A permutation of r objects taken from n different objects without repetition is an arrangement of the objects in a specific

order

For example, there 12 permutations for the letters A, B, C and D taken 2 at a time.

These are : AB BA CA DA AC BC CB DB AD BD CD DC

Permutation of r objects taken

from n different object

Using the multiplication principle

The number of permutations of 4 objects taken two at a time = 4 x 3 = 12. Similarly, the number of permutations of 10 objects taken 3 at a time = 10 X 9 X 8 = 720.In general, the number of permutations of n objects taken r at a time

)1)...(2)(1( rnnnn

123)...1)((

123)...1)()(1)...(3)(2)(1(

xxrnrn

xxrnrnrnnnnn

)!(

!

rn

n

=

=

=

P r = n! (n - r)!

n

The number of permutations of r objects chosen from a set of n different objects is denoted by

Example

Suppose you have 4 different flags. How many different signals could you make using (i) 2 flags(ii) 2 or 3 flagsSolution

(i) n = 4 r = 2 There are 12 different signals using 2

flags from 4 flags.

nP

r= 4

P2 =

4!

2!=4 x 3 x 2!=(4)(3) =

122!

n = 4 r = 2 or n = 4 r = 3

There are 36 different signals using 2 or 3 flags from 4 flags.

4P

2

4P3

+2!

4!==

=

=

____ 4!

1!+ 4x3x2!

2!

+ 4x3x1!

1!

12 + 24

36

(ii) 2 or 3 flags(ii) 2 or 3 flags

Example How many arrangements of the letters of

the word B E G I N are there if(i) 3 letters are used(ii) all of the letters are used

Solution(i) n = 5 r = 3

The arrangements of the letters of the word BEGIN is 60 if 3 letters are used.

5 P3 = = =5!

2!

(5)(4)(3) 60

ii) n = 5 Number of arrangements if all of the letters are used.

Example

A relay team has 5 members. How many ways can a coach arrange 4 of them to run a 4x100 m race.

The order of the four runners is important.Number of arrangements the coach can make

5P

5= = =5! (5)(4)(3)(2)(1) 120

5P4=

= 120

Solution(i) n= 5 r= 3

There are 60 different arrangements.

 

nP

r= = = = =60

5P

35!

2!

(5)(4)(3)(2)(1)

2!(5)(4)(3)

Example How many three-digit numbers can be made from the integers 2, 3, 4, 5, 6 if (i) each integer is used only once?(ii) there is no restriction on the number of times each integer can be used?

Number of ways of making the three-digit numbers

= 5 x 5 x 5

= 125

(Repetition is allowed)

ii) there is no restriction on the number of times each ii) there is no restriction on the number of times each integer can be used?integer can be used?

Solution:Solution:

Example

Find the number of arrangements of 4 digits taken from the set { 1, 2, 3, 4}. In how many ways can these numbers be arranged so that(a) The numbers begin with digit ‘1’(b) The numbers do not begin with digit ‘1’

SolutionSolution

Number of arrangements of 4 digits = 4! = 24

(a) If the arrangements begin with digit ‘1’, then the number of ways the 3 remaining digits can be arranged

= 3!

= 6

(b) The number of arrangements that do not begin with digit ‘1’

Example

Four sisters and two brothers are arranged in different ways in a straight line for several photographs to be taken. How many different arrangements are possible if

(a) there are no restrictions(b) the two brothers must be separated

= 24 - 6

= 8

Solution

(a) 6! = 720

(b)(b) First,First, find the numbers of arrangements with the find the numbers of arrangements with the two two brothers standing next to each otherbrothers standing next to each other. In these . In these arrangements, the two brothers move together as one unit arrangements, the two brothers move together as one unit and this is equivalent to the arrangement of 5 objects except and this is equivalent to the arrangement of 5 objects except that they are able to switch positions with each other.that they are able to switch positions with each other.

Number of arrangements with two brothers next to each Number of arrangements with two brothers next to each other other

= 5! x 2! = 120 x 2 = 240

Number of arrangements with the two brothers separated Number of arrangements with the two brothers separated == 720 – 240 = 480 720 – 240 = 480

Example

Arrange 6 boys and 3 girls in a straight line so that the girls are separated. In how many ways can this be done?

Solution

Consider this arrangement : o B o B o B o B o B o B oLet the 6 B’s represent the 6 boys and the ‘o’ represent the spaces for the girls.

Number of arrangements for the boys = 6!

Number of arrangements for the girls 7

P3

= (7 spaces available for the 3 girls)

= 210

Total number of arrangements of 6 boys and 3 girls where the girls are separated

= 151200

= 6! x 210

Solution

a) The NOP =

b) The NOP =

EXAMPLE

There are 10 students out of whom six are females. How many possible arrangements are there if

a) they are arranged in a row?

b) males always sit on one side and female on the other side?

= 362880010!

2! x x6! 4! = 34560

Example A witness to a hit-and-run accident told the police that the plat number contained the letters PDW followed by 3 digits, the first of which is 5. If the witness cannot recall the last 2 digits, but is certain that all 3 digits are different, find the minimum number of automobile registrations that the police may have to check.

5

Solution

 

P D W

The NOP = 1 x 9x 8 = 72 ways

Example

In how many ways can 4 girls and 5 boys sit in a row if the boys and girls must sit alternate to each other?

Solution

B   B   B   B   B

The NOP = 5! x 4! = 2880 ways

Example

Four digit numbers are to be formed from the digits 0, 1, 2, 3, 4, 5, 6 without repetition .

How many numbers can be formed if each number

a) is less than 5000

b) begins with digit 4 or 6

c) is between 2000 and 6000

d) is an odd number

Solution 

The NOP =a)

b) or

The NOP = 2 ( 1 x 6 x 5 x 4 ) = 240 ways

c) The NOP = 5 x 5 x 4 x 3 = 300 ways

4 6

4 x 6 x 5 x 4 = 480 ways

EExampleHHow many four-digit even numbers can be formed from the digits, 1, 2, 3, 4, 5, 6, 7 to make up between 2000 and 6000 Aaa) without repetition

Solution:Solution:

consider the last position by two parts : “0” and “not 0”

ends with 0 not ends with 0

or0

The NOP = (4 x 6 x 5 x 1 ) + ( 3 x 6 x 5 x 3)

= 390 ways

The NOP = 4 x 8 x 8 x 4 = 1024 ways.

b) With repetition

Example Three married couples have bought 6 seats in the same row for

a concert In how many different ways can they be seated a) with no restrictions b) If each couple is to sit together c) If all the men sit together to the right of all the women

SolutionSolution

a)

The NOP = 6! = 720

H1 W1H2 W2

H3 W3

b)

The NOP = 3! X 2! X 2! X 2! = 48 ways48 ways

W1 W2 W3 H1H2H3

The NOP = 1 x 3! X 3! = 36 ways.

c) If all the men sit together to the right of all the womenc) If all the men sit together to the right of all the women