⋅ 2 k = 3n

Example – 23

Prove logically the following assertion: n

Cr + n −1Cr + n − 2Cr + ...... + r Cr = n +1Cr +1

Solution: The right hand side says that we have to select (r + 1) people out of a group of (n +1). To do so, we list down the following (mutually exclusive) cases which exhaust all the possible cases: (1)

The group contains A1

Mathematics / P & C, Binomial Theorem

:

n

Cr ways (since we have to select r people more apart from A1 from n that are available for selection)

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(2)

n −1

:

Cr ways (since we have to select r people more apart from A2 from (n – 1) that are available for selection)

:

n−2

Cr ways (since we have to select r people more apart from A3 from (n – 2) that are available for selection)

(n – r) The group does not contain A1, A2 ....... An–r–1 but contains An–r

:

r +1

Cr ways (since we have to select r people more apart from An–r from the (r + 1) that are available for selection

(n – r + 1) The group does not contain

:

r

(3)

The group does not contain A1 but contains A2

The group does not contain A1 and A2 but contains A3

'

A1 , A2 ... An − r but contains An − r +1

Cr ways (since we have to select r

more people apart from rn − r +1 from the remaining r that are available for selection)

Convince yourself that all these cases are mutually exclusive and exhaust all the possibilities. Thus, the total number of (r + 1) - groups from (n + 1) people is n +1

Cr +1 = nCr + n −1Cr + n− 2Cr + ... + r Cr

Example – 24

Consider n points in a plane such that no three of them are collinear. These n points are joined in all possible ways by straight lines. Of these straight lines so formed, no two are parallel and no three are concurrent. Find the number of the points of intersection of these lines exclusive of the original n points. Solution: To gain more insight into the problem, let us consider the case when n = 4.

A B

1

3 C D 2 Fig - 15 Mathematics / P & C, Binomial Theorem

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We originally have 4 points in the plane labelled as A, B, C and D. When we draw all the straight lines possible (by joining every possible pair of points), we see that 3 new intersection points are generated, labelled as 1, 2 and 3. As we now discuss the general case, refer to this figure for help. Since we have n points, the number of straight lines that can be generated is equal to the number of pairs of points that can be selected from n points, which is equal to nC2 . Every pair of straight lines so generated will intersect at some point. Thus, the total number of intersection points if we count this way should be equal to the number of pairs of straight lines possible from the nC2 lines, which is (

n

C2 )

C2 . However, observe that in this number, the original intersection points have also been counted, and that too multiple times. Lets determine how many times a particular point, say A has been counted. (n – 1) lines pass through A. Thus n −1C2 pairs of straight lines are possible from these (n – 1) lines. Thus, A has been counted n −1C2 times. This means that the n original points have been counted n × n−1C2 times. To calculate the new intersection points, we subtract this number from the (supposed) number of intersection points we obtained earlier. Thus, the actual number of new intersection points is C2 − n × n −1C2

( n C2 )

n( n − 1) n( n − 1) − 1 2 2 − n × (n − 1)(n − 2) = 2 2 n(n − 1)(n 2 − n − 2) n( n − 1)(n − 2) − 8 2 2 n(n − 1){(n − n − 2) − 4(n − 2)} = 8 =

n(n − 1)(n 2 − 5n + 6) 8 n( n − 1)( n − 2)(n − 3) = 8 Verify that this formula works for n = 4 =

Example – 25

How many 4 letters strings can be formed from the letters of the work INEFFECTIVE ? Solution: Observe that some letters repeat more than once: Letter E F I C T V N Frequency 3 2 2 1 1 1 1 This means that our string of 4 letters could contain repeated letters. (It’s thus obvious that we cannot straightway use n Pr to evaluate the number of strings. Mathematics / P & C, Binomial Theorem

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In such a case, we divide the types of strings that we can form into different mutually exclusive cases: Case 1:

All 4 letters are different In this case, we have 7 letters to choose from and we have to arrange them in 4 places. The number of such strings will be 7 P4 = 840

Case 2:

2 letters are same, the other 2 are different We have 3 types of letters (E, F and I) that can be repeated twice. Thus, the twicerepeated letter can be selected in 3 ways. Once that is done, the rest of the 2 letters can be selected from amongst the remaining 6 options in 6C2 = 15 ways. Once we’ 4! ve formed the combination of the 4 letters we can permute them in = 12 ways. The 2! “2!” occurs in the denominator because of the twice-repeated letter. Thus, the total number of such strings will be 3 ×15 ×12 = 540 .

Case 3:

2 letters are the same, the other 2 are also the same For example, the string “EFEF” will be such a string. There are 2 letters now that we want to be twice-repeated; observe that there are only 3 types of letters (E, F and I) that can be twice repeated. Thus, the letters that will occur in the string can be selected in 3C2 = 3 ways. Once that is done, we can 4! = 6 ways. permute the 4 letters in 2!2! The total number of such strings is therefore 3 × 6 = 18.

Case 4:

3 letters are the same, the 4th is different Only “E” can be repeated thrice so that E must occur in this string. The 4th letter can 4! be selected in 6 ways. Then, the combination so formed can be permuted in = 4 3! ways. The number of such strings is 6 × 4 = 24. Verify that these four cases are mutually exclusive and they exhaust all the possibilities. The total number of strings is 840 + 540 + 18 + 24 = 1422

Example – 26

(a) (b)

In how many ways can 3 girls and 9 boys be seated in 2 vans, each having numbered seats with 3 seats in the front and 4 seats at the back? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats?

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Solution: (a) We have 12 people to seat and 14 available seats. We first select the 12 seats on which to seat these people. This can be done in 14C12 ways. Once that is done, we can permute the 12 people amongst these 12 seats in 12! ways. Thus, the total number of seating arrangements is 14C12 × 12! . (b) Since the 3 girls need to be seated together in one back row, we first select the back row. This can be done in 2 ways since there are 2 vans. In this row, 3 adjacent seats can be selected in 2 ways. "#$#%

"#$#% selection of 3 adjacent seats can be done in 2 ways.

In the 3 adjacent seats, the 3 girls can be permuted in 3! ways. Thus, the girls can be seated in 2×2×3! = 24 ways. We now have 11 remaining seats. The 9 boys can be seated in these 11 seats in 11P9 ways. Thus, the total number of ways to seat the entire group is 24 × 11P9 = 24 ×

11! = 12 × 11! = 12! . 2!

Example – 27

We have 3n objects, of which n are identical and the rest are all different. In how many ways can we select n objects from this group ? Solution: To select n objects, we select k objects from the identical ones and the remaining (n – k) from the different ones. ( k will vary from 0 to n) As discussed somewhere earlier, from the group containing identical objects, there will always be only 1 way of selection. From the group of 2n nonidentical objects, (n – k) objects can be selected in 2nCn −k ways. Thus, the group of n objects containing k identical objects and the remaining (n – k) as nonidentical objects can be formed in 1× 2 nCn −k = 2 nCn − k ways. The total number of ways is S where n

S = ∑ 2 nCn − k = 2 nCn + 2 nCn −1 + ... 2 nC0 k =0

A closed-form expression for S can be obtained by the following manipulation : 2 S = 2( 2 nCn + 2 nCn −1 + ... + 2 nC0 ) 2 nCn + 2 nCn −1 + ... + 2 nC0 = + 2 nCn + 2 nCn −1 + ... + 2 nC0 Mathematics / P & C, Binomial Theorem

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2 nCn + 2 nCn −1 + ... + 2 nC0 = + 2 nCn + 2 nCn +1 + ... + 2 nC2 n

For all the terms in the lower row, we can use 2n 2n Cr = C2 n− r

= 2 nCn + ( 2 nC0 + 2 nC1 + 2 nC2 + ... + 2 nC2 n ) = 2 n Cn + 2 2 n

⇒

S=

(This step uses the result of example - 7)

1 2n ⋅ Cn + 22 n −1 2

Example – 28

5 balls are to be placed in 3 boxes. Each box can hold all the 5 balls. In how many ways can we place the balls into the boxes so that no box remains empty, if (a) balls and boxes are all different (b) balls are identical but boxes are different (c) balls are different but boxes are identical (d) balls as well as boxes are identical (e) balls as well as boxes are identical, but the boxes are kept in a row. Solution: One of the constraints that should always be satisfied is that no box should remain empty. Thus, each box should get at least one ball. This means that the distribution of balls can have the configuration ( 1, 1, 3) or (1, 2, 2). Only in these two configurations does no box remain empty. Readers might observe some similarity between this problem and Example - 14 where we discussed division of a deck of cards into groups. Observe carefully which part in this problem corresponds to which case in Example - 14

(a) In this case, the balls and the boxes are all different. If you observe carefully, you will note that this case is equivalent to dealing cards to players. Here, we are dealing balls (all different) into boxes (all different). In the card game, we were dealing cards (all different) to players (all different). Suppose we distribute the balls in the configuration (3, 1, 1). We first divide the group of balls into this configuration. This can be done in 5 C3 ways since we just need to select a group of 3 balls and our division will be accomplished. Once the division of the group of balls into 3subgroups in the configuration (3, 1, 1) has been done, we can permute the 3 sub-groups among the 3 different boxes in 3! ways. Thus, the number of ways to achieve the (3, 1, 1) configuration is 5 C3 × 3! = 60 We now find the number of ways to achieve the (1, 2, 2) configuration . Mathematics / P & C, Binomial Theorem

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We first select 2 balls out of the 5 which can be done in 5 C2 ways. We then select 2 balls from the remaining 3, which can be done in 3 C2 ways. Thus simple division into the configuration C 2 × 3C 2 ways. Division by 2! is required since two subgroups 2! are of the same size and right now we are just dividing into sub groups so the order to the subgroups doesn’t matter. Once division has been accomplished, we permute the 3 subgroups so formed amongst the 3 different boxes is 3! ways. 5

(2, 2, 1) can be achieved in

5

Thus, the number of ways to achieve the (2, 2, 1) configuration is

C2 × 3C2 × 3! = 90 2!

