Permutation and Combination Notes for IIT JEE, Download PDF!
View our expert-backed Permutation and Combination PDF for IIT JEE Notes to clear all your doubts.
Permutation and Combination is one the most frequently asked questions in JEE Main/JEE Advanced. Every year it is certain to find 2-3 questions from the topic. Revise the short notes on Permutation and Combination and improve your scores. Download the P&C short notes PDF from the link given below.
1.The fundamental principle of counting
1.1 Product Rule
If a task A can be completed in m ways and another task B can be completed in n ways, then both the tasks can be completed in (m*n) ways
1.2 Sum Rule
If task A can be completed in m ways and another task B can be completed in n ways, then either of the tasks can be completed in ( m X n ) ways
Example: Four-digit numbers are to be formed using the digits 0,1,2,3,4,5. Find the number of such numbers if
(a) Repetition is not allowed
(b) Repetition is allowed
(c) At least one digit is repeated
Answer: Obviously 0 will not be at the left because then it won’t be a 4 digit number
(a) Since repetition is not allowed, so the number of numbers that can be made=
(b) Since repetition is allowed, so the number of numbers that can be made=
(c) Numbers with at least one digit repeated plus the numbers with no repeated digits will give us the total number of numbers that can be made using these 6 digits.
So, the number of numbers with at least one digit repeated +number of numbers with no repetition=Total number of numbers
Or the number of numbers with at least one digit repeated=Total number of numbers-Number of numbers with no repetition=1080-300=780
2. Permutation and Arrangement
The meaning of the word permutation is the arrangement. While arranging objects the order of their appearance becomes paramount.
2.1 Line permutation
If there are ‘n’ different objects, then all or few of them can be arranged in a line with or without repetition as follows
Taking r (0≤r≤n) objects, with repetition, it can be done in nr ways
Taking r (0≤r≤n) objects, without repetition, it can be done in P (n,r) ways
2.2 The arrangement of ‘n’ different objects
(a) With repetition, n different objects can be arranged in a line in nn ways
(b) Without repetition, n different objects can be arranged in a line in n! ways
2.3 The arrangement of 'n' similar objects
n similar objects can be arranged in P(n,r) / n! ways. If p objects are alike, q objects are alike, r objects are alike, then p + q + r = n, then they can be arranged in a line in ways
2.4 Circular Permutation
(a) If there are n different objects, then all or few of them can be arranged around a loop as follows.
Taking r (0≤r≤n) objects, without repetition, it can be done in P(n,r) / r ways; where clockwise and anticlockwise orders are treated as different arrangements
Taking r (0≤r≤n) objects, without repetition, it can be done in ways; where clockwise and anticlockwise orders are treated as same arrangements.
If positions are marked, then the circular arrangement should be treated as a line arrangement.
Example: If there 2 girls and 1 boy sitting in a roundtable then in how many ways can they sit?
Answer: Let the girls be A and B and the boy be C. AB can sit together. C can sit on the side of A which is also the side of B. So this is 1 way to sit.
If A and B sit separately, that means C sits in the middle of A and B. This is another way to sit.
So there are only 2 ways to sit.
From formula n=3 r=3.
No of ways=(3! / (3-3) !) / 3 = (3! /0! ) / 3 =6 / 3 =2
So, this verifies our analytical method
3. Combination and Selection
The meaning of the word ‘combination’ is the selection. While selecting objects, the order of appearance is not considered.If there are n different objects, then all or few of them can be selected with or without repetition as follows.
3.1 Selection of n distinct of n distinct objects:
(a) objects, with repetition, can be selected in n+r-1 Cr ways.
(b) objects, without repetition, can be selected in nCr ways.
Where nCr = (n! / (r! (n-r)!)
3.2 Selection of objects where n objects are of 1 kind, m objects of another kind and p objects are different
Example: In how many ways can you select 2 balls from a set of 3 identical blue balls, 4 identical black ball, and 5 other different colored balls.
Case 1: Select 2 balls from 3 blue balls in 1 way
Case 2: Select 2 balls from 4 black ball in 1 way
Case 3: Select 2 balls from 5 different color balls in 5
Case 4: Select 1 ball from blue balls, 1 from black in 1*1=1 way
Case 5: Select 1 blue ball and 1 different colored ball in 1*5=5 ways
Case 6: Select 1 black ball and 1 different colored ball in 1*5 ways=5 ways
So total number of ways=1+1+10+1+5+5=23 ways
nC1+ nC2+…...+ nCn=2n - 1
nPr = r! nCr
4. Group Formation
(a) The total number of ways of dividing (m + n) different things into two groups of m and n items respectively, is given by
m + n C n = (m+n) ! /m! n!
(b) The total number of ways of dividing 2n different things into two groups of n items in each group respectively is given by
2n C n / 2!
(c) The total number of ways of dividing (m + n + p) different things into three unequal groups m, n, p is ((m+n+p)! / m! n! p! )
(d) The total number of ways of distributing (a + b + c) different items into 3 people and then distributed among three persons, then the number of ways of doing this is ((a+b+c)!3!) / a!b!c!
5. Multinomial Theorem
Consider the equation x1+x2+x3+x4+...+xn=r where r is a given non-negative integer. The number of integral solutions of the form (x1,x2,x3,...xn ) satisfying the conditions
Where ai , bi ,(i=1,2,3…., n) are all integers is given by:
The coefficient of λr in the expansion of
(a) The number of non-negative integral solutions of the form (x1 ,x2 , x3 ,..., xn ) satisfying the equation is x1 + x2 + x3 +...+ xn =r is n+r-1Cr
(b) The number of positive integral solutions of the form (x1 ,x2 , x3 ,..., xn ) satisfying the equation is x1 + x2 + x3 +...+ xn =r is r-1Cn-1
Out of n different objects, each of which has a specified position, if all are arranged so that none of them occupies its original position, then it is called the derangement of objects and it can be done in,
7. Application in number theory
(a) Any natural number N can be factorized as follows.
Where p1,p2 ,p3 ,…. are distinct prime numbers and a1, a2 ,... an are non-negative integers.
Any divisor of N must be of the following form
Where 0≤bi≤ai for i=1,2,3…n
(c) The number of divisors of N including 1 and N =(a1+1)(a2+1)...(an+1)
(d) The number of divisors of N excluding 1 and N=(a1+1)(a2+1)...(an+1) - 2
(e) Sum of the divisors =
More from us:
All the best!