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

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

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

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Permutation of r objects taken

from n different object

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

Using the multiplication principle

The number of permutations of4objects taken two at a time = 4 x3=12. Similarly, the number of permutations of10 objects taken 3 at a time =10X 9 X 8=720.

In general, the number of permutations of n objects taken r at a time

=

=

=

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

n

n!

P

=

r

(n - r)!

Example

Suppose you have 4 different flags. How many different signals could you make using

(i) 2 flags

(ii) 2 or 3 flags

4

P

2

Solution

(i) n = 4r = 2

There are 12 different signals using 2

flags from 4 flags.

4!

n

4 x 3 x 2!

=

(4)(3) = 12

P

=

=

=

r

2!

2!

(ii) 2 or 3 flags

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

4!

4

4!

4

4x3x1!

____

4x3x2!

2!

+

+

=

+

P

=

P

3

2

1!

2!

1!

12 + 24

=

=

36

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

- Example
- How many arrangements of the letters of the word B E G I N are there if
- (i) 3 letters are used
- all of the letters are used

Solution

(i) n = 5r = 3

5!

5

P

=

=

60

=

(5)(4)(3)

3

2!

The arrangements of the letters of the word BEGIN is 60 if

3 letters are used.

ii) n = 5

Number of arrangements if all of the letters are used.

5

P

=

5!

=

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

=

120

5

Example

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

5

The order of the four runners is important.

Number of arrangements the coach can make

P

4

=

=

120

5

P

3

permutation with conditions

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

Solution

(i) n= 5 r= 3

n

5!

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

P

=

=

=

=

(5)(4)(3)

=60

r

2!

2!

There are 60 different arrangements.

ii) there is no restriction on the number of times each

integer can be used?

Solution:

- Number of ways of making the three-digit numbers

=

5

x

5

x

5

(Repetition is allowed)

=

125

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’

Solution

Number of arrangements of 4 digits = 4! = 24

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

=

3!

=

6

- The number of arrangements that do not begin
- with digit ‘1’

=

24 - 6

=

8

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

Solution

(a) 6! = 720

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

= 120 x 2

= 240

= 5! x 2!

Number of arrangements with the two brothers separated

= 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 oB o B o B oB o

Let 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

=

(7 spaces available for the 3 girls)

P

3

= 210

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

= 6! x 210

= 151200

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?

Solution

a) The NOP =

10!

= 3628800

b) The NOP =

2!

x

6!

x

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.

Solution

5

P D W

The NOP =

1 x 9x 8

= 72 ways

B

B

B

B

B

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

The NOP

= 5! x 4!

= 2880 ways

Solution

a)

The NOP =

4 x 6 x 5 x 4

= 480 ways

or

b)

4

6

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

= 240 ways

c)

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

EExample

HHow 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

A

aa) without repetition

Solution:

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

ends with 0

not ends with 0

0

or

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

= 390 ways

b) With repetition

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

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

Solution

a)

The NOP = 6! = 720

b)

H2 W2

H3 W3

H1 W1

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

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

W1 W2 W3 H1H2H3

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