The total number of ways is 60 + 90 = 150 To make things more clear, let us list down in detail the various configurations possible for the 3 different boxes, A, B & C: Box - A Box - B Box - C Number of ways 1 1 3 20 1 3 1 20 3 1 1 20 1 2 2 30 2 1 2 30 2 2 1 30 Total:150 (b) The balls are now identical so it doesn’t matter which ball goes into which box. What matters is only the configuration of the distribution. By simple enumeration, only 6 configurations exist for this case. (The notation [a, b, c] implies that Box - A has a balls, Box - B has b balls and Box - C has c balls.)

[1,1,3] [1, 2, 2]

[1,3,1] [2,1, 2]

[3,1,1] [2, 2,1]

Thus, 6 possible ways exist for this case (c) The boxes are identical. This means that it does not matter which sub-group of balls you put in which box. What matters is only the division of the group of balls. This case is akin to the one where you have to divide a deck of cards into sub-groups. (you aren’t required to distribute those sub-groups to players). For the configuration (1, 1, 3), the number of ways of division is 5C3 = 10 (we just choose 3 balls out the 5 and the division is automatically accomplished. For the configuration (1, 2, 2), 5 C 2 × 3C 2 = 15 (division by 2! is required since two the number of ways of division is 2! sub-groups are of the same size, and here the order of the group doesn’t matter) Thus, the total number of ways is 10 + 15 = 25 Mathematics / P & C, Binomial Theorem

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(d) This case is quite straightforward. The balls are identical. The boxes are identical too. The only 2 possible configuration are (1, 1, 3) and (1, 2, 2). There can be no permutation of these configurations since the boxes are indistinguishable. Thus, only 2 ways of division exist. (e) If we keep the boxes in a row, we have inherently ordered them and made them non-identical, since the boxes can be numbered now. Therefore, in this case, the balls are identical but the boxes are different so this question becomes the same as the one in part - (b)

Example – 29

PIRATES OF THE CARRIBEAN

In the Caribbean Sea, 13 pirates, while plundering an English ship, come upon a chest full of gold. Since the pirates have found the chest simultaneously, no one can claim the chest as his own. To protect the chest from the avarices of the pirates, the pirate leader Captain Jack Sparrow suggests a scheme. “We must put a certain number of locks on this chest and distribute their keys amongst ourselves in such a way that it can be opened only when 7 or more then 7 pirates decide to open it. In other words, only when a majority of the 13 pirates agree should the chest be able to be opened.” How can this scheme be implemented? What would be the minimum number of locks required and how must their keys be distributed? Solution: This interesting problem can be easily solved by the application of the elementary concepts developed in this chapter. Let us rephrase our problem slightly. We want that only a group of 7 or more pirates should be able to open the chest. This means that whenever such a majority group decides to open the chest, they should have amongst themselves the keys to all the locks on the chest. Suppose on the other hand, that only 6 pirates decide to open the chest. Then there should be at least one lock on the chest whose key(s) are not with anyone amongst that group of 6. Thus, that single lock will prevent the minority group of 6 pirates from opening the chest. This is the approach we now follow. For every possible group of 6 pirates, we put a lock on the chest and distribute 7 keys of the lock amongst the remaining 7 pirates. That lock will prevent our group of 6 pirates from opening the chest. Such a lock will exist for every group of 6 pirates. Thus, whenever any group of 6 pirates decides to open the chest, they will be prevented by one lock whose keys are with the other 7 pirates (the case when even less than 6 pirates decide to open the chest is automatically solved because then there will be more than one lock to prevent that group from opening the chest) Also, whenever any group of 7 pirates decides to open the chest, there is no lock whose key is not amongst one of the 7 members of that group. Thus, any group of 7 pirates (or more) will be able to open the chest. 13 13 Thus, what we need to do is put C6 ( = C7 ) locks on the chest and for each lock, we select a

group of 7 pirates, make 7 keys of that lock and give one key to each member of this group. Mathematics / P & C, Binomial Theorem

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Example – 30

Prove that (n!)! is divisible by (n!)(n–1)! Solution: We can equivalently show that

( n !)! n −1 ! ( n !)( )

is an integer.

To keep things easier, let us take n = 6. Thus, we need to show that

(6!)! 5! (6!)

is an integer. (You must

appreciate the magnitude of the numbers we are dealing with here!) Consider the following sequence of symbols:

a 1a 1a 1a 1a 1a 1a 2 a 2 a 2 a 2 a 2 a 2 ....................a120 a120 a120 a120 a120 a120 This is a sequence of symbols whose length is 720, because each symbol ai occurs 6 times in this sequence and there are in total 120 symbols, and thus the total number of symbols is 6 × 120 = 720. Let us find the number of permutations of this string. Since there are 120 ‘types’ of symbols, each symbol being repeated 6 times, we will have to divide by 6! a total of 120 times, i.e, the number of permutations of this string will be (720 )! 6! × 6! × 6! × .......6! "## #$### % 120 times

=

(6!)! 5! (6!)

Since the number of permutations of any string must obviously be an integer, the term = must be an integer! We can now easily generalise this result.

(6!)! 5! (6!)

Example – 31

n persons are seated around a circular table. In how many ways can we select 3 persons such that no 2 of them are adjacent to each other on the table? Solution: We will solve this problem by first considering the total number of selections possible and then subtracting from this, the number of cases where 2 persons or all the 3 persons are adjacent. Observe the following figure for reference. Pn

P1

P2 P3

Pn-1

P4

Fig - 16 Mathematics / P & C, Binomial Theorem

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The total number of ways of selecting 3 persons out of these n is nC3 . Let us count the cases where only 2 of the 3 persons are adjacent. 2 adjacent people can be selected as follows:

{P1P2 } or {P2 P3} or {P3 P4 }...............or {Pn P1} Thus, there are n ways in which 2 adjacent persons can be selected. Once these 2 have been selected (for example, say we select {P2 P3}), we can select the third non-adjacent person in (n – 4) ways (since for example, if we selected {P2 P3}, we cannot select P1 or P4). Thus the cases where only 2 of the 3 people are adjacent are n (n – 4) in number. We now count the cases where all 3 persons are adjacent. 3 persons can be adjacent in the following manner:

{P1P2 P3} or {P2 P3 P4 } or ..................or {Pn PP 1 2} Thus, observe that there are n ways of selecting 3 persons who are adjacent. Finally, the number of ways to select 3 persons such that no 2 are adjacent is n

C3 − n ( n − 4 ) − n

=

n ( n − 1)( n − 2 ) − n (n − 4) − n 6

=

1 n ( n − 4 )( n − 5 ) 6

Example – 32

We have 21 identical balls available with us which we need to be distributed amongst 3 boys A, B and C such that A always gets an even number of balls. How many ways of doing so are possible? Solution: The only constraint is that A should get an even number of balls. There’s no constraint on the minimum number of balls a boy should get. This means that a boy can also not be given any ball. We can represent the number of balls given to A by 2x since A must get an even number of balls. If we represent the number of balls given to B and C by y and z, we should have 2 x + y + z = 21

... (1)

This means that to find the number of distributions possible, we find the number of non-negative integral solutions to the equation (1). Note that x can take a maximum value of 10 and a minimum value of 0. We rearrange (1) so that we get an integral equation with y and z as variables, treating x as a constant y + z = 21 − 2 x Mathematics / P & C, Binomial Theorem

.... (2) www.locuseducation.org

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The number of non-negative integer solutions of (2) is 21− 2 x + 2 −1

C1 =

22 − 2 x

C1 = 22 − 2 x

We now add the number of solutions so obtained for all the possible values of x. The total number 10

of solutions is therefore ∑ ( 22 − 2 x ) = 132. x=0

Note: An alert reader must have noticed that we can form arbitrarily complex integer equations. For example, what do we do if we intend to find out the number of non-negative integer solutions to the equation α1 x1 + α 2 x2 + ...........α n xn = r where the αi ' s are integers that are not necessarily equal to unity. In addition, what if we impose constraints on the xi' s themselves say, we define an upper and lower bound for xi , i.e li ≤ xi ≤ ui For example, how do we distribute 20 apples among 4 boys such that each boy gets more than 2 apples but less that 8 apples, in addition to the constraint that one of the boys, say A, must always get an even number of apples? Think about it. We will revisit the problem of such general integral equations in Example - 39 and in the chapter on Binomial Theorem. There we’ll learn to solve such problems using the Multinomial theorem. Example – 33

Find the sum of the divisors of 120. Generalise the result for an arbitrary natural number N. Solution: The divisors of 120 are listed out below: {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120} The sum of these divisors is 360. We have to determine an elegant way to deduce this sum because we cannot repeat everytime the procedure of listing down all the factors and summing them. For this purpose, we again resort to the use of the prime factorization form. 120 = 23 ⋅ 3' ⋅ 51

The sum of the divisors will be S=

∑

2i ⋅ 3 j ⋅ 5k

0≤ i ≤3 0 ≤ j ≤1 0 ≤ k ≤1

This notation is simply a shorthand which implies that we vary the integral indices i, j and k (in their respective allowed ranges) and this way we will have listed down all the factors and hence evaluated the required sum. Mathematics / P & C, Binomial Theorem

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To generate the expression for the sum S, we can alternatively use the following method: S = (1 + 21 + 22 + 23 )(1 + 31 )(1 + 51 )

Did you realize the trick? Writing S this way also gives rise to all the factors. You are urged to convince yourself about this by expanding this expression and observing that all possible factors will be generated. Thus, S can now be simply evaluated as follows: S = (1 + 21 + 22 + 23 ) (1 + 31 ) (1 + 51 )

= 15 × 4 × 6 = 360 This is the same result that we got earlier! To do the general case, assume that the prime factorization form of N is N = p1α1 p2α 2 ........ pnα n

where the α i' are all positive integers and the pi' s are all primes. The sum Sf of all the factors will be

(

S f = (1 + p1 + p12 + ..... p1α1 ) (1 + p2 + p22 + ..... + p2α 2 )......... 1 + pn + pn2 + ..... pnα n

)

Example – 34

Find the sum of all the five-digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 if no digit is repeated. Solution: This problem can be solved very easily if we view it from an individual digit’s perspective. Suppose that we only consider the digit “4”. How many numbers will there be with “4” in the units place?

