## Permutation I

Subject: Mathematics

#### Overview

The note gives a knowledge on permutation. It teaches you how to denote permutation in symbol, finding the permutation using factorial notation, different cases of permutation and solving some problems.
##### Permutation I

APermutationis a process of arranging all the members of a set into some sequence or order, or if the set is already ordered, rearranging (reordering) its elements. Each arrangement in this process is also called permutation.Let us say that there are 3 alphabets A, B, C then the arrangement of these three alphabets such that 2 are taken at a time are AB, AC, BA, BC, CA, CB. Each one of these arrangements or order is a permutation.In permutation, the combination AB and BA are different like two different digits 23 and 32.

Denoting permutation :The number of permutations of 'n' different elements taken 'r'' at a time is denoted by the symbolnPr ornPr or P(n,r).

Theorem:The total number of permutations of n different elements takingrat a time when (n>r) is

nPr=n(n-1) (n-2) . . . . . . (n -r) (n-r+1).

Proof:The number of permutations of n different elements taken r at a time is the same as the number of ways of filling up r position by r elements taken from n elements (objects). The first position can be filled up by any of n objects, and this can be done in n ways.After filling up the first place, the second place can be filled up by any of the remaining (n-1) elements, and this can be done in (n-1) ways.

Likewise, the third position can be filled up by (n-2) ways, and so on. The rth can be filled by (n-(r-1)) i.e n-r+1 ways. Therefore, by the fundamental principle of counting, the number ways or filling up the r positions is n(n-1)(n-2). . . . (n-r+1).

Therefore,

nPr=n(n-1)(n-2) . . . . . . . . . . . . (n-r+1) . . . . . . . . . .(1)

Case 1:Let us putr=n, then from the above equation (1),

nPn= n (n - 1) (n - 2) (n - 3). . . . . . . . . .(n - n + 2) (n - n + 1)

= n (n - 1) (n - 2) . . . . . . . . 3.2.1 = n! . . . . . . . . . . . . . (2)

We can use factorial notation to denote permutation, the expressions fornPr may be defined as

nPr= n( n - 1) (n - 2) (n - 3) . . . . . . . . . . . . . . . . (n - r + 1)

$$\;=\frac{ n (n - 1) (n - 2) (n - 3). . . . . . . (n - r + 1) (n - r) (n - r -1) . . . . . . . .3.2. 1} {(n - r) (n -r - 1) . . . . . . . 3.2.1}$$

$$\;=\frac{n!}{(n-r)!} \text{ . . . . . . .. . . . .(3)}$$

let us suppose again thatr = n - 1,

$$\;P(n,n-1)\;=\;\frac{n!}{(n-n-1)!}$$

$$\;=\frac{n!}{1!}\;\;\;\;=\;n!\;\text{. . . . . . . . .(4)}$$

So from the eqn 1, 2, 3 and 4 we can see that greatest value of premutation is whenr = n or n-1.

Example: Find the value of P(5,2).

Solution :

Here, n = 5 and r = 2.

$$\therefore\;P(5,2)\;=\;\frac{5!}{(5-2)!}\;=\;\frac{5!}{3!}$$

$$\;=\;20$$

$$\therefore\;P(5,2)\;=\;20\;$$

### Different cases of permutation

Permutation of objects when all are taken at a time but,

• Out of nobjects, m objects are of one type and (n-m) are of another kind. Then the permutation of the objects is $$\frac{n!}{m! (n-m)! }$$
• Out of the n objects taken all at a time, a of them are of one type ,b of them are of other and there is still others left i.e. (n – a – b ).Then the permutation of the objects taken all at one once is $$\frac{n!}{a! b! (n – a – b )! }$$
• Out of the n objects taken all at a time , when a objects are of one type and rest of them (b) are all different. Then the permutation of taking all of them at once is given by $$\frac{n!}{a! b! }$$
• Out of the n objects taken all at a time, when p are of one type q are of another type and r are remaining but all different. Then the permutation of taking all of them at once is given by $$\frac{n!}{p! q! r!}$$

Example:find the number of arrangements that can be made out of the letter of the word "SPONSORED".

Solution:

In the word "SPONSORED" the letter S is repeated two times, the letter O is also repeated two times and remaining all are different. The total number of words here is 9. So, p=2, q=2, r=5 and n=9.

The total number of arrangements (permutation)$$\;=\frac{n!}{p! q! r!}\;=\;\frac{9!} {2! 2! 5!}\;=\;756$$

• The permutation of n different objects, taking r at a time, when repetition is allowed , is nr.

Example:There are 10 letter boxes in a post office. in how many ways can a man post 5 distinct letter?

Solution:

each letter can be posted in 10 ways. Hence the number of ways in which he can send the 5 letters (permutation) = 105 = 100000

Alternative method: Hereevery letter can be posted in 10 ways. There are 5 letters. By using the principle of counting, the number of ways the letters can be posted is 10 x 10 x 10 x 10 x 10 = 100000

Example:IfnP5 = 20 xnP3 then find the value of n .

solution:

here,nP5 =20 xnP3

$$\;or,\;\frac{n!}{(n-5)!}\;=\;20\;x\;\frac{n!}{(n-3)!}$$

or, (n - 3 )! = 20( n - 5)!

or, ( n - 5) ! (n - 4 )(n - 3 ) = 20 (n - 5)!

or, (n - 4) ( n - 3) = 20

or, n2 - 7n + 12 - 20 = 0

or,n2 - 7n - 8 = 0

or, n(n - 8) + 1(n - 8) = 0

or, (n - 8) (n + 1) = 0

Either,n = 8 or n = -1

But the numbers used in permutation must be natural.

$$\therefore\;n\;=\;8$$

Taken reference from

( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )

##### Things to remember
• The number of permutations of 'n' different elements taken 'r'' at a time is denoted by the symbolnPr ornPr or P(n,r).
• $$\;P(n,r)\;=\frac{n!}{(n-r)!}$$
• Out of nobjects, m objects are of one type and (n-m) are of another kind. Then the permutation of the objects is $$\frac{n!}{m! (n-m)! }$$
• Out of the n objects taken all at a time, a of them are of one type ,b of them are of other and there is still others left i.e. (n – a – b ).Then the permutation of the objects taken all at one once is $$\frac{n!}{a! b! (n – a – b )! }$$
• Out of the n objects taken all at a time , when a objects are of one type and rest of them (b) are all different. Then the permutation of taking all of them at once is given by $$\frac{n!}{a! b! }$$
• Out of the n objects taken all at a time, when p are of one type q are of another type and r are remaining but all different. Then the permutation of taking all of them at once is given by $$\frac{n!}{p! q! r!}$$
• The permutation of n different objects, taking r at a time, when repetition is allowed , is nr.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.