Jee 2014 Booklet5 Hwt Solutions Permutations and Combinations

Vidyamandir Classes

Solutions to Home Practice Test-5/Mathematics Permutation and Combination 1.(A)

Since

m

C r 1  n



 2.(D)

m

Cr 

HWT - 1

m 1

Cr we write the given inequality as

C6  n C7

n

C6

n

C7

1

7 1 n6





n > 13

the least value of n is 14.

The required number of ways = the number of ways in which 8 girls can sit in a row – the number of ways in which two sisters sit together = 8!   2   7!  30240

3.(B)

The word MATHEMATICS contains 11 letters viz. M, M, A, A T, T, H, E, I, C, S. The number of words that begin with T and end with T is

9!  90720 2!2! 4.(B)

We have : 15  3  5

 100   100  E5 100 !      20  4  24   5   52 

Now,

E3 100!  48



Exponent of 15 in 100! = min (24, 48) = 24

and

5.(B)

The mint has to perform two jobs, viz. (i) Selecting the number of days in the February month (there can be 28 days or 29 days), and (ii) Selecting the first day of the February month. The first job can be completed in 2 ways while the second can be performed in 7 ways by selecting any one of the seven days of a week.

6.(D)

We have E 

31! 31

2

7.(C)

 32 ! 



1 31

2

 32 



1 36

 

 236  23

2

12

 812

Thus, x  12 The non-zero perfect square digits are 1, 4 and 9. 1 can occur at units place in 3  3 = 9 ways



Sum due to 1 at units place is 1  9 . Similarly, sum due to 1 at tens place is 1  10  9 and sum due to 1 at hundreds place is 1  100  9 . We can deal with the digits 4 and 9 in a similar way.

Thus, sum of the desired numbers is (1 + 4 + 9) (1 + 10 + 100) (9) = 13986 8.(C)

Letters appearing in the word SACHIN are A, C, H, I, N, S The words beginning with letters A, C, H, I and N appear before the word SACHIN. There are 5(5!) = 600 words beginning with A, C, H, I and N Word SACHIN is the first word beginning with S. Therefore, SACHIN appears at serial number 601.

9.(D)

Let A and B be two subsets of S. If x  S , then x will not belong to A  B or x belongs to at most one of A, B. This can happen in 3 ways. Thus, there are 34  81 subsets of S for which A  B   . Out of these there is just one way for which A  B  

10.(B)





1 4 3  1  1  41 2 Since a true/false type question can be answered in 2 ways either by marking it true or false. So, there are 2 ways of answering each of the 5 questions. So, total number of different sequences of answers

As, we, are interested in unordered pairs of disjoint sets, the number of such unordered pairs is

 2  2  2  2  2  25  32

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HWT-5/Mathematics

Vidyamandir Classes Out of these 32 sequences of answers there is only one sequence of answering all the five questions correctly. But no student has written all the correct answers and different students have given different sequence of answers. So, Maximum number of students in the class = Number of sequences except one sequences in which all answers are correct = 32  1  31

Permutation and Combination 1.(A)

HWT - 2

For each of the first  n  1 elements a1 , a2 , . . . ., an 1 we have two choices: either ai 1  i  n  1 lies in the subset or ai doesn’t lie in the subset. For the last element we have just one choice. If even number of elements have already been selected, we do not include an in the subset, otherwise (when odd number of elements have been selected), we include it in the subset. Thus, the number of subsets of A  a1 , a2 , . . .., an  which contain even number of elements is equal to 2n1 .

2.(B)

The number of triangles that can be formed by using the vertices of a regular polygon of n sides is n C3 . That is, Tn  n C3 Now,

Tn 1  Tn  21



n 1



n

C3  n C3  21  

C2  n C3  n C3  21

C r  n C r 1  n C r  

n 1

1 n  n  1  21  n  6 or 7 2 As n is a positive integer, n = 7. 

3.(A)

A number between 100 and 1000 has three digits. So, we have to form all possible 3-digit numbers with distinct digits. We cannot have 0 at the hundred’s place. So, the hundred’s place can be filled with any of the 9 digits 1, 2, 3, . . . . , 9. So, there are 9 ways of filling the hundred’s place. Now, 9 digits are left including 0. So, ten’s place can be filled with any of the remaining 9 digits in 9 ways. Now, the unit’s place can be filled with in any of the remaining 8 digits. So, there are 8 ways of filling the unit’s place. Hence, the total number of required numbers  9  9  8  648 .