' '

4 4 4

There are 24 numbers with 4 in the units place because the remaining four digits can be permuted among the remaining 4 places in 4!=24 ways

From these 24 numbers, what is the total contribution of the digit “4” to the sum we are required to calculate? Since “4” is at the units place and it occurs 24 times, its contribution will be 4 × 24 = 96 Mathematics / P & C, Binomial Theorem

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Similarly, there will be 24 numbers where “4” is at the tens place. The total contribution of “4” from these 24 numbers will be 4 × 240 = 960 Proceeding in this way, we see that the total contribution of the digit “4” from all the 120 numbers that can be possibly formed is:

4 ( 24 + 240 + 2400 + 24000 + 240000 ) = 4 × 24 × (1 + 10 + 100 + 1000 + 10000 ) = 4 × 24 × 11111

This is the contribution to the sum from only the digit “4”. To calculate the entire sum S, we calculate the contributions from all the five digits. Thus, the sum is

S = (1 + 2 + 3 + 4 + 5 ) × 24 ×11111 = 3999960

Example – 35

DEARRANGEMENTS

Find the number of ways in which 5 different letters can be taken out of their 5 addressed envelopes and put back into the envelopes in such a way that all letters are in the wrong envelope. Solution:

The problem of rearranging objects so that each object is assigned to a place different from its original place is referred to as the problem of dearrangments. Here, we need to find out the number of dearrangements possible with 5 letters and 5 envelopes. Let us denote by Dn the number of dearrangements possible with n things. We will attempts a general solution, that is for an arbitrary n, and then substitute n = 5 Denote by Li the ith letter and by Ei, the original envelope of the ith letter. Consider the envelope E1. It can be assigned a wrong letter in (n – 1) ways. Suppose that we assign the letter L2 to E1. E1 E2 ............ En ( L1 L2 ............ Ln The dearrangements that can now arise can be divided into two mutually exclusive classes: (i) Those in which L1 is assigned to E2 (ii) Those in which L1 is not assigned to E2 If L1 is assigned to E2, we have the following configuration: E1 E2 ............ En ( ! L1 L2 ............ Ln In this case, we have a remaining of (n – 2) letters which can be dearranged in Dn–2 ways.

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Suppose the other case now, where we do not assign L1 to E2. In this case, we have (n – 1) letters to dearrange (L1 will also be counted as a letter to be dearranged since it is being assigned to an envelope other than E2). This can be done in Dn–1 ways. Thus, if we give L2 to E1, the total dearrangements possible are Dn −1 + Dn − 2 . Since E1 can assigned a wrong letter in (n –1) ways, the overall total number of dearrangements Dn is

Dn = ( n − 1)( Dn −1 + Dn− 2 ) We have thus related the nth order 'dearrangements–coefficient' Dn to lower order coefficients. We now have to somehow use this relation to obtain Dn only in terms of n. This is how we do it:

Dn = ( n − 1)( Dn −1 + Dn− 2 )

(

Dn − nDn−1 = ( −1) Dn−1 − (

⇒

n −1)

Dn− 2

)

... (1)

But if we substitute n → ( n − 1) we get

Dn−1 − ( n − 1) Dn− 2 = ( −1) ( Dn− 2 − ( n − 2 ) Dn−3 )

... (2)

We substitute (2) back into (1) to get Dn − nDn −1 = ( −1) ( Dn − 2 − ( n − 2 ) Dn −3 ) 2

= ( −1) ( Dn −3 − ( n − 3) Dn − 4 ) {By the same process} 3

' = ( −1)

n−2

( D2 − 2 D1 )

It is obvious that D1 = 0 since one letter cannot be dearranged while D2 = 1 because two letters can be dearranged in only one way : by exchanging them. Thus, Dn − nDn −1 = ( −1)

n−2

... (3)

We still have not obtained a relation involving only Dn. We do it using (3) repeatedly Dn − nDn −1 = ( −1)

= ( −1)

n−2

n

( −1) Dn Dn −1 − = n ! ( n − 1)! n!

n

⇒

Mathematics / P & C, Binomial Theorem

(Division by n!)

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Now we substitute n → ( n − 1) , n → ( n − 2 ) .... successively and add the corresponding sides of all the relations so obtained to get

( −1) + ...... + ( −1) Dn D1 ( −1) ( −1) − = + + n ! 1! n! 2! ( n − 1)! ( n − 2 )! n

n −1

n−2

2

Since D1 = 0, we finally get n 1 1 1 −1) ( Dn = n ! − + − ........ n! 2! 3! 4!

n 1 1 1 −1) ( = n !1 − + − + ... 1! 2! 3! n!

The first two terms "1"and " 1 " 1! are just added to make the series look more sequenced

This is the number of de arrangements possible with n things For n = 5, we have 1 1 1 1 1 D5 = 5!1 − + − + − 1! 2! 3! 4! 5!

= 44 Thus, there are 44 ways to rearrange the letters back into their envelopes so that each letter goes to a wrong envelope. Since n = 5 is a small number, we could have worked out an alternative solution as follows: Total No. of arrangements No. of ways in which all letters go to correct envelopes - No. of ways in No. of dearrangemets = which one letter goes to incorrect envelope No. of ways in which two letters go to incorrect envelopes -.............................and so on

You are urged to work out the solution by this way yourself. Example – 36

Prove that for n ≥ 4, the sum 1! + 2! + 3! + ... + n ! cannot be the square of a positive integer. Solution: The approach lies in realizing the general properties of a perfect square. Note that a square will have, at its unit place, a digit from amongst the set {1, 4, 5, 6, 9}. It cannot have a ‘3’ or an ‘8’ as its unit digit. Work it out yourself. Mathematics / P & C, Binomial Theorem

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Now, for the given sum, the first four terms add to 33:

1!+ 2!+ 3!+ 4! = 1 + 2 + 6 + 24 = 33 For any integer greater than 4, its factorial will always have a ‘0’ at its unit place. Can you see why? This means that the given sum will have a ‘3’ at its unit place, so it cannot be a perfect square! Example – 37

A set P contains x elements while Q is another set which contains y elements. How many functions f : P → Q exist which are (a) one - one (b) onto? Solution: Obviously, the total number of functions possible is y x (a) If the function is to be one-one, y, the number of elements in the co-domain of f, must be greater than or equal to x : y≥x In that case, the number of one-one functions will simply be y Px , since you are creating a mapping from x elements in P to x elements in Q, out of a total y elements in Q. (b) For the function to be onto, the only constraint that needs to be satisfied is that each element in the co-domain Q must have a pre-image, which means that x must be greater than or equal to y. The remaining part is left to the reader as an exercise. Example – 38

Four non-identical dice are rolled simultaneously. How many rolls are possible in which 6 shows up on at least one dice? Solution: Its easiest to proceed by considering the complementary case: Total number of rolls possible : 64

(Why ?)

Rolls in which no 6 shows up : 54 (Why ?) Our answer is therefore 64 – 54. Now, a very important issue which we leave to you to resolve: what if the four dice were identical? Example – 39

In an examination, the maximum marks for each of three papers is n and that for the fourth paper is 2n. Find the number of ways in which a candidate can get 3n marks. Mathematics / P & C, Binomial Theorem

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Solution: We need to find the number of solutions to the non-negative integral solution x1 + x2 + x3 + x4 = 3n,

... (1)

where x1 , x2 , x3 all lie in [0, n], while x4 can vary upto 2n, i.e., x4 ∈ [0, 2n], where all the xi' s are integers. Let us construct the polynomial P ( x ) given by P( x) = (1 + x + x 2 + ... + x n )3 (1 + x + x + ... + x 2 n ) Now, think hard ! The first bracket in P ( x ) is actually three identical brackets (1 + x + x 2 + ... + x n ). When P ( x ) is expanded , each of these three brackets can contribute at the most an nth power in x, whereas the fourth bracket can contribute upto 2n in powers of x. When P ( x ) is expanded and the terms written out, note that x 3n will also be formed. But what will be the coefficient of x3n ? It will be precisely the number of solutions to the equation (1) under consideration ! Why ? Because, each time x 3n is formed is a unique combination of power contribution from each of the four brackets. For example, P( x) = (1 + x + ... xn )(1 + x + ... xn )(1 + x + ... + xn )(1 + x + ... + x2n ) ↓ Take 1

↓ Take 1

↓ Take x n

↓ Take x2 n

"##########$##########% Product formed = x3 n

In this example, we take the term 1( or x0) from the first two brackets, x n from the third and x 2n from the fourth. This is a unique combination in which x 3n can be generated, but this is only one such. There will be many more ways in which x 3n can be generated. How many ? Precisely, the coefficient of x 3n when P ( x ) is expanded ! To find out the coefficient, you’ll have to wait until you are familiar with the Binomial / Multinomial expansion.

Example – 40

In how many ways can one fill an m × n table with ±1 such that the product of the entries in each row and each column equals –1? Solution: Denote by aij the entries in the table. For 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1, we let aij = ±1 in an arbitrary way. This can be done in 2( m −1)( n −1) ways. The values for amj with 1 ≤ j ≤ n − 1, and for ain with 1 ≤ i ≤ m − 1, are uniquely determined by the condition that the product of the entries in each row and each column equals –1. The value of amn is also uniquely determined, but it is necessary that n −1

m−1

j =1

i =1

∏ amj = ∏ ain ⋅ Mathematics / P & C, Binomial Theorem

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If we denote by m −1 n −1

P = ∏∏ aij i =1

j =1

we observe that n −1

P∏ amj = (−1) n −1 j =1

and m −1

P∏ ain = ( −1) m −1 i =1

Hence (*) holds if and only if m and n have the same parity.

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CONCLUSION : A MAGIC TRICK

For those of you interested in cards, here is a very interesting card trick that has its basis in an elementary application of this chapter’s concepts. A magician sends out his assistant into the audience with a well-shuffled standard deck of cards. The assistant asks 5 different people chosen at random from this audience to pick a card each from the deck. The magician is on-stage and cannot see the cards drawn from the deck. The magician then asks the assistant to reveal 4 of the 5 drawn cards while keeping the fifth card secret. The magician then thinks for a while and tells the audience the number and suit on the fifth secret card ! We have to figure exactly how the assistant conveys the information about the fifth card to the magician. You can assume that no one is cheating!

