# Permutations And Combinations Class 11 Notes Mathematics Chapter 7 - CBSE

## What Are Permutations And Combinations ?

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## Introduction

The basic idea of permutation and combinations are based on two essential ideas of counting :

Fundamental Principle of Multiplication

According to this principle, if an event can occur in m different ways and another event can occur in n different ways, then the total number of occurrence of the events in the given occur is (m × n).

For example: In a class there are 12 boys and 10 girls. Teacher wants to select a boy and a girl for a function. So, by principle of multiplication, the teacher can select a boy and a girl in 12 × 10 = 120 ways.

If there are two events performing independently in 'a' and 'b' ways respectively, then either of the two events can occur in (a + b) ways.

For example: In a class, there are 16 boys and 10 girls. The teacher wants to select either a boy or a girl as a representative. Then, by fundamental principle of addition teacher can make the selection in (16 + 10) = 26 ways.

## Factorial Notation

In earlier classes, we come across the products of number in the form 1 × 2, 2 × 1 etc.

For making it convenients, we use a notation called factorial i.e.,

1! = 1

2! = 2 × 1

3! = 3 × 2 × 1

i.e., n!, represents the product of first 'n' natural numbers.

Thus, we have

n! = n(n – 1) × (n – 2) ........... 3 × 2 × 1

n! = n × (n – 1)!

0! = 1

## Permutation

The all possible arrangements which can be make out of a given number of things by taking some or all at a time, are called permutations.

In permutation, the order of arrangement of
objects is taken into considertion. When the order is changed, a different permutation is obtained. The number of permutation of n different objects taken r at a time when, 0 < r ≤ n is given by :

$$^{n}\text{P}_{r}=\frac{n!}{(n-r)!}$$

Example: The permutations the letters P, Q, R taking all at a time are: PQR, PRQ, QPR, QRP, RPQ, RQP. Thus, the 3 letters P, Q, R taking at all a time gives 6 permutations.

## Permutation When All Objects Are Distinct

Theorem 1

The number of permutation of n different objects taken 'r' at a time, where 0 < r ≤ n and the object do not repeat is n(n – 1)(n – 2) ...... (n – r + 1), which is denoted by nPr or P(n, r)

$$\text{i.e.\space}\text{P(n,r)} = ^{n}\text{P}_{r} =\frac{n!}{(n-r)!},\\0 \leq r \leq n$$

Proof: We know that

P(n, r) = nPr = n(n – 1)(n – 2)(n – 3) ..... [n – (r– 1)]

On multiplying numerator and denominator by

(n – r)(n – r – 1) .... 3 × 2 × 1,

$$^{n}\text{P}_{r}=\\\frac{n(n-1)(n-2).....[n-(r-1)]×(n-r)[n -(r+1)].....3.2.1}{(n-r)\lbrack n-(r+1)\rbrack......3.2.1}\\$$

$$=\\\frac{\lbrack n(n-1)(n-2)\rbrack......3×2×1}{\lbrack(n-r)(n-r-1)\rbrack.....3×2×1}\\=\frac{n!}{(n-r)!}\\\text{Hence Proved.}\\\text{When r = 0, then}\\\text{P}_{0}=\frac{n!}{(n-0)!}=\frac{n!}{n!}=1$$

When r = n, then

$$^{n}\text{P}_{n}=\frac{n!}{(n-n)!}=\frac{n!}{0!} = n!$$

## Permutation Of Objects Not All Different

Theorem 2

Let there are n objects, of which 'm' objects are of one kind and the remaining (n−m) objects are of another kind. Then, total number of mutually distinguishable permutations that can be formed from these objects is :

$$\frac{n!}{m!(m-n)!}$$

Theorem 3

The number of permutations of n objects where P1 objects are of one kind, P2 are of second kind and the rest, if any, are of different kind is :

$$\frac{n!}{\text{P}_{1}!\text{P}_{2}!\text{P}_{k}!}$$

## Permutation With Repetitions

Theorem

The number of permutations of n different objects, taken 'r' at a time when each may be repeated any number of times in each arrangement, is nr.

Combination

It is an unordered collection of some or all of the objects in a set. Each of different groups or selections that can be made out of a given number of things by taking some or all of them at a time, irrespective of their arrangement.

The number of combinations of 'r' objects that can be selected from 'n' different objects where [0 < r ≤ n'] is given by :

$$^{n}\text{C}_{r}=\frac{n!}{r!(n-r)!}$$

and  nCa  =  nCb

Here, n = a + b

Example: All combinations of four persons A, B, C, D taken two at a time are

$$^{4}\text{C}_{2} =\frac{\angle 4}{\angle 2 \angle 4-2} =\\\frac{4×3×\angle 2}{\angle 2 × \angle 2} = 6$$

i.e., AB, AC, AD, BC, BD, CD

## Difference Between Permutation And Combination

Combination

Only a group is made and the order in which they are kept is of no use.

Permutation

Not only group is formed but also an arrangement in definite order is considered.

## Important Properties

Property 1

nPr = nCr × r! ⟹ 0 < r < n

Proof : We have

$$^{n}\text{P}_{r} = \frac{n!}{(n-r)!} \\=\frac{n!×r!}{(n-r)!× r!}\\=\begin{bmatrix}\frac{n!}{(n-r)!× r!}\end{bmatrix}× r!\\=\space^{n}\text{C}_{r}× r!\\\text{Hence Proved.}$$

Property 2

nCr = nCn – r , 0 ≤ r ≤ n

Proof : We have

$$^{n}\text{C}_{n-r}=\frac{n!}{(n-r)!\lbrack n - (n-r)\rbrack!}\\\frac{n!}{(n-r)!r!} =^{n}\text{C}_{r}\\\text{Hence Proved.}$$

Property 3

nCr + nCr – 1 = n + 1Cr

Proof : LHS = nCr + nCr – 1

$$^{n}\text{P}_{r} =\frac{n!}{(n-r)!r!} +\\\frac{n!}{(r-1)!(n-r+1)!}\\=\frac{n!(n-r+1)}{r!(n-r+1)(n-r)!} +\\ \frac{n!r}{r(r-1)!(n-r+1)}\\=\frac{n! (n-r+1)}{r!(n-r+1)}+\\\frac{n!r}{r!(n-r+1)}\\=n!\bigg[\frac{n-r+1+r}{r!(n-r+1)!}\bigg]$$

$$= \frac{(n+1)n!}{r!(n-r + 1)!}\\=\frac{(n+1)!}{r!(n+1-r)!} = ^{n+1}\text{C}_{r}$$

Property 4

If nCx = nCy, and x ∈ y, then prove that x + y = n

Proof : We have

nCx = nCy = nCn–y [∴ nCr = nCn–r ]

⟹ x = y or x = n –y

⟹ x = n – y

⟹ x + y = n Hence Proved.