4.(D)

Number of groups having 4 boys and 1 girl



 C  C g 4

4

g

1

and number of groups having 3 boys and 2 girls



 C  4

3

g



C2  2 g  g  1

Thus, the number of dolls distributed  g 1   2   2 g  g  1   4 g 2  3 g

We are given 4 g 2  3g  85  g  5 5.(B)

Letters of the word ENDEANOEL are 3E’s, 2N’s, 1D, 1O, 1L, 1A Letters D, L, N can be permuted at first 4 places in

 6.(C)

required number of ways

4! 5! ways and the remaining letters can be permuted in ways. 2! 3!

4! 5!  = 2(5!) 2! 3!

The signals can be made by using at a time one or two or three or four or five flags. The total number of signals when r flags are used at a time from 5 flags is equal to the number of arrangements of 5, taking r at a time i.e. 5Pr. Since r can take values 1, 2, 3, 4, 5. Hence, by the fundamental principle of addition, the total number of signals

 5 P1  5 P2  5 P3  5 P4  5 P5  5 5 4  5 4  3 5 4  3 2  5 4  3 2 1 = 5 + 20 + 60 + 120 + 120 = 325 7.(B)

We can choose 4 novels out of 6 in 6C4 ways and 1 dictionary out of 3 in 3C1 ways. We can arrange 4 novels and 1 dictionary in the middle in 4! Ways. Thus, required number of ways



 C   C   4!  1080  1000 6

4

3

1

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HWT-5/Mathematics

Vidyamandir Classes 8.(C)

Let us denote the four married couples by C1, C2, C3 and C4. We consider each couple as one unit. We can permute four units in 4! ways. Each couple can be seated in 2! ways. Thus, the required number of ways is (4!) (2) (2) (2) (2) = 384

9.(B)

The total number of numbers formed with the digits 2, 3, 4, 5 taken all at a time = Number of arrangements of 4 digits, taken all at a time =

4

P4  4!  24 .

To find the sum of these 24 numbers, we will find the sum of digits at unit’s, ten’s hundred’s places in all these numbers. Consider the digits in the unit’s places in all these numbers. Each of the digits 2, 3, 4, 5 occurs 3! (= 6) times in the unit’s place. So, total for the digits in the unit’s place in all the numbers = (2 + 3 + 4 + 5)  3! = 84. Since each of the digits 2, 3, 4, 5 occurs 3! times in any one of the remaining places. So, the sum of the digits in the ten’s hundred’s and thousand’s places in all the numbers = (2 + 3 + 4 + 5)  3! = 84





 84 100  101  102  103  93324 10.(B)

As n C4 , n C5 and n C6 are in A.P.,

2 

 C  n

2

5

n n

C4

n



C5

C4  n C6 n

C6

n

C5

5 n5  n4 6



2



n  7 , 14



5! n  5  ! 5! n  5  ! n! n!  4! n  4  ! n! 6! n  6  ! n! 

n 2  21n  98  0

Permutation and Combination 1.(B)

HWT - 3

The 5 boys can be seated in a row in 5 P5  5! . In each of these arrangements 6 places are created, shown by the cross-marks, as given below : BBBBB Since no two girls are to sit together, so we may arrange 3 girls in 6 places. This can be done in 6

6

P3 ways i.e. 3 girls can be seated in

P3 ways.

Hence, the total number of seating arrangements  5 P5  6 P3  5!  6  5  4  14400 2.(D)

We can permute M, I, I, I, I, P, P in

7! ways. Corresponding to each arrangement of these seven letters, we have 8 places where S can 4!2!

be arranged as shown below with X. X X X X X X X X We can choose 4 places out of 8 in 8 C4 ways. Thus, the required number of ways  3.(A) 4.(A)

7!    7  C   C   C   4!2!  8

4

8

4

6

4

For r  6 ; r! is divisible by 240. Thus, when x is divided by 240, the remainder is 1! + 2! + . . . + 5! = 153 This can be done in the following ways : (i)

paint the central triangle with one of the three colours ;

(ii)

paint each of the remaining triangles with any one of the two remaining colours.

