# Permutation: Definition, Formulas and Example

In this post we will try to understand the basics of Permutation and the concepts which are relevant for Class 11 NCERT Math. All the concepts and formulas are explained with the help of examples and solved questions for your easy understanding.

## What is Permutation?

Permutation basically tell us number of possible arrangement of numbers or letters in a given combination.

For example,
Suppose you want to find number of possible three letter word derived from the word “GRAPH”

When we try to find the words manually, it would be time consuming.
Some of the possible words are:
GRA, GRP, GRH, GAR, ARG. . . . . . etc.

Thus finding individual words is very difficult and also increase chance of error. Instead of doing this manually, we will take help of the concept of Permutations.

Number of possible 3 letter word from the word “GRAPH” is given as:
P(n, r) = P(5, 3) = \frac{5!}{( 5-3) !} =60

Where
n = number of letter in main word “GRAPH”
r = number of letter in target word

Hence there are 60 possible combinations

### Explanation of Permutation Concept

Let me tell you how the formula work using following graphical representation

So when we use the above mentioned formula of permutation we are basically:
(a) Selecting the letters from the main word
(b) Then recombining all the letter to cover all the word combination
(c) The total number of possible combinations will be expressed by the permutation formula

## Symbol and formula of Permutation

Permutation can be expressed as

Where
n = number of given objects
p = number of required objects

Permutation formula 01
Formula for P (n, r) = \frac{n!}{( n-r) !}

Where ! represents factorial of any given number.

## Formula for Rearrangement of words

Suppose you are provided with letter “BLOCK” and you have to find the number of possible combination after rearranging the letters.

If we try to find the solution manually we have to perform following manipulation
BLOCK, BLOKC, BLKCO, BLKOC, BLCOK, BLCKO
LOCKB, LOCBK, LOBKC, LOBCK, LCOKB, LCBKO

– , – , – , – , -, –
and the process goes on and on

This method is complex and time taking, hence not recommended for your exams

Permutation Formula 02
The solution of above question is provided by formula = r!
Where r = number of element in the object

For this questions the answer is = 5!

Example 01
Find the number of Permutation of the word “DELTA”

Solution
Number of object in the word = 5
Number of possible permutation = 5! = 120
Hence the number of words formed after rearranging the letters = 120

Example 02
Find the number of Permutation of the word “BLUE”

Solution
Number of object in word = 4
Number of possible permutation = 4!

## Formula when similar objects are present

Suppose you need to know number of possible words made from “APPLE”
You can observe that there are two “P” in the apple which if used with permutation formula can cause duplicity

Hence same elements cause duplicity.
To remove that we have to modify our formula.

Modified Formula for this case
P(APPLE) = \frac{5!}{2!}

In the above formula, we have divided the equation with 2! because P1 and P2 in APPLE are the same element and cause duplicity in calculation.
Since P1 & P2 can be arranged in 2! ways, so we divided the main equation with 2!

Formula Number 03
The number of permutation of n objects when p objects are of same kind is given as
Permutation = \frac{n!}{p!}

Example 01
Find the number of possible permutation of word “FOOTBALL”

Solution
There are total of 8 objects in the word ” FOOTBALL”, out of which there are 2 O’s and 2 L’s which are common element and cause duplicity in calculation.

P (FOOTBALL) = \frac{9!}{2!\times 2!}

Example 02
In how many ways can 4 red, 3 yellow and 2 green disc can be arranged in a row?

Solution
Total Number of elements = 4 + 3 + 2 = 9
Out of these elements 4 Red, 3 Yellow and 2 Green are common element which can cause duplicity in calculation.

The number of arrangement is = \frac{9!}{4!\times 3!\times 2!}