# Permutations and Combinations exercise solution.pdf

November 25, 2017 | Author: Chai Usajai Usajai | Category: Vertex (Geometry), Triangle, Permutation, Polygon, Odds

#### Description

Permutation

1.

Numbers that can be formed = 4P3 × 5P2 = 4 × 3 × 2 × 1 × 5 ×4 = 480

2.

If the thousandth digit is

3, 4 or 5, the number of 4 – digit numbers

If the thousandth digit is

2, the number of 4 – digit numbers greater than

= 3 × 4P2

(the hundredth place is 3, 4 or 5) +

4P2

3×5P3

=

=

3 × 5 ×4 × 3 = 180

2100 can be formed

(the hundredth place is 1) = 48

∴ Total number of required 4 – digit number = 180 + 48 = 228

3.

(i)

The number of

(ii)

There are

5

9 – digit

numbers that can be formed

odd numbers and

Therefore there are

4/9

4.

Assume that the Together with

3 5

even numbers in

of the numbers in

Of all such even numbers there are The possible numbers

4

1/2

set

=

362880

1, 2, …, 9

are even. 4.

9! × 4/9 × 1/2 = 80640

=

Chinese books, there are 6

9!

of them are divisible by

English books as if it were

Total permutation of

(i)

=

1

set.

6 sets.

= 6!

But the English books may permutate among themselves, number of permutations Therefore total number of arrangements

5.

= 6! × 3!

= 5!

= 120

Since each digit place from unit place to ten-thousandth place, each number

1, 2, 3, 4, 5

hence to become the place value of the 5–digit number, the mean place value is 120

numbers is

The sum = 120 × 33333

= 3999960

6.

Total permutation of

33333 × 55

8

books

may have the same c

(1+2+3+4+5)/5

=3

33333.

If repetition is allowed, the number of 5–digit numbers formed The required sum =

3!

= 4320

If repetitions are not allowed, number of 5 – digit numbers formed

Therefore the mean of the

=

= 55

= 3125

= 3 × 55 × 11111 = 104165625

= 8!

If two particular must be placed together, the number of arrangements

=

2 × 7!

Therefore if two books must be separated, the number of arrangements = 8! – 2 × 7! = 30240

1

7.

The circular permutation for

5

gentlemen

= (5 – 1)! = 4!

After the gentlemen has taken their seats, the ladies must sit between two of the gentlemen, the permutation of

5

=

= 4! × 5!

Therefore the total arrangements

8.

(a)

If the order of

a, o, i

5! = 2880

can be changed the number of ways

If the order cannot be changed, the number of ways

9.

= 9! / 3!

= 60480

(b)

The number of ways in which the order of consonants cannot be changed

(c)

The number of possible ways

Consider the L.H.S.

=

9! / (3!6!)

of the central mark.

=

Total number of arrangements in The counters of

R.H.S.

L.H.S.

There must be

=

m white counters and =

n

red counters in order to

m + n.

(m + n)! / (m!n!)

L.H.S., there is only

1

way of setting the

=

We may consider all the letters

are the same and all the letters

x

x

and

y

R.H.S.

( m + n )!

therefore the total number of ways

2n

= 504

must be symmetric about the central mark.

Given any arrangement of

After taken such

= 9! / 6!

84

have symmetry and the total number of counters on L.H.S.

10.

= 9! = 962880

m! n!

y

are the same.

letters out to form an arrangement, we then number the

x

and

y

letters in ascending orders, Since there are 11.

(i)

2n

letters, n

x-letters and

n

y-letters, the total number of arrangements

=

( 2 n )! n! n!

Place a nought on the left-most end. For the rest, either we place a nought or we place a unit consisting of a cross followed by a nought (x, o). As there are

q

number of such units, the number of remaining noughts is

Total number of arrangements =

(ii)

[( p − q − 1) + q ]! ( p − q − 1)! q!

Give each job a nought and each man a cross. If a cross is followed by be a cross.

k

=

( p − 1)! ( p − q − 1)! q!

(p – q – 1).

.

The question is reduced to part (i) as in the followings:

noughts, it means that the man gets the

k

jobs.

The leftmost end must

After placing this cross (1 man used), every thing is the same as in (i) and we have the same

number of arrangements. i.e.

( p − 1)! ( p − q − 1)! q!

.

2

Combination 1.

(a)

Number of straight lines that can be formed from n Since

p

points

points are in a straight line, therefore only

1

Number of triangles that can be formed from n Since

p

.

= =

n

n

pC2

lines.

C2 − p C2 + 1 .

C3 .

points are in a straight, they can’t form any triangle.

