JEE Main Study Notes for Permutation and Combinations: Permutation and Combination are the ways in which objects from a set may be selected, generally, without replacement, to form a subset.

• These study notes for JEE Main 2022 include topics such as the fundamental principle of counting, Permutation as an arrangement and combination as selection, The meaning of P (n,r) and C (n,r), and Simple applications.
• Permutation and combination section usually carries 1-2% weightage in JEE Main Mathematics syllabus.
• In JEE Main Question Paper around 4-5 can be expected from Permutation and Combination section.

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What are Permutation and Combination?

Combination

The combination involves choosing one or more objects from the specified objects (might be similar or distinct). The words selection, collection, or committee can also substitute combinations.

For example, the choosing of 5 players (in any order) is the combination of the top 5 cricket players from the team of 11 players.

The order in which they are chosen here is not significant. We can also say that in the case of a combination, the selection order is not the concern.

Permutation

Permutation implies the arrangement of similar or different objects taken some or all at once. So we can analyze the word 'arrangement' that was used in the permutation description. The 'arrangement' here signifies choice as well as order. That means in this case, the order in which the objects are chosen was taken care of as well.

For example – The number of 5 digits that can be formed using 0, 1, 2, 3, 4, and 5 digits.

In this example, we simply do not need to pick the 5 digits from the given 6 digits, but we also need to see the number of possible cases for the different arrangements. So, the 34251, 21034, 42351 numbers are all distinct cases.

What is Factorial Notation in Mathematics?

As the definitions or the formulae of permutation and combination mandate the usage of factorial notation, it is important to understand what it is.

In Mathematics, the factorial is expressed by the symbol ‘!’. If we have to show 5 factorials, it will be represented as 5! Typically, the factorial of any positive number n will be expressed by n!.

Mathematically,

 

where n is any positive integer.

So, 4! = 3! x 4 = 2! x 3 x 4 = 1! x 2 x 3 x 4 = 0! x 1 x 2 x 3 x 4 = 1 x 2 x 3 x 4

In a similar manner, for any positive integer ‘n’,

n! = n x (n – 1) x (n – 2) x …….. x 3 x 2 x 1.

Thus, we can also define the factorial of any positive integer ‘n’ as ‘the product of all the positive integers less than or equal to n’.

Given below are the factorial of a few regularly used numbers -

0! = 1

1! = 1

2! = 2 x 1 = 2

3! = 3 x 2 x 1 = 6

4! = 4 x 3 x 2 x 1 = 24

5! = 5 x 4 x 3 x 2 x 1 = 120 and so on.

Must Read:

• JEE Main Study Notes for Matrices & Determinants
• JEE Main Study Notes for Mathematical Reasoning

Difference Between Permutation and Combination

Theorem of Permutation

Every arrangement can be made by taking, some or all of a number of things is called a Permutation.

Theorem 1: To find the number of permutations of ‘n’ things taken ‘r’ at a time

The problem is equivalent to filling ‘r’ places with ‘n’ different things.

The first place can be filled in 'n' ways, as any one of the 'n' things may be taken.

Having filled the first place, there remain 'n – 1' things so the second place can be filled in (n – 1) ways.

Hence, the first two places can be filled in n (n – 1) ways.

After filling the first two places there are 'n – 2' objects left to fill the third place, and so on.

Hence, the number of ways of filling 'r' places with 'n' things

= 

The above formula for nPr involves the following conditions :

• All the things are distinct.
• Repetition of things is not allowed in any of the arrangements.
• No arrangement is repeated.
• The arrangement is linear.

Theorem 2: The number of permutations of ‘n’ things taken all at a time

This will be given by the above formula after taking r = n.

Thus, the required number of ways = nPn = n!

Theorem 3: Number of circular permutations of ‘n’ distinct objects

Let, ‘x’ is the required number of permutations. Since each circular permutation corresponds to ‘n’ linear permutations depending on where we start.

So, xn is the total number of linear permutations

But, The linear permutations number is n!

Therefore,

Thus the total number of circular permutations of 'n' distinct things is (n – 1)!.

If no distinction is made between anti-clockwise and clockwise arrangements, then the number of permutations is (n – 1)!

