## Mathematics | Notes

Permutation Theorem – 1 | Proof and Example

For: Science Class : 12

Permutation is the arrangement of objects with respect to some order. The permutation on n objects taken r at a time is denoted by P(n, r) or nPr.

**Theorem-1**

The total no of permutation of a set of n objects taken r at a time is given by:

P(n, r) = n.(n – 1).(n – 2).(n – 3) ………….. (n – r + 1)

[OR]

P(n, r) = n!/(n – r)! [where : ! is sign for factorial]

**Proof:**

The total number of permutation of a set on n objects taken r at a time is equivalent to filling of those n objects in r positions.

- To fill up the first position there are n choices. So, the first position can be filled up by n ways.
- For the second position (n – 1) objects left there to fill up. So, second position can be filled up by (n – 1) ways.
- Similarly, the third position can be filled up by (n – 2) ways.
- Continuing this process up to r^th position so, the r^th position can be filled up by {n – (r – 1)} ways.
- Hence, the total number of permutation of set on n objects taken r at a time is

n. (n -1).(n – 2)………{n – (r – 1)} ways

I.e. P(n, r) = n.(n -1).(n – 2)………{n – (r – 1)}

or, P(n, r) = [n.(n -1).(n – 2)………{n – (r – 1)}.(n – r).{(n – (r + 1)}……..3, 2, 1] / [(n – r). {n – (r + 1)}….. 3, 2, 1]

or, P(n, r) = n! / (n – r)!

**Note:**

P(n, n) = n! / (n – n)! = n! / 0! = n! / 1 [Because 0! = 1] = n!

i.e. P(n, n) = n!

**Permutation of object not all different**

The number of permutation of a set of n objects taking all at a time when P objects are of 1st kind, Q objects are of 2nd kind, and R objects are of 3rd kind is given by n! / (P! Q! R!)

**For Example:**

Q.1. How many ways the letters of word ‘CALCULUS’ can be arranged.

CALCULUS

Total number of objects = 8

Number of C’s = 2

Number of L’s = 2

Number of U’s = 2

Therefore, total Number of permutation = 8!/ (2! 2! 2!)

= [8 × 7 × 6 × 5 × 4 × 3 × 2 × 1] / [2 × 1 × 2 × 1 × 2 × 1]

= 5040

## One Comment

Quite interesting. I really understood it clearly. Thanks for this. I hope to get all the old questions and all the chapters’ brief and easy explaination.