The trick explained Somehow, the assistant has been able to convey information about the fifth card to the magician, even though he reveals only 4 of the 5 cards. The information must be embedded in the order in which the assistant reveals the 4 cards. Thus, the magician and assistant must have pre-decided on a scheme whereby the magician can deduce the secret card from the order of the 4 revealed cards. Here is how this can be done. Since 5 cards are drawn, at least 2 cards must be of the same suit. If there are more than 2 cards of the same suit, the assistant simply chooses any 2 of them. Now, one of these 2 cards will be revealed first by the assistant while the other card will become the secret card. So, the magician knows from the suit of the first revealed card what the suit of the secret card is. The first issue is that, of the 2 cards of the same suit, how does the assistant decide which one to reveal and which one to keep secret? For that, the assistant refers to the following diagram (mentally of course!): A

K

2 3

Q 4 J 5 10 6 9 8

7

Fig - 17 Mathematics / P & C, Binomial Theorem

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In this circular arrangement, observe that the difference between any two members is at the most 6. For example, the difference between “7” and “Q” is 6 while that between “J” and “2” is 4

A

K

4

2

ste

ps

A

K

2

3

3

Q

Q 4

4 J

5

5 10

10

6

6

6

steps

J

9

9 7

8

7

8 Fig - 18

Thus, of the 2 cards of the same suit, the assistant reveals that card from which it is possible to reach the other card in at most 6 steps. For example, if the assistant finds a “J” of Hearts and a “2” of Hearts in the 5 drawn cards, he will reveal the “J” of Hearts first and keep the “2” of Hearts secret since in our circular arrangement it is possible to reach from “J” to “2” in 4 steps. Thus, till now the scheme is that the secret card is of the same suit as the first revealed card and lies within at the most 6 steps from the first card, that is, the offset of the secret card from the first card is at the most 6. We now require a scheme whereby the assistant can communicate the offset to the magician. For this purpose, they pre-decide upon an order of all the 52 cards of the deck. For example, suppose they order the deck like follows:

1

2

3

4 ... 14

↓ ↓ ↓ ↓ A of 2 of 3 of 4 of clubs clubs clubs clubs

15

16 ... 27 ...

↓ ↓ ↓ A of 2 of 3 of spades spades spades

↓ A of hearts

40

...

52

↓ ↓ A of K of diamonds diamonds

→ Increasing order

Fig 19 Now, after the assistant has revealed the first card, he has 3 more cards to reveal. Those 3 cards can be ordered according to the order of Fig 19. Thus, one of those 3 cards will be the ‘smallest’, one will be the ‘middle’ card and one will be the ‘largest’. Label the 3 cards as S, M and L.

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To communicate the offset to the magician, the assistant reveals the 3 cards in a particular order as shown in the table below Offset Order in which the assistant reveals the 3 cards 1 SML 2 SLM 3 MSL 4 MLS 5 6

LSM LMS

Since the maximum offset can be 6, the 6 permutations of S, M and L cover all the possible offsets! Once the offset is communicated, the magician simply adds it to the first card to deduce the fifth, secret card! Let us try this out with a particular example. Suppose that the following 5 cards are drawn: “A of spades” “2 of Hearts” “3 of clubs” “7 of Hearts” “8 of clubs”. The assistant picks 2 cards of the same suit. Here we have 2 options for the 2 cards. We can choose either. Suppose the assistant chooses “2 of Hearts” and “7 of Hearts”. The offset of “7” from “2” is 5 which is less than 6. Therefore, “2 of Hearts” is the first card that the assistant reveals. To communicate the offset, the assistant orders the three other cards (to be revealed) according to Fig. 19 as follows: "3 of clubs" < "8 of clubs" < "A of spades" S M L

Since the offset required is 5, the assistant reveals these three cards (according to the Table on the previous page) in the order L S M. Thus, this is what the assistant says: “2 of Hearts” “A of spades” “3 of clubs” “8 of clubs” The magician thinks for a while and says, “The fifth card is the 7 of Hearts !” And the audience gasps!

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P & C [ ASSIGNMENT ]

[ OBJECTIVE ] LEVEL - I Q. 1

The number of ways in which 4 balls can be selected from a bag containing 4 identical and 4 different balls, is (a) 120 (b) 80 (c) 14 (d) 16

Q. 2

The maximum number of intersection points of n straight lines, none of which are parallel to each other, is n

C2 (d) n C2 − n 2 There are 4 bulbs in a room. The number of ways of illuminating the room, is (a) 16 (b) 15 (c) 24 (d) 4 n

n

(b) P2

(a) C2 Q. 3 Q. 4

There are n different books and each book has p copies. The number of selection of books from these is (a) ( p + 1) − 1 n

Q. 5

(c)

(b) ( p + 1) − 2 n

(c) n ( p + 1) − 1

(d) none of these

(c) 19 C16

(d) none of these

The value of 18 C2 +17 C2 +16 C2 +15 C2 +14 C3 is (a) 17 C3

(b) 18 C3

Q. 6

The number of ways in which 6 men can be made to sit round a table in 6 numbered seats is (a) 5! (b) 6! (c) 6 × 6! (d) none of these

Q. 7

The number of diagonals that can be formed by joining the vertices of an n-sided polygon, is (a) n C2 − n

Q. 8

(b) n P3

(c) n ( n − 1)

(d) n C2 − ( n + 1)

The maximum number of intersection points of n circles, is n

C2 (d) n C2 − n 2 In an examination there are 5 multiple choice questions. Each question has four choices of which only one is correct. The number of ways in which an examinee can get at least one answer wrong, is n

Q. 9

n

(a) C2

(b) P2

(c)

(a) 45 − 1

(b) 54 − 1

(c) 5!4!− 1

(d) 5 C4 − 1

(c) 47

(d) 65

Q. 10 The exponent of 4 in 200!, is (a) 86 (b) 101

Q. 11 The number of different messages by signals with three dots and two dashes is (a) 120 (b) 12 (c) 10 (d) 20 Q. 12 If n Cr −1 = 36, n Cr = 84 , n Cr +1 = 126 then the value of r is (a) 1

(b) 2

Mathematics / P & C, Binomial Theorem

(c) 3

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Q. 13 Five persons including one lady are to deliver lectures to an audience. The organiser can arrange the presentation of their lectures, so that the lady is always in the middle, in (a) 5 P5 ways

(b) 4.4 P4 ways

(d) 5 C4 ways

(c) 4! ways

Q. 14 The number of ways in which a set A where n(A) = 12 can be partitioned in three subsets P, Q, R each of 4 elements if P ∪ Q ∪ R = A , P ∩ Q = φ , Q ∩ R = φ , R ∩ P = φ , is 12!

12!

(a) 4! ( )

(b) 3! 4 ( )

3

(c)

12! 1 . 3 ( 4!) 3!

(d) none of these

Q. 15 In a test of 10 multiple-choice questions of one correct answer, each having 4 alternative answers, the number of ways to put ticks at random for the answers to all the questions is (a) 410

(c) 410 − 4

(b) 10 4

(d) 10 4 − 10

Q. 16 The number of positive integral solutions of the equation x + y + z = 50, is (a) 51 C2

(b)

50

(c)

C3

Q. 17 The number of zeros at the end of 50!, is (a) 17 (b) 5

49

(d)

C2

(c) 12

48

C3

(d) 47

Q. 18 The number of ways in which 9 different objects can be distributed to three different people giving 2, 2 and 5 objects to each of them, is (a) 328 (b) 984 (c) 756 (d) 2268 Q. 19 If m parallel straight lines intersect n parallel straight lines, then the number of parallelograms thus formed, is (a) m n

(b)

m+n

C2

(d) m C2 × n C2

(c) mn

Q. 20 There are 2 points on a line, 3 points on another line and 4 points on yet another line. The total number of triangles that can be formed by joining these points, is (a) 30 (b) 205 (c) 79 (d) 85 [ LEVEL - II ] Q. 1 Q. 2

The number of divisors of 1008 of the form 4n + 2( n ≥ 0, n ∈ N) , is (a) 5 (b) 6 (c) 11

Numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 all at a time. The number of such numbers in which the odd digits occupy even places, is (a) 162

Q. 3

(d) 7

(b) 175

(c) 180

(d) 160

The number of ways of distributing 10 different toys among 4 children C1, C2, C3, C4 such that C1 and C2 get 2 toys each and C3 and C4 get 3 toys each, is 10! (a) 2! 2 3! 2 ( )( )

10! 4! (b) 2! 2 3! 2 × 2! 2 (c) ( )( ) ( )

Mathematics / P & C, Binomial Theorem

(

10

C2 )

2

(

10

C3 )

2

(d) none of these

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Q. 4

55

There are n different colours and p balls of each colours. The number of ways in which one or more than one ball can be chosen, is (a) ( n + 1) − 1

(b) ( p + 1) − 1

p

Q. 5

Q. 6

n

(c) p Cn − 1

(d)

p +1

Cn − 1

A four digit number is formed using the digits 1, 2, 3, 4, 5. The total number of numbers which have at least one digit repeated, is (a) 625 (b) 500 (c) 120 (d) 505 The number of integral solutions of the equation x1 + x2 + x3 = 0 ( x1 , x2 , x3 ≥ −3) , is 9! (b) 3! 3 (c) 11 C3 (d) 11 C2 ( ) The number of natural numbers less than 2000 that can be formed using the digits 0, 1, 2, 3, 4 without repeating any digit, is

(a) (3!)