Check:

Points to Remember for Permutation

• The number of permutations of n distinct objects when a certain object is not taken into the arrangement is represented by n-1Pr.
• The number of permutations of n distinct objects when a particular object is always taken in the arrangement is represented by r.n-1Pr-1.
• Suppose there is a need to calculate the number of permutations of n distinct objects, from which r have to be chosen and every object has the probability of occurring one, two, or three… up to r times in any arrangement is represented by (n)r.
• A circular permutation is utilized when an arrangement has to be made in the shape of a circle or ring.
• When ‘n’ distinct or dissimilar objects have to be arranged in a circle such that the clockwise and anticlockwise arrangements are varied, then the number of these arrangements is expressed as (n – 1)!
• Suppose r things are chosen at a time out of n different things and arranged in a circle, the number of ways of doing this is expressed as nCr(r-1)!.
• When clockwise and anti-clockwise are taken to be the same, the total number of circular permutations is represented by (n-1)!/2.
• Suppose n people have to be seated around a round table in a way that no person has a similar neighbor, it is represented by ½ (n – 1)!
• The number of necklaces formed with n beads of varied colors = ½ (n – 1)!

• nP0 =1
• nP1 = n
• nPn = n!/(n-n)! = n! /0! = n!/1 = n!

Check:

• JEE Main Study Notes for Algebra
• JEE Main Study Noes for Sequence & Series

Theorem of Combinations

Theorem 1: To find the number of combinations of 'n' dissimilar things taken 'r' at a time

Let the required number of selections be 'x'.

Consider one of these 'x' ways. there are 'r' things in this selection which can be permuted in r! ways.

Thus, each of the 'x' selections gives r! permutations. Consequently, the total number of permutations of 'n' things taken 'r' at a time is x (r!).

But this number is also nPr.

Thus, the total number of combinations of 'n' dissimilar things taken 'r' at a time is nCr.

The number of combinations of 'n' dissimilar things taken all at a time = nCn = 1.

Theorem 2: The number of selections of some or all out of (p + q + r + ..... ) things out of which p are alike of one kind, q - alike of the second kind, and so on :

The total number of required ways

= (p + 1) (q + 1) (r + 1) ...... – 1

Theorem 3:

The number of combinations of n distinct things chosen r at a time when p particular things are usually excluded is represented as -

n-pCr

For example, take one similar to that which we considered in the previous theorem.

Calculate the number of ways of a combination of 11 players out of 20 players where Ravindra Jadeja and Balaji have to be always excluded.

Here as well, you have been given 20 players of which you have to choose 11 players. However, this time, 2 particular players have to be excluded. Therefore, you now have the option of a total of 18 players out of which 11 players are to be selected.

Hence, the total number of selections = 20 - 2C11 = 18C11

Also Check:

• JEE Main Study Notes for Probability
• JEE Main Study notes for Integral Calculus

Points to Remember for Combination

• Choosing when both similar and different objects are given -

The number of choices, taking a minimum of one out of a1+ a2+ a3 + ... an+ k objects, where a1 are similar - of one kind, a2 are similar - of one kind, and so on ... an are similar - of the nth kind, and k are different = {[(a1+ 1)(a2+ 1)a3+ 1) ... (an + 1)]2k} - 1.

• The combination of n distinct things chosen some or all of n things at one time is represented by 2n – 1.
• The combination of n things chosen some or all at one time when p of the things are similar and of one kind, q of the things are similar and of a different kind and r of the things are similar and of a third kind = [(p + 1) (q + 1)(r + 1)….] – 1.
• The number of ways to choose some or all out of (p+q+t) things where p are similar and of the first kind, q are similar and of the second kind, and the remaining t are distinct is = (p+1)(q+1)2t – 1.
• The combination of choosing s1 things from a set of n1 objects and s2 things from a set of n2 objects where the combination of s1 things and s2 things are not dependent is expressed as n1Cs1 x n2Cs2
• The total number of ways in which n identical items could be distributed between p people such that every person might obtain any number of items is given by n+p-1Cp-1.
• The total number of ways in which n identical items can be distributed between p persons in a way that every one of them receives a minimum of one item is given by n-1Cp-1
• A few results related to nCr are given below -

1. nCr = nCn -r
2. If nCr = nCk , then r = k or n-r = k
3. nCr + nCr -1 = n+1Cr
4. nCr = n/r n-1Cr-1
5. nCr /nCr -1 = (n-r+1)/ r
6. If n is even, nCr is the highest for r = n/2.
7. If n is odd, nCr is the highest for r = (n-1)/2,(n+1)/2.