Thus, the required number of ways  3  2  2  2  24 5.(D)

There are 7 boys and 3 girls. Seven boys can be arranged in a row in

7

P7  7! ways. Now, we have 8 places in which we can arrange

8

3 girls in P3 ways. Hence, by the fundamental principle of counting, the number of arrangements  7!  8 P3  7!  336 . 6.(B)

Six ‘+’ signs can be arranged in just one way. There are seven places for ‘–’ signs as shown in the following figure marked with X. X+X+X+X+X+X+X We can choose 4 places out of 7 in 7 C4  35 ways.

VMC/Solutions/Permutation & Combination

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HWT-5/Mathematics

Vidyamandir Classes 7.(B)

Let the number of yellow balls be x, that of black be 2x and that of green be y. Then x  2 x  y  20 

3 x  y  20

or

y  20  3 x

As 0  y  20 , we get 0  20  3 x  20

8.(A)



0  3 x  20



The number of ways of selecting the balls is 7

The number of subsets of A containing exactly three elements is is

n 1



n

C3 whereas the number of three subsets of A that contain a1 ,

C2 . We are given, n 1

9.(A)

0 x6

or

C2 

20 100

 C n

3



 n  1  n  2   1 n  n  1  n  2  2 5 6

n = 15

In dictionary the words at each stage are arranged in alphabetical order. Starting with the letter A, and arranging the other four letters GAIN, we obtain 4! = 24 words. Thus, there are 24 words which start with A. These are the first 24 words. 4! 24 Then, starting with G, and arranging the other four letters A, A, I, N in different ways, we obtain   12 words. 2! 2 Thus, there are 12 words, which start with G. Now, we start with I. The remaining 4 letters A, G, A, N can be arranged in

4!  12 ways. So, there are 12 words, which start with I. 2!

Thus, we have so far constructed 48 words. The 49th word is NAAGI and hence the 50th word is NAAIG. 10.(A)

We can choose one denomination in can be chosen in

48

13

C1 ways, then 3 cards of this denomination can be chosen in

C1 ways. Thus, the total number of choices is



13

C1

Permutation and Combination 1.(C)

 C  4

3

48

4

C3 ways and one remaining card



C1  13  4  48  2496

HWT - 4

Letters appearing in the word COCHIN are C, C, H, I, N, O Words appearing before COCHIN are of the form CX..... Where X is one of the letters C, H, I, N and the four remaining places can be filled by the remaining four letters. Thus, the number of words before COCHIN is (4) (4!) = 96

2.(C)

Numbers p and q must be of the form

where

p  r a sb t c ,

q  r s  t 

0  a,   2

and at least one of a,  is 2

0  b,   4

and at least one of b,  is 4

0  c,   2

and at least one of c,  is 2

Possible values of  a,   and  c,   are (0, 2), (1, 2), (2, 2), (2, 0), (2, 1). Possible values of  b,   are (0, 4), (1, 4), (2, 4), (3, 4), (4, 4), (4, 0), (4, 1), (4, 2), (4, 3) Thus, number of possible ordered pairs (p, q) is 5  9  5  225 3.(B)

Number of diagonals = 15 C2  15  90

4.(B)

This can be done in four mutually exclusive ways as follows :

VMC/Solutions/Permutation & Combination

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HWT-5/Mathematics

Vidyamandir Classes Row R1

Row R2

Row R3

Number of ways

I.

1

3

2

 C   C   C 8

II.

1

III.

2

IV.