3

Q. 7

(a) 78

(b) 102 n

Q. 8

In the identify ∑ k =0

(a) n Ck Q. 9

(c) 88

(d) 92

Ak n! = the value of Ak is x + k x ( x + 1)( x + 2 )... ( x + n ) (b) n Ck +1

(c) ( −1) .n Ck k

(d) ( −1)

k −1 n

. Ck −1

The number of times the digit 1 will occur in natural numbers less than 1000, is (a) 250

(b) 300

(c) 350

(d) 450

Q. 10 The number of ways in which m prizes can be distributed among n players if each player can receive any number of prizes, is (a) m n

(b) n m

(c) m Cn

(d) n cm

Q. 11 There are n identical seats (unmarked) around a circular table. The number of ways in which m ( m < n ) people can be seated, is ( n − 1)! (a) ( m − 1)! (b) n − m ! (c) ( n − m + 1)! (d) n Cm ( m − 1)! ( ) Q. 12 The number of non-negative integral solutions of the equation x + y + 3 z = 10 , is (a) 26

(b) 10 C2 × 9C3

(c) 160

(d)

10! 2!3!

Q. 13 There are p points in space, no four of which are in the same plane with the exception of q points which are all in the same plane. The number of different planes determined by the points is (a)

p ( p − 1)( p − 2 ) q ( q − 1)( q − 2 ) − 6 6

(c)

p ( p − 1)( p − 2 )( p − 3) q ( q − 1)( q − 2 )( q − 3) − +1 24 24

Mathematics / P & C, Binomial Theorem

(b)

p ( p − 1)( p − 2 ) q ( q − 1)( q − 2 ) − +1 6 6 (d) none of these

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Q. 14 The number of ways in which 8 candidates A1 , A2 , A3 ,....., A8 can be ranked such that A3 is always comes before A4 and A4 always comes before A5 , is (a) 2 × 8 P6

(b)

8! 6

(c) 8!

(d)

8! 3

Q. 15 The number of ways in which 8 candidates A1 , A2 ,....., A8 can be ranked such that A4 is always above A5 , is (a) 2 × 8 P6

(b) 8P6

(c) 8!

(d)

8! 3

Q. 16 The number of triangles that can be formed from n points of which m are collinear, is (a) n C3 − ( m + n ) (b) n C3 (c) n C3 − m C3 (d) n − m C3 Q. 17 The number of ways in which a sum of 10 can be obtained by throwing a dice thrice, is (a) 21 (b) 27 (c) 6 (d) 15 Q. 18 There are 6 identical blue balls and 6 identical red balls. The number of ways in which 6 balls out of the given 10 balls can be arranged in a row such that the number of blue balls is equal to the number of red balls, is (a) 6 (b) 64 (c) 20 (d) 9 Q. 19 The sum of the digits in the units place of all numbers formed with the digits 1, 1, 2, 3 when taken all at a time, is (a) 21 (b) 17 (c) 32 (d) 28 Q. 20 If n objects are arranged in a row, then the number of ways of selecting three of these objects so that no two adjacent object is selected, is n! (a) n C3 − n C2 (b) 3! 3 (c) n C3 (d) n − 2 C3 ( )

Mathematics / P & C, Binomial Theorem

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[ SUBJECTIVE ] Q. 1

A letter lock consists of 4 rings each marked with 15 different letters. In how many ways is it possible to make an unsuccessful attempt to open the lock?

Q. 2

Show that the greatest value of

Q. 3

There are an unlimited number of identical balls of four different colors. How many arrangements of at the most 8 balls in a row can be made by using them?

Q. 4

How many different numbers greater than 5000 can be formed with the digits 1, 2, 5, 9, 0 when repetition of digits is not allowed?

Q. 5

A polygon has 44 diagonals. How may sides does it have?

Q. 6

There are m points on one straight line AB and n points on another straight line AC, none of them being A. How many triangles can be formed with these points as vertices (excluding A)?

Q. 7

In how many ways can five different rings be worn in four fingers with at least one ring in each finger?

Q. 8

m men and n women are to be seated in a row so that no two women sit together. If m > n, show that

2n

Cr for 0 ≤ r ≤ 2n is

the number of ways in which they can be seated is Q. 9

2n

Cn

m !( m + 1)! . ( m − n + 1)!

How many four digit numbers with distinct digits can be formed using the digits 0, 1, 2, 3, 4, 5 which are (a) divisible by 3? (b) divisible by 6?

Q. 10 How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS? Q. 11 Find the total number of ways of selecting 5 letters from the letters of the word INDEPENDENT. Q. 12 Two numbers a and b are chosen from the set {1, 2, 3, ....., 3n}. In how many ways can these integers be selected such that (a) a2 – b2 is divisible by 3?

(b) a3 + b3 is divisible by 3?

Q. 13 What is the number of non-negative integral solutions of the equation x1 + x2 + x3 + 4 x4 = 20? Q. 14 An exam consists of four papers. Each paper has a maximum of m marks. Show that the number of m +1 ways in which a student can get 2m marks in the exam is ( 2m 2 + 4m + 3 ) 3

Mathematics / P & C, Binomial Theorem

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Q. 15 Find the total number of positive unequal integral solutions of the equation x + y + z + w = 20 Q. 16 In how many ways can 10 persons take seats in a row of 24 fixed seats so that no two persons take consecutive seats? Q. 17 There are 15 seats in a row numbered 1 to 15. In how many ways can 4 persons sit in such a way that seat number 6 is always occupied and no two persons sit in adjacent seats? Q. 18 In how many ways can 2n + 1 identical balls be placed in distinct boxes so that any two boxes together will contain more balls than the third? Q. 19 12 seats are to be occupied by 4 people. Find the number of possible arrangements if (i) no two persons sit side by side. (ii) there should be at least two empty seats between any two persons. (iii) each person has exactly one neighbour. Q. 20 The sides of a triangle are a, b, c inches where a, b, c are integers and a ≤ b ≤ c . If c is given, show (c + 1) 2 c (c + 2) if c that the number of different triangles that can be formed is if c is even and 4 4 is odd.

Mathematics / P & C, Binomial Theorem

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59

P & C / ANSWER

[ TRY YOURSELF - I ] 1. 161280 6. 180

2. 120 7. 1200

3. 10 8. 246

4. 120960 9. 26

5. 360 10. 35

[ ASSIGNMENT ] OBJECTIVE LEVEL - I

LEVEL - II

1. d

2. a

1. b

2. c

3. b

4. a

3. a

4. b

5. c

6. b

5. d

6. d

7. a

8. b

7. d

8. c

9. a

10. d

9. b

10. b

11. c

12. c

11. b

12. a

13. c

14. c

13. b

14. b

15. a

16. c

15. d

16. c

17. c

18. d

17. b

18. c

19. d

20. c

19. a

20. d

SUBJECTIVE 1. 154 – 1

3. 87380

4. 48

5. 11

6.

mn (m + n − 2) 2

7. 480

9. 96, 52

10. 2454

11. 72

12.

5n 2 − 3n 3n 2 − n , 2 2

13. 536

15. 552

16.

17. 3072

18.

n ( n + 1) 2

19.

9

15

P10

P4 , 6P4 , 4! × 9C2

Mathematics / P & C, Binomial Theorem

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60

APPENDIX : BINOMIAL THEOREM

Binomial theorem is something that has been known to mathematicians since many centuries ago. In this introduction, we’ll trace the origins of this theorem to the coefficients we obtain when we expand any binomial term raised to an integral power. Consider the following expansions, which can be verified by direct multiplication:

(x + y)

=

1

(x + y)

=

x+ y

(x + y)

=

x 2 + 2 xy + y 2

(x + y)

3

=

x 3 + 3 x 2 y + 3 xy 2 + y 3

(x + y)

4

=

x 4 + 4 x 3 y + 6 x 2 y 2 + 4 xy 3 + y 4

(x + y)

5

=

x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 xy 4 + y 5

(x + y)

6

=

x 6 + 6 x 5 y + 15 x 4 y 2 + 20 x 3 y 3 + 15 x 2 y 4 + 6 xy 5 + y 6

0

1

2

' and so on Do you notice anything special about these expansions, in particular, any general rule or trend these expansions follow that might enable us to expand ( x + y ) directly for a general n ? First of all, notice that the number of n

terms in each expansion is one more than the power of the binomial term. For example, ( x + y ) has 6 terms. 5

However, mathematicians long back also realized another important fact, namely, the relation between the coefficients obtained upon expansion. To see what this relation is, let us write the coefficients in the following ‘triangular’ pattern: n=0

1 1

1

1 3

1

5

1

6

1 3

4

1 1

2

n=1

6 10

4

20

n=3

1

10

15

n=2

5 15

n=4

1

n=5

1 6

1

n=6

Fig - 20 Mathematics / P & C, Binomial Theorem

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61

Do you observe any relation between the various coefficients. If not, consider this same arrangement in a slightly modified form: 1 1

1

1 1 1 1

(1+1) (1+2)

(1+3) (1+4)

1

(2+1)

(3+3)

(4+6)

1

(3+1)

(6+4)

1

(4+1)

1

Fig - 21 The ‘rule’ for constructing this triangular pattern should be pretty obvious now. All edge-numbers are 1. Any other is obtained by adding the number directly above and to the left with the number directly above and to the right, as in Fig-34 Extending this process gives us all the ‘binomial coefficients.’ This geometrical arrangement of the binomial coefficients in a triangle is called Pascal’s triangle. The figure below shows a Pascal triangle containing the coefficients upto n = 15. 1 2

1 3

1 5

1 6

1 7

1 8

1 9

1

26

10 20

35 56

84

4 5 15 35

70 126

1 6 21

56 126

1 1 7 28

84

1 8

26

1 9

1

120 210 252 210 120 45 10 1 165 330 462 462 330 165 55 11 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 1

10

15

28

1

6 10

21

1 3

4

1

Pascal's triangle, upto n = 15

1

1

45

55

1

14

91

364

1001 2002 3003 3432 3003 2002 1001

364

91

1

14

1

Fig - 22

If we denote the (i + 1)th number at the nth level by Tn, i , then we have Tn , i +1 = Tn −1, i + Tn −1, i +1 ... (1) Later on, when we actually write Tn , i +1 in terms of combinational notation (in fact, we’ll see that Tn , i +1 corresponds to n Ci ), we will immediately understand that (1) is equivalent to n Ci = n −1Ci −1 + n −1Ci which, as we already know from the last chapter on P & C, is true. Mathematics / P & C, Binomial Theorem

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Section -1 BINOMIAL THEOREM, POSITIVE INTEGRAL INDEX