Check: JEE Main Study Notes on Differential Calculus

Defragments

If any changes are made in the things that are Defragment.

If n things form arrangements in a row, the number of ways in which they can be defragged so that no one of them occupies its original place is

Some Special Formulae

Conditional Permutations

• Number of permutations of n things taken r at a time, when a particular object is to be always included in each =
• Number of permutations of n things taken r at a time, when a particular object is never taken in any arrangement =
• Number of permutations of n different things taking all at a time, when m specified things always come together
  =
• Number of permutations of n different things taking all at a time, when m specified things never come together
  =
• Number of permutations of n different things taken r at a time, in which two specified objects always occur together =
• Number of ways of arranging n objects on a circle taking r at a time
  • =, if clockwise and anticlockwise arrangements are distinct
  • =, if clockwise and anticlockwise arrangements cannot be distinguished.

Conditional Combinations

• Number of combinations of n distinct things taking r  at a time, when k  particular objects always occur = 
• Number of combinations of n distinct objects taking  at a time, when k particular objects never occur =.
• Number of selections of r things from n things when p particular things are not together in any selection
  = nCr – n – pCr – p
• Suppose a number N, is expressed in prime factorization as following
  • N =
  • where p1, p2, p3, ........, pm are prime integers and are any positive integers. Then
  • Total number of all divisors of N = 
  • Sum of all divisors = 
    •  
    • =
• Number of selection or r consecutive things out of n things in a row = n – r + 1
• Number of selection of r consecutive things out of n things along a circle
  • = 
• Number of selection of r things (r ≤ n) out of n identical things = 1.
• Number of selection of zero or more things out of n identical things = n + 1.
• Number of selection of one or more things out of n identical things = n.

Sample Questions on Permutation & Combination

Question: The total number of positive integral solutions for x, y, z in a way that xyz = 24, is -

(a) 30

(b) 60

(c) 90

(d) 120

Solution: Given,

xyz = 24

xyz = 23 × 31

The number of ways of arranging 'n' identical balls into 'r' distinct boxes is (n + r − 1)C(r− 1)

We are to group 4 numbers into three groups.

The number of integral positive solutions

= (3 + 3 − 1)C(3 − 1) × (1+ 3 − 1)C(3 − 1)

= 5C2 × 3C2 = 30

Therefore, the answer is (a) 30.

Question: Find the number of permutations and combinations, if n = 15 and r = 3.

Answer: n = 15, r = 3 (Given)

Using the formulas for permutation and combination, we get:

Permutation, P = n!/(n – r)!

= 15!/(15 – 3)!

= 15!/12!

= (15 x 14 x 13 x 12!)/12!

= 15 x 14 x 13

= 2730

Also, Combination, C = n!/(n – r)!r!

= 15!/(15 – 3)!3!

= 15!/12!3!

= (15 x 14 x 13 x 12! )/12!3!

= 15 x 14 x 13/6

= 2730/6

= 455

Question: In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?

Answer: The word ‘OPTICAL’ has 7 letters. It has the vowels’ O’, ‘I’, ‘A’ in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA).

Hence we can assume total letters as 5 and all these letters are different.

Number of ways to arrange these letters

= 5!

= 5×4×3×2×1

= 120

All the 3 vowels (OIA) are different.

Number of ways to arrange these vowels among themselves

= 3!

= 3×2×1

= 6

Hence, the required number of ways:

= 120×6

= 720

Question: How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’ if repetition of letters is not allowed?

Answer: The word ‘LOGARITHMS’ has 10 different letters.

Hence, the number of 3-letter words (with or without meaning) formed by using these letters

=10P3

= 10×9×8

= 720

Tricks to Solve Questions from Permutation and Combination

1. Attend the classes of the coaching institute you have enrolled in. This will help you grasp the basic concepts of Permutation and Combination.
2. Read from R. D. SHARMA Class 12th Mathematics and NCERT books. Attempt to finish all the Permutation and Combination questions from this book in two days.
3. Solve as many JEE Main papers for thorough practice and to gain an idea of the kind of questions that can be asked from this chapter.

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