2

4

2

2

3

1

4

4

2

2

4

2

2

2

1

2

2

 C   C   C 8

1

2

2

4

3

Total 5.(A)

2

 C  C  C 6

2

3

4

 C  C  C 4

1

2

1

2

1

26

The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in  4  1 ! = 3! ways. Now, 4 vacant seats can be occupied by 4 women in 4! ways. 5

6.(B)

We have

47

C4 



52  j

C3 .

j 1

7.(D)



47

C4 

51



47

C3 

47







48







49







50







51

47

C3 

C4 

48

C3 

50

48

47

49

49

50



C3 

48

C3 

50

49

50

C3 

47

C3 

51

C3 

C3 

50

C3 

C3 

50

C3 

C3 

C3 

49

C4 

50

C3 

C3 C3 51

C3

51

C3

C3 

51

C3

 n Cr  1  n Cr  

n 1

Cr  

 n Cr  1  n Cr  

n 1

 n Cr  1  n Cr  

n 1

 n Cr  1  n Cr  

n 1

Cr  

51

C3

C3 

51

C3

Cr  

51

C3



C4 

C3 

51

C3 

48



C3 

50

49



C4 

49

C3 

C4 

C3 

48

50

C3 

C4 

C4 

49

C3 

C4 

50

48

C4 

49

C3 

52

51

C3

 n Cr 1  n Cr  

C4

Cr  

Cr  

n 1

First arrange 6 men around the table in 5 ways. Now, there are 6 gaps between these men. Select any 5 gaps and arrange the women in these gaps in 6 C5  5 ways. So, the required number of ways 5  6C5  5  6  5 .

8.(C)

9.(D)

7 x

Px  3 is defined when x  3  7  x  x  5 and x  3  0  x  3 .



The possible values of



The range of values of

7 x 7 x

Px  3 are when x  3, 4 or 5. Px  3 is



4



P0 , 3 P1 , 2 P2 or {1, 3, 2}.

The sum of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 is 45. So, basically we have to remove 2 digits from the above list such that there sum is 9 because only in those cases we will get an 8-digit number divisible by 9. When we remove 0 and 9. 8 numbers can be formed. We have 7  7 numbers. So, total of numbers = 8   4  7  7   36  7 

10.(B)

There are 9 women and 8 men. A committee of 12, consisting of at least 5 women, can be formed by choosing : (i) 5 women and 7 men (ii) 6 women and 6 men (iii) 7 women and 5 men (iv) 8 women and 4 men (v) 9 women and 3 men Total number of ways of forming the committee 

 9C5  8C7  9C6  8C6  9C7  8C5  9C8  8C4  9C9  8C3  126  8  84  28  36  56  9  70  1  56  6062

Permutation and Combination VMC/Solutions/Permutation & Combination

HWT - 5 25

HWT-5/Mathematics

Vidyamandir Classes 21

C18 ways and 3 wrong results in 23 ways. Thus, required number of ways is

1.(B)

18 correct results can be predicted in

2.(C)

The required number of triangles is

3.(D)

Let  x  x   x  denote the fractional part of x. Note that 0   x  1 . We can write the given equation as

15

21

C18 23 .

C3  4 C3  455  4  451

x  x  3 x  3 x  y  y  3 y  3 y  11 5         x y 3 3 2  2  2 2 4  4  6 4



 x   3x   y   3 y        0 3  2  2  4 

As each number on the L.H.S. lies in the interval 0  x  1 , we must have  x   3x   y   3 y      0 3  2  2  4 

4.(B)

x 3x y 3y , , and must be integers. 3 2 2 4





x = 6, 12, 18, 24,



Number of ordered pairs (x, y) equals 4  7  28

y = 4, 8, 12, 16, 20, 24, 28

Let there be n sides of the polygon. We know that the number of diagonals of n sided pdygon is

n  n  3 2

n  n  3 2

. Therefore,

 44  n 2  3n  88  0   n  11 n  8  0  n = 11

 n  0

Hence, there are 11 sides of the polygon. 5.(D)

We have : 2160  24  33  51 The total number of divisors is same as the number ways of selecting some or all out of four 2’s, three 3’s and one 5. The number of such ways is (4 + 1) (3 + 1) (1 + 1) – 1 = 39.