Section - 1

Let us now consider more formally the binomial theorem. We need to expand ( x + y ) , where x, y are two arbitrary quantities, but n is a positive integer. n

In particular, if we write

(x + y)

n

= ( x + y )( x + y )( x + y ) ... ( x + y )

( n times )

... (1)

we need to find out the coefficient of xi y j . Note that i + j must always equal n, so that we can write a general term of the expansion (without the coefficient) as x r y n − r so that 0 ≤ r ≤ n . Now, to find the coefficient of x r y n − r , note that we need the quantity x, r times, while y is needed n – r times. Thus, in (1), x r y n − r will be formed whenever x is ‘contributed’ by r of the binomial terms, while y is ‘contributed’ by the remaining n - r of the binomial terms. For example, in the expansion of ( x + y ) , to form x 2 y 3 , we 5

need x from 2 terms and y from 3: take x 5

( x + y ) = (x + y )(x + y )(x + y )(x + y )(x + y ) take y 2

One of the ways to form x y

3

Fig - 23

How many ways are there to form x 2 y 3 ? In other words, how many times will x 2 y 3 be formed? The number of times x 2 y 3 is formed is what is the coefficient of x 2 y 3 . That number, which would be immediately obvious to the alert reader, is simply 5 C2 . Why? Because this is the number of ways in which we can select any 2 binomial terms from 5. These 2 terms will contribute x. The remaining will automatically contribute y. In the general case of ( x + y ) , we see that the coefficient of x r y n − r would be n Cr . (which is infact the same n

as n Cn − r ). Thus, the general binomial expansion is n

= nC0 x n + nC1 x n- 1 y + nC2 x n- 2 y 2 + ... + nCn y n

(x + y )

⇒

(x + y)

n

Mathematics / P & C, Binomial Theorem

n

= ∑ n Cr x n − r y r r =0

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63

The coefficients n Ci are called the binomial coefficients, for a reason that should now be obvious. Note that the (i + 1) coefficient in this expansion is n Ci , which now explains the relation th

Tn , i +1 = Tn −1,i + Tn −1, i +1

we observed in the Pascal triangle; this relation simply corresponds to n

Ci = n −1Ci −1 + n −1Ci

Also, the binomial coefficients of terms equidistant from the beginning and the end are equal, because we have n Cr = n Cn − r . The general term of expansion, n Cr x n − r y r , is the (r + 1)th term from the beginning of the expansion and is conventionally denoted by Tr +1 , i.e. Tr +1 = nCr x n − r y r Since we have (n + 1) terms in the general expansion, we see that if n is even, there will be an odd number of terms, and thus there will be only one middle term, which would be n Cn / 2 x n / 2 y n / 2 . For example,

(x + y)

4

= x 4 + 4 x3 y + 6 x 2 y 2 + 4 xy 3 + y 4 only one middle term

On the other hand, if n is odd , then there will be an even number of terms in the expansion, and thus there will be two middle terms, namely n C n −1 x

n +1 2

y

n −1 2

n

and C n −1 x

2

n −1 2

y

n +1 2

2

For example;

(x + y)

5

= x5 + 5 x 4 y + 10 x3 y 2 + 10 x 2 y 3 + 5 xy 4 + y 5 Two middle terms

Example – 1

1 Find the middle term(s) in the expansion of x − x

9

Solution: Since n = 9, there will be 10 terms in the expansion, which means that there will be 2 middle terms in the expansion, the 5th and the 6th: T5 = C4 ( x )

−1 = 126 x x

T6 = 9C5 ( x )

−1 −126 = x x 5

9

9−4

9 −5

Mathematics / P & C, Binomial Theorem

4

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Example – 2 10

2x2 1 Is there any term in the expansion of − that will be independent of x? x 5

Solution: The general term in the expansion is 10 − r

2x2 Tr +1 = Cr 5 10

10 − r

2 = 10Cr 5

10 − r

2 = 10Cr 5

−1 x

r

( −1) x

20 − 2 r −

( −1)

20 −

r

r

x

r 2

5r 2

Thus, for the term that is independent of x, we have 20 −

⇒

5r =0 2

r =8

Thus, the term free of x is the 9th term given by 2

36 8 2 T9 = C8 ( −1) = 5 5 10

Example – 3

Evaluate the sum n C0 + n C1 + n C2 + ... + n Cn . Solution: We have already evaluated this sum in the chapter on P & C. That approach was as follows: this sum basically counts the number of all sub-groups of a set of size n; this can also be counted by focusing on each element of the set, which has two corresponding choices - you either include it into your sub-group or you don’t, which means that the total number of ways to form sub-groups is 2 × 2 × 2 .... n times = 2n. The sum of the binomial coefficients therefore equals 2n. Here, we evaluate the same sum using a binomial approach. Consider the following expansion:

(1 + x )

n

= nC0 + nC1 x + n C2 x 2 + nC3 x 3 + ... + nCn x n

If we put x = 1, we simply obtain 2n = n C0 + n C1 + n C2 + ... n Cn Mathematics / P & C, Binomial Theorem

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65

Thus, the same result is obtainable from both a combinatorial and a binomial approach. We can also derive another useful result by putting x = –1 in the above relation, so that we obtain 0 = nC0 − nC1 + nC2 − nC3 + ... + ( −1) nCn n

⇒

n

C0 + n C2 + n C4 + ... = n C1 + n C3 + n C5 + ...

This states the sum of the even-numbered coefficients is equal to the sum of the odd-numbered coefficients. Can you prove this using a combinatorial approach? As an exercise, prove the following relations: n

C0 2n + n C1 2n −1 + n C2 2n − 2 + ... n Cn = 3n

n

C0 − nC1 x + nC2 x 2 − ... + ( −1) nCn x n = (1 − x ) n

n ( −1) nCn = 1 C nC C C0 − 1 + 2 2 − 3 3 + ... + 2 2 2 2n 2n n

n

n

n

Example – 4

(a) What is the greatest coefficient in the expansion of ( x + y ) ? n

(b) What is the greatest term in the expression of ( x + y ) ? n

n Solution: (a) For this part, we basically need to only determine max ( Cr ) for 0 ≤ r ≤ n ; x and y have no role to play in this part. To find the greatest coefficient, consider the following ratio:

n! ( r + 1)!( n − r − 1)! C q = n r +1 = n! Cr r !( n − r )! n

=

n−r r +1

Thus, q >1

Mathematics / P & C, Binomial Theorem

⇒

n−r >1 r +1

⇒

n − r > r +1

⇒

r<

n −1 2 www.locuseducation.org

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66

Similarly, q nCr

whenever

n

Cr +1 < nCr

whenever

n −1 2 n −1 r> 2

r<

... (1)

If n is odd, we have n

C n −1 > nC n −3 2

n

and

2

C n +3 < nC n +1 2

2

Also, since n

C n −1 = n C n +1 2

2

we see that for odd n, the two middle coefficients are the greatest. This can be verified by considering the following expansion:

(x + y)

5

= x5 + 5 x 4 y + 10 x3 y 2 + 10 x 2 y3 + 5 xy 4 + y5 ! ( The two middle coefficients are the greatest for odd n

If n is even, (1) gives n

C n > nC n 2

n

and

2

−1

C n < nC n 2

+1

2

In this case therefore, the greatest coefficient is the single middle coefficient n C n . Lets verify 2

this again:

(x + y)

6

Mathematics / P & C, Binomial Theorem

= x6 + 6 x5 y + 15 x 4 y 2 + 20 x3 y3 + 15 x 2 y 4 + 6 xy 5 + y 6 ! The single middle coefficient is the greatest for even n

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67

(b) To find the greatest term, we must also consider x and y. We again follow the approach of part (a): n Tr +1 Cr x n − r y r q= = n Tr Cr −1 x n − r +1 y r −1

=

( n − r + 1) . y r

x

Observe that q >1

If

⇒

( n − r + 1) ⋅ y > 1

⇒

( n + 1) y x + y − r⋅ >0 x + y r x

r

x

( n + 1) y

is an integer m , which must lie in (0, n ], we see that there are two greatest terms x+ y Tm and Tm + 1. (Why). Here’s the explanation: We have q >1

Now, if

q m

1 ≤ r < m and Tr > Tr +1 for m + 1 ≤ r ≤ n

⇒

Tr < Tr +1 for

⇒

Tm −1 < Tm , Tm +1 > Tm + 2 , Tm = Tm +1

⇒

Tm and Tm +1 are the two greatest terms

( n + 1) y x+ y

for 1 ≤ r < m and

( n + 1) y = m. is a non-integer, assume x+ y

We now have q > 1 for 1 ≤ r ≤ m and q < 1 for r > m

⇒

Tr < Tr +1 for 1 ≤ r ≤ m and Tr > Tr +1 for r > m

⇒

Tm < Tm +1 , Tm +1 > Tm + 2

⇒

Tm +1 is the greatest term.

Mathematics / P & C, Binomial Theorem

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Example – 5

How will you expand the multinomial expression ( x1 + x2 + ... + xm ) ? n

Solution: We will approach this problem using combinatorics. Note that a general term of the expansion would be of the form (without the coefficient) x1n1 x2n2 x3n 3 ....xmnm

... (1)

where the various powers must always sum to n (why?). i.e., n1 + n2 + n3 + ... + nm = n Now, to evaluate the coefficient of the term in (1), we consider the multinomial expression in expanded form: x1 + x2 + ... + xm )( x1 + x2 + ... + xm ).............. ( x1 + x2 + ...xm ) ("########## #$########### % n times

To generate the term in (1), we must get x1 from n1 terms, x2 from n2 terms and so on. Let us find the number of ways in which this can be done. First select those n1 multinomials that will contribute x1 : this can be done in n Cn1 ways. Now, from the remaining ( n − n1 ) multinomials, select those n2 multinomials that will contribute x2 : this

can be done in (

n − n1 )

Cn2 ways. Continuing this process, we see that the number of ways to get

x1 from n1 , x2 from n2 ... and so on, that is, the number of times the term in (1) will be generated in the expansion is n

Cn1 × (

n − n1 )

Cn2 × (

n − n1 − n2 )

Cn3 × ...

