# Permutation and Combination – Definition, Formulae, and Examples

Permutation and Combination are used for lists (order necessary) or groups( order does not matter).

A permutation is used for lists and Combination is used for groups.

## Permutation

An arrangement of a finite number of objects taken some or all at a time is called a permutation. For example the permutation of n different objects taking r (where r ≤ n) at a time is denoted by ^{n}P_{r }or P(n,r)

^{n}P_{r} = n!/(n-r)!

**Theorem:**

^{n}P_{r }= n (n-1)(n-2)…(n-r+1) = n!/(n-r)!

**Proof:**

Consider we have r places to fill n different objects. First place can be filled-in n different ways. When first place has been filled we are left (n-1) objects. Now second place can be filled in(n-1) different ways. First two places can be filled in n(n-1) ways.

After filling the second place we are left (n-2) different objects. Now third place can be filled in (n-3) different ways. First three places can be filled in n(n-1)(n-2) ways. Proceeding in this way by using fundamental law of counting, total number of arrangements in which r place can be filled up is given by

^{n}P_{r }= n(n-1)(n-2)(n-3)…r factors = n(n-1)(n-1)(n-3)…[n-(r-1)]

^{n}P_{r }= n(n-1)(n-2)(n-3)…(n-r+1)

Multiply and divide R.H.S by (n-r)(n-r-1)…3.2.1

^{n}P_{r }= n(n-1)(n-2)(n-3) x (n-r)(n-r-1)…3.2.1/ (n-r)(n-r-1)…3.2.1

^{n}P_{r }= n(n-1)…(n-r+1)(n-r)(n-r-1)…3.2.1/(n-r)(n-r-1)…3.2.1

^{n}**P _{r } = n!/(n-r)!**

**Examples:**

^{16}P_{4} = 16!/(16-4)! = 16!/12! = 16.15.14.13.12!/12! = 16.15.14.13 = 43,680

^{14}P_{3} = 14!/(14-3)! = 14!/11!

= 14.13.12.11!/11! = 14.13.12

= 2,184

## Permutation with related objects

### Theorem:

The number of permutations of n objects taken all at a time when one kind is q alike, the second kind is r alike, and third is s alike is given by

**Proof:**

Arrangements of q like objects = ^{q}P_{q }= q!

Arrangement of s like objects = ^{s}P_{s} = s!

Let x be the required number of permutations of n objects

Then total permutations = x. q!r!s!

But a number of permutations of n objects = n!

px. q!r!s! = n! => x = n!/q!r!s!

or

**Example:**

ASSESSEMENT

S is repeated 4 times

E is repeated 3 times

A,M,N and T come only 1 time.

**Combination**

When the selection of objects is done neglecting its order, this is called combination.

Combinations of n different objects taken r at a time is denoted by ^{n}C_{r} and defines as:

^{n}C_{r }= ^{n}P_{r}/r! or ^{n}C_{r }= n!/r!(n-r)! ^{n}P_{r }= n!/(n-r)!

**For r = n**^{ n}C_{n} = n!/n!(n-n)! = n!/n!0! = 1

**For r = 0 ^{n}**C

_{0 }= n!/0!(n-0)! = n!/0!n! = 1

**Prove that **^{n}C_{r }= n!/r!(n-r)!

^{n}C

_{r }= n!/r!(n-r)!

**Proof**

There are ^{n}C_{r} combinations of n different objects taken r at a time.

Each combination consists of r different objects which can be permuted among themselves in r! permutation. Total number of permutation of n objects taken r at a time

^{n}C_{r }x r! = ^{n}P_{r }=> ^{n}C_{r }= ^{n}P_{r}/r! or ^{n}C_{r }= n!/r!(n-r)!

## Difference between permutation and combination

- An arrangement where the order is important is called a permutation.
- An arrangement where the order is not important is called combination.

**Examples:**

^{10}C_{3 }= 10!/3!(10-3)! = 10!/3!7!

= 10.9.8.7!/3.2.1.7! = 120

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