6.(D)

Using prime factorization of 1050, we can write the given equation as :

x1 x2 x3 x4 x5  2  3  52  7 We can assign 2, 3 or 7 to any of 5 variables. We can assign entire 52 to just one variable in 5 ways or can assign 52  5  5 to two variables in 5 C2 ways. Thus, 52 can be assigned in 5

C1  5C2  5  10  15 ways

Thus, the required number of solutions is 5  5  5  15  1875 7.(C)

We have 36 84



r! n  r  ! 3 n!  7  r  1! n  r  1! n!

r 3  n  r 1 7



10r  3n  3



r 1 2  nr 3

n

C r 1 n



n

Similarly,

n



Cr

Cr

Cr  1



84 126

5r  3  2 n  Solving we obtain r = 3 8.(A)

Number of ways  5C1  5C2  5C3  25

9.(D)

For each book we have (p + 1) choices. As there are n different books, number of ways of selecting books   p  1   p  1  . . . .n times =  p  1 . n

But when we choose 0 copies of each book, we don’t choose any book at all.  10.(A)

The required number of ways   p  1  1 . n

Total number of non-negative integral solutions of the given equation is same as the number of ways of distributing 100 items among 4 persons such that each person can receive any number of items. Hence, total number of solutions 

100  4 1

C 4 1 

103

C3

Permutation and Combination VMC/Solutions/Permutation & Combination

HWT - 6 26

HWT-5/Mathematics

Vidyamandir Classes 1.(C)

A, I, I, I, O can occur at odd places in 5

2.(D)

4

Cn Cn



3.(A)

 1

5

Cn

6

Cn

5 6n  1 5n 6

5! 4! ways, and the remaining letters N, D, L, L can be arranged at the remaining places in ways. 3! 2!

n 2  17 n  30  0





n=2

The total number of points is 15. From these 15 points we can obtain

15

C3 triangles. However, if all the 3 points are chosen on the same

straight line, we do not get a triangle. Therefore, the required number of triangles

 15C3  3

 C   425 5

3

4.(D)

We can arrange remaining 17 boys in 17! ways. For 3 particular boys we have 18 positions. We can arrange 3 particulars boys at these 17!  18! places in 18 P3 ways. Thus, the required number of way is 17!  18 P3  15!

5.(C)

Each object can be put either in box B1 (say) or in box B2 (say). So, there are two choices for each of the n objects. Therefore the number of choices for n distinct objects is 2  2  . . .  2  2n . In one way B1 is empty and in one other way B2 is empty. n  times

6.(C)

We have : 1.1! + 2.2! + 3.3! + . . . . . + n.n! n





r .  r! 

r 1

n

 r 1

n

 r  1 r!  r!    r  1!  r! r 1

  2!  1!   3!  2!  . . . .   n  1 !  n!   n  1 !  1!   n  1 !  1 7.(B)

Use 20! = 218 38 54 72 (11) (13) (17) (19)

8.(C)

When exactly three digits are identical and the remaining two are different, then the number of such numbers  5!  3 C1    60  3! 

 

When three digits are identical and the remaining two are also identical, then the number of ways 

 C   3!5!2!   30 3

2

Thus, the number of such numbers = 90 9.(B)

10.(C)

x1  x2  x3  8, 1  x1 , x2 , x3  6





The required answer is the co-efficient of x8 in x  x 2  . . . .  x 6

or

the co-efficient of x5 in 1  x  . . . .  x 5



  1  x  1  x  3

6

3

3



3



The required answer is 7 C5  21

For any integer i  r , we have r

Cr 



r 1



r2



r3

r 1

Cr  r  2 Cr  . . . . .  i Cr

Cr  1  r  1Cr  . . . . .  i Cr

r

Cr 

Cr  1  

r 1

Cr  1  r  2 Cr  . . . . .  i Cr

Cr  1  r  3Cr  . . . . .  i Cr

 . . . . . i

Thus,

 



k

Cr 

i 1

Cr  1

i 1

Cr  1

. . . .(i)

k r n





k m

m 1

n

k

Cr 



k

Cr 

kr



k

Cr

rr

 n  1Cr  1  m  1  1Cr  1

[Using (i)]

 n  1Cr  1  m Cr  1

Permutation and Combination VMC/Solutions/Permutation & Combination

HWT - 7 27

HWT-5/Mathematics

Vidyamandir Classes 1.(A)

A, E, G, R, D, N As vowels will be in alphabetical order A will come before E. First select 4 out of 6 places and arrange G, R, D and N in them in 6

C4  4 ways. For the remaining two places we have only 1 way of arranging A and E in them due to alphabetical order.