=

( n − n1 )! × ( n − n1 − n2 )! × ... n! × n1 !( n − n1 )! n2 !( n − n1 − n2 )! n3 !( n − n1 − n2 − n3 )!

=

n! n1 ! n2 !...nm !

This is what is known as the general multinomial coefficient. The multinomial expansion can now be written compactly as

( x1 + x2 + ... + xm )

n

Mathematics / P & C, Binomial Theorem

=∑

n! x1n1 x2n2 ... xmnm n1 ! n2 !...nm !

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69

where the summation is carried out over all possible combinations of the ni 's such that ∑ ni = n . For example, in ( x1 + x2 + x3 ) , let us consider some terms in the expansion: 4

Multinomial

Term

Coefficient 4! = 12 2!1!1!

x12 x2 x 3

4! =4 3!1!

x13 x2 ( x1+x2 + x3)

4

4! 4!

x24

=1

4! = 12 1!1!2!

x1 x2 x 23 etc

Example – 6 10

Find the coefficient of x4 in the expansion of 1 + x − 22 . x Solution: From the previous example, the general term in the expansion will be n −2 10! (1) n1 ( x ) 2 2 n1 ! n2 ! n3 ! x

n3

10 x n2 −2 n3 (−2) n3 n1 ! n2 ! n3 !

=

where n1 + n2 + n3 must be 10. Now, x4 is generated whenever n2 − 2n3 = 4. The possible values of the triplet (n1 , n2 , n3 ) can now simply be listed out: (n1 , n2 , n3 ) ≡ (6, 4, 0), (3, 6, 1), (0, 8, 2) Thus, the (total) coefficient of x4 is 10! 10! 10! ( −2) 0 + ( −2)1 + ( −2) 2 6! 4! 0! 3! 6!1! 0!8! 2!

= – 1290 (verify) Mathematics / P & C, Binomial Theorem

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Section - 2

70

-1 DIFFERENTIATIONSection & INTEGRATION TECHNIQUES

The techniques of calculus enable us to sum a lot of series involving binomial coefficients. This is the subject of this section. Suppose that we have to evaluate the sum S given by S = n C1 + 2 n C2 + 3 n C3 + ...... + n n Cn

From now on, to avoid clutter, we’ll write n Cr as simply Cr, where the upper index n should be understood to be present. Thus, S = C1 + 2C2 + ..... + n Cn

=∑ rCr This series can be generated using a manipulation involving differentiation, as follows: Consider the binomial expansion (1 + x ) n = C0 + C1 x + C2 x 2 + ...... + Cn x n

If we differentiate both sides with respect to x, look at what we’ll obtain: n(1 + x ) n −1 = C1 + 2C2 x + 3C3 x 2 + ..... + nCn x n −1

Now, all that remains is to substitute x = 1, upon which we obtain: n ⋅ 2n −1 = C1 + 2C2 + 3C3 + ..... + n Cn

This is what we were looking for. Thus, S = n ⋅ 2n −1 Had we substituted x = −1 , we would’ve obtained 0 = C1 − 2C2 + 3C3 ....... + ( −1) n −1 n Cn

Thus, we have evaluated another interesting sum. Suppose that we now wish to evaluate S1 given by S1 = C0 +

C C1 C2 + + .... + n 2 3 n +1

The alert reader would immediately realize that integration needs to be applied here. How exactly to do so is now described. Consider again the expansion. (1 + x ) n = C0 + C1 x + C2 x 2 + ..... + Cn x n

If we integrate this with respect to x, between some limits say a to b, we obtain b

b

b

b

(1 + x)n+1 x2 x3 x n+1 b = C0 x a + C1 + C2 + .... + Cn n +1 a 2 a 3 a n +1 a Mathematics / P & C, Binomial Theorem

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To generate the sum S1 , a little thought will show that we need to use a = 0, b = 1, so that we obtain C C C 2n +1 − 1 = C0 + 1 + 2 + ..... + n n +1 n +1 2 3 2n +1 − 1 Thus, S1 equals n +1 Try some other values for a and b and hence generate other series on your own. Be as varied as you can in choosing these limits.

Example – 7

Find the sum S given by S = 12 ⋅ C1 + 22 ⋅ C2 + 32 ⋅ C3 + .... + n 2 ⋅ Cn

Solution: We have to plan an approach wherein we are able to generate r2 with Cr. We can generate one r with every Cr, as we did earlier, and which is now repeated here: (1 + x)n = C0 + C1x + C2 x2 + ...... + Cn xn

Differentiating both sides with respect to x, we have n(1 + x)n−1 = C1 + 2C2 x + 3C3 x 2 + ...... + nCn x n−1

Now we have reached the stage where we have an r with every Cr. We need to think how to get the other r. If we differentiate once again, we’ll have r(r – 1) with every Cr instead of r2(understand this point carefully). To ‘make-up’ for the power that falls one short of the required value, we simply multiply by x on both sides of the relation above to obtain: nx (1 + x ) n −1 = C1 x + 2C2 x 2 + 3C3 x 3 + .... + nCn x n

It should be evident now that the next step is differentiation: n ( n − 1) x (1 + x ) n − 2 + n (1 + x ) n −1 = C1 + 2 2 ⋅ C2 x + 32 ⋅ C3 x 2 + .... + n 2 ⋅ Cn x n −1

Now we simply substitute x = 1 to obtain n ( n − 1) ⋅ 2 n − 2 + n ⋅ 2 n −1 = C 1 + 2 2 ⋅ C 2 + 3 2 ⋅ C 3 + ..... + n 2 ⋅ C n

The required sum S is thus S = n(n − 1) ⋅ 2n − 2 + n ⋅ 2n −1 = n ⋅ 2n −2 {(n − 1) + 2}

= n(n + 1) ⋅ 2n− 2 Mathematics / P & C, Binomial Theorem

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Example – 8

Evaluate the following sums: (a) S1 =

C0 C2 C4 + + + ....... 1 3 5

(b) S 2 =

C1 C3 C5 + + + ....... 2 4 6

Solution: The first sum contains only the even-numbered binomial coefficients, while the second contains only odd-numbered ones. Recall that we have already evaluated the sum S given by C C1 C2 2n +1 − 1 + + ...... + n = n +1 n +1 2 3 Note that S is the sum of S1 and S2, i.e., S = C0 +

2n +1 − 1 S1 + S2 = n +1 Thus, if we determine S1, S2 is automatically determined, and vice-versa. Let us try to determine S1 first. (a) Consider again the general expansion (1 + x ) n = C0 + C1 x + C2 x 2 + .... + Cn x n

Integrating with respect to x, we have (we have not yet decided the limits) b

b

b

b

(1 + x)n+1 x2 x3 xn+1 b = C0 x a + C1 + C2 + ..... + Cn n +1 a 2 a 3 a n +1 a Since we are trying to determine S1 which contains only the even-numbered terms, we have to choose the limits of integration such that the odd-numbered terms vanish. This is easily achievable by setting a = – 1 and b = 1 (understand this carefully). Thus, we have

2n+1 C C = 2 C0 + 2 + 4 + .... n +1 3 5 which implies that 2n S1 = n +1 (b) S2 is now simply given by S 2 = S − S1 =

2 n +1 − 1 2n − n +1 n +1

=

2n − 1 n +1

Mathematics / P & C, Binomial Theorem

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Section - TECHNIQUES 1 MISCELLANEOUS

Section - 3

Not all questions can be subjected to the method(s) described earlier. For example, consider the sum S given by S = C0C1 + C1C2 + C2C3 + ...... + Cn −1Cn Let us first go through a combinatorial approach, using the observation that C0 = Cn , C1 = Cn−1 and so on, so that S can be rewritten as S = CnC1 + Cn −1C2 + Cn − 2C3 + ...... + C1Cn Consider a general term of this sum, which is of the form Cn − r Cr +1 . We can think of this as the number of ways of selecting (n – r) boys from a group of n boys and (r + 1) girls from a group of n girls. The total number of people we are thus selecting is (n − r ) + (r + 1) = (n + 1) . Therefore, S represents the total number of ways of selecting (n + 1) people out of a group of 2n, so that S is simply

2n

Cn +1 .

Now to a binomial approach. This will involve generating the general term Cr Cr +1 somehow, which is the same as Cn − r Cr +1 . Consider the general expansion of (1 + x) n . (1 + x ) n = C0 + C1 x + C2 x 2 + ..... + Cn x n

...(1)

We have to have the terms CnC1 , Cn−1C2 and so on, which suggests that we write (1) twice, but in the second expansion we reverse the terms, multiply, and see what terms contain the (combinations of) coefficients we require. n

(1 + x) = C0 n

2

+ C1x n

n

+ C2x n–1

+ ..........+ Cnx n–2

(1 + x) = Cnx + Cn – 1x + Cn – 2x

+ .....+ C1x + C0

Multiplying, we find on the left hand side we have (1 + x) 2 n , while on the right hand side, the terms containing the (combinations of) coefficients we want will always be of the form (

) x n+1 , that is, the power of x will be

(n + 1). No other terms will contain x n +1 , verify this for yourself. Thus, the sum Cn C1 + Cn−1C2 + ........ + C1Cn is actually the total coefficient of x n +1 on the right hand side, and from the left hand side we know that the coefficient of x n +1 would be simply 2 n Cn +1 . Thus, S = 2 n Cn +1 A very similar approach could have been: n

2

n

(1 + x) = C0 + C 1x + C2x + ........ + Cnx 1 1+ x

n

Mathematics / P & C, Binomial Theorem

1 1 = C0 + C1 1 + C2 2 + ........ + Cn n x x x www.locuseducation.org

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Thus,

74

S

n

=

1 n Coefficient of x in (1 + x ) 1 + x

=

Coefficient of x in

=

Coefficient of x n +1 in (1 + x ) 2 n = 2 nCn +1

(1 + x) 2 n xn

Example – 9

Find the sum S given by S = C02 + C12 + C22 + ...... + Cn2

Solution: Note that S can be rewritten as S = C0Cn + C1Cn−1 + C2Cn− 2 + ...... + Cn C0 Using a combinatorial approach, the sum should be immediately obvious to the alert reader as 2n Cn . In brief, this is because the right hand side represents, as an example, the total number of ways of selecting n people from a group of n boys and n girls, etc. Now, we discuss the binomial expansion approach:

(1 + x ) n = C 0 *

+ C1 x

+ C 2 x 2 + C 3 x 3 + ..... + C n x n

*

*

)

(1 + x ) n = C n x n + C n −1 x n −1 + C n − 2 x n − 2 + ..... + C 0

Thus, we observe that the required sum is the coefficient of xn in (1 + x) 2 n , which is simply

2n

Cn .