Therefore

p C3

triangles formed from p

Therefore total number of triangles

2.

points

n C2

line can be formed instead of

Therefore total number of straight lines that can be formed (b)

=

=

n

points must be neglected.

C3 − p C3 .

Number of triangles that can be formed by joining any three vertices of a polygon with

n

sides

= n C3 .

By taking any one special vertex on the polygon, the no. of triangles with common sides with the polygon = n – 2 . However, there are two triangles with two sides in common with the sides of the polygon. duplicated when we consider other vertices. duplication

These two triangles are

Therefore number of triangles formed from one special vertex without

= n – 2 – 1 = n – 3.

Total number of triangles with sides common to the polygon

= n(n – 3).

∴ Number of triangles that can be formed with no sides in common with the polygon = 3.

n C3

– n(n – 3) =

1 6

By choosing any two lines from the form a parallelogram,

4.

n ( n − 4)( n − 5) . 10

( 10C2 )

and by choosing any two lines

Therefore the total number of parallelograms

The number of ways of choosing the committee

=

When

Mr. Y is a member, number of combination

When

Miss X

When both

8 C3

Mr. Y and Miss X

× 7 C3

=

is a member, number of combination

7 C3

=

=

=

56 × 35

× 6 C3

7 C2

10C2

× 6 C4

=

=

Number of ways in a committee can be formed

700 + 315 + 525 =

6 C2

=

For any two circles, there are

2

1960 =

21 × 15 =

7 C3

700 =

× 6 C4

315 =

35 × 15

=

525

= 1540

× 7 C3 × 3 C1

=

945

Number of ways so that a particular Frenchman belong to the committee = 6.

( 7C2 ) , we can

× 7C2 = 135

35 × 20

=

are not members, number of combination

Therefore total number of combinations 5.

parallel lines

5 C1

× 6 C1 × 3 C1

=

90

greatest number of intersection points.

Therefore the greatest number of points of intersection between circles themselves

= 2 × nC2 = n(n – 1)

The greatest number of points of intersection between

2.

1

line and

Therefore the greatest number of points of intersection between

1

circle is

m lines and

n

circles

= 2mn 3

The number of points of intersection between

2

lines is

1.

Therefore the greatest number of points of intersection between Total number of intersections

=

1 2

m lines = mC2 = m(m – 1)/2

m( m − 1) + n ( 2 m + n − 1) .

Miscellaneous 1.

Choose

0

thing from those alike,

Choose

1

thing from those alike,

Choose

2

things from those alike,

::

::

Choose ∴ 2.

n

::

Total number of selections

=

B

::

things from those different, number of ways

n Cn

+ nCn-1 + nCn-2 + … + nC0

B.

are not in the selection, the number of ways

Total number of selections = n-2Cr-1 + n-2Cr-1 + n-2Cr =

The rest must be placed in

cards,

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

3

=

cards to place in

The number of ways of labeling

10C5.

B , the number of ways

batches of

p

=

5 C3 .

=

9! 4!3!2!

= 2520 .

= 1260

is an integer because it is the number of ways of dividing up

( p!)n n!

.

C.

Total number of selections = 10C5 × 5C3

( np)!

= n-2Cr.

postcards to place in A , the number of ways 5

= n C0 .

2n .

=

= n-1Cr-1.

5

= nCn-2.

::

is in the selection, A is not in the selection, the number of ways

Then we choose from the remaining

6.

::

= n-1Cr-1.

First we choose

::

is not in the selection, the number of ways

B

B

= nCn-1.

things from those different, number of ways

:: 0

= n Cn .

things from those different, number of ways

(n – 2)

things from those alike,

If A and

5.

(n – 1)

Let the two particular objects be A and

If

4.

things from those different, number of ways

::

If A is in the selection,

3.

n

np

different objects into unordered

elements each.

(2n + 1)(2n + 3)(2n + 5)…(4n – 3)(4n – 1) n

=

Π

( 2 n + 2i − 1) =

i =1

n

Π

( 2 n + 2i − 1)( 2i )

i =1

2i

=

n

1

n

n

2( 2 n + 2i − 1)

i =1

i

Π2Π Π i =1

i

i =1

n

1 = ⎛⎜ ⎞⎟ n! ⎝2⎠

n

2i( n + i ) 2 ( 2 n + 2i − 1)

i =1

i 2 ( n + i) 2

Π

n

n

1 = ⎛⎜ ⎞⎟ n! ⎝2⎠

Π i( n + i)(2n + 2i)(2n + 2i − 1) i =1

n

n

i =1

i =1

Π i( n + i ) Π i( n + i )

n

1 n!( 4 n )! = ⎛⎜ ⎞⎟ . ⎝ 2 ⎠ ( 2 n )!( 2 n!)

4