 2.(A)

n

The total number of ways = 6 C4  4  1  360

Cr  1  n Cr  1  2  n Cr 

 n  1Cr 

n 1

Cr  1

 n  2 Cr  1 3.(D)



n

 

Cr  1  n Cr 

n

Cr  n Cr  1

(As n Cr  n Cr  1 

n 1

(As n Cr  n Cr  1 

n 1



Cr ) Cr )

Required number of possible outcomes = Total number of possible outcomes – Number of possible outcomes in which 5 does not appear on any dice.

 63  53  216  125  91 4.(D)

x  6!  6!  6!  6!  2  6! and y  5!  6! 2

 5.(D)

As 0 + 1 + 2 + 3 + 4 + 5 = 15, for a five digit number either do not use 0 or 3. When 0 is not used, the number of such numbers is 5! And when 3 is not used, the number of such number is 5!  4! .

 6.(A)

8.(B)

Required number of ways = 5!   5!  4!  216

We can choose 10 players from  22  2  players in



7.(D)

x  2  6  x  12 y y

required number of ways 

C10 ways and one player from 2 players in 2 C1 ways.

20!  2  369512 10! 10!

We have

an 10n  n  1 ! n  1    an 1 n! 10n 1 10

Note that

a a a a a1 a a a  1, 2  1, 3  1, 4  1, . . . . 8  1, 9  1, 10  1, 11  1, . . . a2 a3 a4 a5 a9 a10 a11 a12



an is greatest if n = 9 or 10.

The number of points of intersection

 9.(B)

20

 C   2   n

2

m

C2

 1   C   C   2  n  n  1  12 m  m  1  2mn n

1

m

1

Since 3 does not occur in 1000. So, we have to count the number of times 3 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form abc where 0  a, b, c  9 . Clearly, Number of times 3 occurs = (No. of numbers in which 3 occurs exactly at one place) + 2 (No. of numbers in which 3 occurs exactly at two places) + 3(No. of numbers in which 3 occurs exactly at three places) Now, No. of numbers in which 3 occurs exactly at one place : Since 3 can occur at one place in 3C1 ways and each of the remaining two places can be filled in 9 ways. So, number of numbers in which 3 occurs exactly at one place = 3 C1  9  9 . No. of numbers in which 3 occurs exactly at two places : Since 3 can occur exactly at two places in 3C2 ways and the remaining place can be filled in 9 ways. So, number of numbers in which 3 occurs exactly at two places = 3 C2  9 . No. of numbers in which 3 occurs at all the three places = 1 Since 3 can occur in all the three digits in one way only. So, number of numbers in which 3 occurs at all the three places is one. Hence, Numbers of times 3 occurs  3C1  9  9  2

10.(B)

C 3

2



 9  3  1  300

Number of ways is 12C6 for M, 6C4 for P and 2C2 for C. Thus, the required number of ways





12

C6

 C  C  6

4

2

2

Permutation and Combination

VMC/Solutions/Permutation & Combination

HWT - 8

28

HWT-5/Mathematics

Vidyamandir Classes 6

1.(B)

50

C4 



56  r

C3

r 1

 50 C4   2.(A)

3.(C)



51



55

C4 

C3 

54

 

C3 

51

C3  . . .  52

50

C3  . . . 

  C  C  C  ...  C  C   C  C  C  ... C  = . . . 

C3  55

50

52

3

50

4

52

4

51

3

3

55

3

53

3

55

3

3

56

C4

The number of points in which the lines can intersect = the number of ways of choosing two lines out of 25 25  24 25 C2   300 2 For every object we have 2 choices : either put it in the subset or not.  The total number of subset = 2  2  2  . . . . n times = 2n

2n  n C0  n C1  . . . .  n Cn 4.(B)

At the first place from left we can put only 1. At every other place we can put either 0 or 1. So, the total of numbers  1  2  2  2  . . . .9 times = 29.