Example – 10

Find the sum S given by S = n C0 +

n +1

C1 +

n +2

C2 + ........ +

n+r

Cr

Solution: We have already evaluated this sum in P & C; here we’ll use a binomial approach. Note that n+ r

⇒

∑

n+r

Cr = Coeff. of x n in (1 + x) n+ r Cr = ∑ (Coeff. of x n in (1 + x) n + r )

= Coeff. of x n in ∑ (1 + x)n + r

Mathematics / P & C, Binomial Theorem

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Thus, S = Coeff. of x n in (1 + x )n + (1 + x )n +1 + ...... + (1 + x )n + r

{

= Coeff. of x n in (1 + x) n 1 + (1 + x ) + (1 + x) 2 + ...... + (1 + x ) r = Coeff. of x in n

(1 + x)n {(1 + x) r +1 − 1} x

= Coeff. of x n +1 in {(1 + x ) n + r +1 − (1 + x) n } =

⇒

S=

n + r +1

Cr

n + r +1

Cr , which is the same result we obtained in P & C.

Mathematics / P & C, Binomial Theorem

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TRY YOURSELF - I

Q 1.

Find the sum of all the rational terms in the expansion of (31/ 4 + 41/ 3 ) .

Q. 2

Prove that three consecutive terms in a binomial expansion can never be in G.P.

12

( ) is even (b) Show that the integral part of (8 + 3 7 ) is odd 67

Q. 3* (a) Show that the integral part of 5 5 + 11

n

Q. 4

Use the binomial theorem to show that 7103 when divided by 5 leaves a remainder 3.

Q. 5

Find the coefficient of x301 in the expansion of (1 + x )500 + x (1 + x)499 + x 2 (1 + x )498 + ...... + x 500 2n

Q. 6* Suppose (1 + x + x ) is written in expanded form, i.e., in the form 2 n

∑ax r =0

r

r

. Show that

ar = a2n−r 32

Q. 7* Use the binomial theorem to show that 3232 when divided by 7 leaves the remainder 4. Q. 8

Prove the following relations: (a) n C0 ⋅ 2 nCn − n C1 ⋅ 2 n −1Cn + n C2 ⋅ 2 n −2Cn − ....... + n Cn (−1)n ⋅ n Cn = 1 (b) n C0 ⋅ 2 nCn − nC1 ⋅ 2 n − 2Cn + nC2 ⋅ 2 n − 4Cn − ..... = 2n (c) C0Cr + C1Cr +1 + C2Cr + 2 + ...... + Cn − r Cn = 2 nCn + r (d)* m C1 n Cm − mC2 2 n Cm + mC3

3n

Cm − ..... + (−1) m −1 ⋅ mCm

(e) (C 0 + C1 )(C1 + C 2 )..... (C n −1 + C n ) = Q. 9

nm

Cm = ( −1) m +1 ⋅ n m

( n + 1) n C1C 2 .....C n n!

Find the sums of the following series. (a) C0 − 2C1 + 3C2 − 4C3 + .....

(b)*

C0 C1 C2 C3 − + − + ..... 3 4 5 6

Q. 10 Using the binomial theorem, show that r

n

Cr = ∑ k C j ⋅ n − k Cr − j j =0

Q. 11 Find the sum of the series n

∑ r =0

1 3r 7r 15r (−1)r n Cr r + r + r + r + ....m terms 2 4 8 16

Q. 12* Show that n

Cm + 2 ⋅ n −1Cm + ....... + (n − m + 1) ⋅ mCm =

Mathematics / P & C, Binomial Theorem

n+2

Cm + 2 www.locuseducation.org

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Section - RATIONAL 1 BINOMIAL THEOREM, INDEX

Section - 4

In the previous section, we discussed the expansion of ( x + y ) n , where n is a natural number. We’ll extend that discussion to a more general scenario now. In particular, we’ll consider the expansion of (1 + x) n , where n is a rational number and | x | < 1. Note that any binomial of the form (a + b )n can be reduced to this form.:

b ( a + b) = a 1 + a

n

n

b = a n 1 + a

(we are assuming | a | > | b |)

n

= a n (1 + x) n where | x | < 1 The general binomial theorem states that (1 + x )n = 1 + nx + ........ +

n(n − 1) 2 n(n − 1)(n − 2) 3 x + x + ........ 2! 3! n( n − 1)( n − 2)......(n − r + 1) r x + .......∞ r!

That is, there are an infinite number of terms in the expansion with the general term given by Tr +1 =

n( n − 1)(n − 2).......(n − r + 1) r x r!

For an approximate proof of this expansion, we proceed as follows: assuming that the expansion contains an infinite number of terms, we have:

(1 + x )

= a0 + a1 x + a2 x 2 + a3 x 3 + ... + an x n + ...∞

n

Putting x = 0 gives a0 = 1. Now differentiating once gives n (1 + x )

n −1

= a1 + 2a2 x + 3a3 x 2 + .....∞

Putting x = 0 gives a1 = n. Proceeding in this way, we find that the rth coefficient is given by an =

n ( n − 1)( n − 2 )... ( n − r + 1) r!

Note that if n is a natural number, then this expansion reduces to the expansion obtained earlier, because Tr +1 becomes n Cr x r , and the expansion terminates for r > n . For the general Tr +1 , we obviously cannot use n Cr since that is defined only for natural n. One very important point that we are emphasizing again is that the general expansion holds only for | x | < 1 . Mathematics / P & C, Binomial Theorem

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Let us denote the genral binomial coefficient by Vr. Thus, we have Vr =

n(n − 1)(n − 2)......(n − r + 1) r! ∞

and

(1 + x) n = ∑ Vr x r r =0

Let us discuss some particularly interesting expansions. In all cases, | x | < 1: (1) (1 + x) −1 :

Since

n = −1, we see that

Vr =

( −1)(( −1) − 1)((−1) − 2)......(( −1) − r + 1) r!

= (−1)r so that the expansion is (1 + x )−1 = 1 − x + x 2 − x 3 + ......∞ (2) (1 − x) −1 :

Again, Vr = ( −1) r and thus (1 − x) −1 = 1 + x + x 2 + x 3 + ......∞

(3) (1 + x) −2 :

We have n = – 2; Vr =

( −2)((−2) − 1)((−2) − 2)....((−2) − r + 1) r!

(−1) r (r + 1)! = r! = (−1)r ⋅ (r + 1) Thus, (1 + x)−2 = 1 − 2x + 3x2 − 4x3 + ......∞ (4) (1 − x) −2 :

Again, Vr = ( −1) r ⋅ ( r + 1) so that (1 − x) −2 = 1 + 2 x + 3 x 2 + 4 x 3 + ........∞

Example – 11

(a) For | x | < 1 , expand (1 − x) −3 (b) Find the coefficient of x n in the expansion of (1+ 3x + 6x2 +10x3 + .......∞)−n Mathematics / P & C, Binomial Theorem

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Solution: (a) We have Vr =

( −3) (( −3) − 1)((−3) − 2))......(( −3) − r + 1) r!

=

(−1) r (r + 2)! 2(r !)

=

(−1) r (r + 1) ( r + 2) 2

Thus, V0 = 1, V1 = −3, V2 = 6, V3 = −10........ so that (1 − x) −3 = 1 + 3 x + 6 x 2 + 10 x3 + .......∞ (b) We use the result of part –(a) in this: (1 + 3 x + 6 x 2 + 10 x 3 .......∞ ) − n = ( (1 − x ) −3 )

−n

= (1 − x)3n The coefficient of x n in this binomial expansion (note: the power is now a positive integer) would be ( −1) n ⋅ 3 nCn . Example – 12

Find the magnitude of the greatest term in the expansion of (1 − 5y )

−2 / 7

1 for y = . 8

Solution: Let us first do the general case: what is the greatest term in the expansion of (1 + x) n , where n is an arbitrary rational number. We have, Tr +1 = Vr x r

and

Tr = Vr −1 x r −1 Tr +1 Vr = ⋅x Tr Vr −1

so that

=

n − r +1 ⋅x r

Now, let us find the conditions for which this ratio exceeds 1. We have

Tr +1 ≥ Tr ⇒ Mathematics / P & C, Binomial Theorem

1 n +1 −1 ≥ r |x|

...(1) www.locuseducation.org

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For this particular problem, (1) becomes −2 +1 1 7 −1 ≥ −5 r 8 5 8 −1 ≥ 7r 5

⇒ ⇒

5 13 ≥ 7r 5

⇒

r≤

⇒

r=0

25 91

Thus, Tr +1 = T1 is the greatest term, with magnitude 1. Example – 13

Find the sum of the series 1+

2 2.5 2 ⋅5 ⋅8 + + + ......∞ 6 6.12 6 ⋅12 ⋅18

if you are told that this corresponds to an expansion of a binomial, of the form (1 + x) n . Solution: We need to determine n and x. For that, we can compare the terms of this series with the corresponding terms in the following general expansion. (1 + x ) n = 1 + nx +

n( n − 1) 2 n( n − 1) ( n − 2) 3 x + x + ......∞ 2! 3!

Thus, nx =

2 6

n( n − 1) 2 2.5 x = 2! 6.12

Solving for n and x from these two equations, we get n = −

2 1 and x = − . Thus, the sum of the 3 2

series is

S = (1 + x )

n

1 = 1 − 2

−

2 3

= 41/ 3 Mathematics / P & C, Binomial Theorem

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Example – 14

Find the sum of the series 1 1 1 + + + ...... to n terms a + b a + 2b a + 3b

for b

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