In this section, the following topics are discussed with examples

**Counting formula for linear permutations****With Repetition****Examples****Circular Permutations****Examples**

### Counting Formula for Linear Permutations

## The permutations and combinations are endless. It is like a game of three dimensional chess.

Sherry Bebitch Jeffe

#### Example 1:

Find the number of ways in which four persons can sit on six chairs.

**Solution: **

^{6}P_{4} = (6 x 5 x 4 x 3 x 2 x 1) /(2 x 1) = 360

## With Repetition

- Number of permutations of n things taken all at a time, if out of n things p are alike of one kind, q are alike of second kind, r are alike of a third kind and the rest n – (p + q + r) are all different is

2. Number of permutations of n different things taken r at a time when each thing may be repeated any number of times is n r.

#### Example 2

Find the number of words that can be formed out of the letters of the word COMMITTEE taken all at a time.

**Solution:**

There are 9 letters in the given word in which two T’s, two M’s and two E’s are identical. Hence the required number of

## Circular Permutations

#### Arrangement Around a Circular Table

In circular arrangements, there is no concept of starting point (i.e. starting point is not defined). Hence number of circular permutations of n different things taken all at a time is

(n – 1)! if clockwise and anti-clockwise order are taken as different.

In the case of four persons A, B, C and D sitting around a circular table, then the two arrangements ABCD (in clock- wise direction) and ADCB (the same order but in anti- clockwise direction) are different.

Hence the number of arrangements (or ways) in which four different persons can sit around a circular table = (4 – 1)! = 3! = 6.

#### Number of Circular Permutations of n Different Things Taken r at a Time

**Case I:** If clockwise and anti-clockwise orders are taken as different, then the required number of circular permutations

**Case II:** If clockwise and anti-clockwise orders are taken as same, then the required number of circular permutations = ^{n}P_{r / 2r}

#### Example 3

In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together ?

**Solution: **

Leaving one seat vacant between two boys, 5 boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls can sit in 5! ways. Hence the required number

= 4! × 5! = 2880 ways.