5.(A)

B, A, A, A, N, N Arrange one b and 3 A’s in

4  4 ways. 3

Now arrange 2 N’s in 5 places created by one B and 3 A’s in 5C2 ways.  6.(D)

Total number of arrangements  4  5C2  40

Let n = 2m + 1. Then, the common difference of the A.P. can be 1, 2, 3, . . . ., m. The number of AP’s with 1, 2, 3, . . . . , m common differences are  2m  1 ,  2m  3 , . . . ., 1 respectively  n  1 So, total number of AP’s =  2m  1  . . . .  1  m 2    2 

2

2

7.(C)

 n  1 Hence, total number of ways    2  Out of 9 men two men can be chosen in 9C2 ways. Since no husband and wife are to play in the same game, so, we have to select two women from the remaining 7 women. This can be done in 7C2 ways. If M1, M2 , W1, W2 are chosen, then a team can be constituted in 4

ways. Thus, the number of ways of arranging the game = 9 C2  7C2  4  3024 8.(D)

The number of triangles = Total number of triangles – No. of triangles having one side common with the octagon – No. of triangles having two sides common with the octagon  8C3  8C1  4C1  8 = 16. 2 n 1

Cn 1  2 n 1Cn  2  . . . 

2 n 1

C2 n  2 2 n  1

9.(B)

Use :

10.(D)

Use unit’s digit of 172009 is same as that of the unit’s digit of 72009.

Permutation and Combination 

HWT - 9

1.(A)

n

2.(B)

Required answer = Total no. of ways – No. of ways in which none of the numbers is 2

C2 = 36

n=9

 64  54  1296  625  671 3.(B)

2n  1 n 1 3   n2 2n  1 5 

 2n  1  2n   3  n  2   n  1 n  5



n=4

4.(B)

Consider the host and the two particular persons as a single unit. So, there are effectively 19 persons. These 19 persons can be arranged in 18 ways and there are 2 ways of arranging the two persons on either side of the host.

5.(C)

Digit at the extreme left can be chosen in 9 ways. (0 is not a possibility). Now the next digit can be chosen in 9 ways as consecutive digits are not same and so on. 

Required number of numbers = 9  9  . . . .  9  9n n times

VMC/Solutions/Permutation & Combination

29

HWT-5/Mathematics

Vidyamandir Classes 6.(B)

 1 1 1 1 1 Answer = 5 1       = 44 1 2 3 4 5 

8.(B)

Answer = m

10.(A)



nr

m

C2 . n C2

Since each ring can be worn on any finger, the required no. of ways = 46.

9.(A)

n 1

Cn  n Cn 

Cn 

n2

Answer = 8 C4  8C5  8 C6  8 C7  8C8  70  56  28  8  1  163

7.(C)

nm

Cn  . . . . . 

Cn

r 0



n 1



n2

 

Cn  1  Cn  1 

n 1

n2

Cn 

n2

Cn  . . . . . 

Cn  . . . . . 

nm

nm

Cn [Use : n Cr  n Cr 1 

Cn

n 1

Cr ]

n  m 1

Cn  1

Permutation and Combination

HWT - 10

C2  66  n  12

1.(B)

n

2.(A)

216  23  33 Odd divisors are multiples of 3



9

Number of odd divisors = 4

 1680

Required number of ways =

5.(D)

Required answer = Total no. of numbers – No. of numbers in which 2 is at extreme left 5 4    60  12  48 2 2 7 Case 1 : Five 1's, one 2, one 3 No. of numbers =  42 5

6.(C)

 3

3

Case 2 : Four 1's, three 2's 7.(C)

4.(D)

No. of numbers =

Answer =

9 2

3.(A)

7  35 4 3

Total no. of subsets = 29 = 512 No. of ways when selecting only one even number = 4 C1  4 No. of ways when selecting only two even numbers = 4 C2  6 No. of ways when selecting only three even numbers = 4 C3  4 No. of ways when selecting only four even numbers = 4 C4  1 Required number of ways = 512   4  6  4  1  1  496 (We subtract 1 due to the null set)

8.(A)

We have A, A, I, I, N, N, E, M, O, T, X Case 1 : 2 alike of one kind, 2 alike of second kind No. of words = 3 C2 

4 2 2

 18

Case 2 : 2 alike of one kind, 2 different 4 3 C1  7 C2   756 2 Case 3 : All different 8 C4  4  1680

Cr 

16

Cr  2  r  7

9.(D)

16

10.(A)

Required no. of ways =

22



r

Pr  3  7 P4  840

C19  1540

VMC/Solutions/Permutation & Combination

30

HWT-5/Mathematics