Permutations And Combinations Class 11 Notes Mathematics Chapter 7 - CBSE
Chapter : 7
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.
Fundamental Principle of Addition
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.
