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Permutation and Combination is a relatively less important chapter in JEE Main Mathematics but questions asked from this chapter are always scoring. Permutation is a quite complex topic where details matter a lot while Combination is easier. **Check JEE Main 2020 Mathematics Syllabus**

Around 1-2 questions are asked in **JEE Main** from Permutation and Combination carrying a total of about 4 marks. Hence, the weightage of this chapter in JEE Main is around 1-2%.

- Questions are mainly asked from the two subtopics - Permutation and Combination.
- Books like R.D Sharma and NCERT are highly recommended for preparation.

Read the article to access the necessary study notes on Permutation and Combination that will help you ace this section.

**Must Read:**

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 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 implies arrangement of the 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 ordering. 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 arrangement. So, the 34251, 21034, 42351 numbers are all distinct cases.

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.

Permutation, already known, is the combination or selection and the arrangement. Therefore, when you are calculating the permutation, you must choose the numbers before their arrangement.

Permutation = Selection x Arrangement

This can be represented by the mathematical relation.

We know that,

Thus, from the two formulas mentioned above, it is very clear that,

nPr = nCr r!

where, nCr represents the selection and r! represents the arrangement of r objects for the r places.

Theorems on Permutations <h3>

Generally, the permutation of ‘n’ distinct objects taken r at a time is expressed and calculated as,

It can even be expressed as P(n, r) or Pnr. In the definition of permutation, r could be any positive integer that is less than or equals to n. Therefore, on the basis of the values of r and if it is less than or equal to n, two different conditions or theorems can exist.

The number of permutations or arrangement of n distinct things taken all at once can be expressed as -

For example, consider 5 seats in a car on which 5 people are to be seated.

In order to calculate the number of scenarios in which 5 people can be seated, permutation will have to be used of 5 people taking all the 5 seats at once (5P5).

The quantity of permutations or arrangement of n distinct things made only r (r < n) at a time could be expressed as -

For example, consider there are 10 chairs in a room on which 15 people are to be seated.

In this situation, 15 people have to be arranged but only on a limited 10 chairs. Therefore, it could be calculated by 15P10.

Another thing that can be learned by studying the example given above is that, you will first have to choose 10 out of 15 people who can be arranged on the 10 available chairs.

Thus, every question on permutation is divided into selection and then arrangement. As learnt earlier, Permutation = Selection x Arrangement

- 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 occuring one, two, or three… up to r times in any arrangement is represented by (n)r.
- 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!

The combination of n distinct objects chosen r at a time is expressed and calculated as -

It could also be expressed as In the definition of combination, r could be any positive integer that is less than or equal to n.

The number of combinations of n different things chosen r at a time is expressed as -

For example, you have to calculate the number of choices of 3 distinctly colored pens from the given 5 pens of all the distinctly colored pens.

Here, you will need to choose 3 pens in any order from the given 5 pens. It could be calculated as -

In this theorem, please note that if r = n, all of the things have to be chosen.

So,

, which is quite evident.

The number of combinations of n distinct things chosen r at a time when p particular things are always included will be represented as -

n-pCr-p

For example, you have to calculate the number of ways of combination of 11 players out of 20 players where Virat kohli, Y. Singh, and M.S. Dhoni is always included.

Here, you are given 20 players out of which only 11 players are to be chosen. You must also understand as given that 3 players out of 20 must be included no matter what. Therefore, it is easy to know that out of 11 players, you already have 3 players. You will only have to choose 8 additional players from the remaining 17 players.

Therefore, the total number of ways to find combinations for the above problem is = 20-3 C11-3 = 17C8

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

n-pCr

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

Calculate the number of ways of 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

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.

**Must Read:**

The number of ways in which (m + n) distinct things could be grouped into two, one consisting m items and the other consisting n items is represented by -

m+nCn = (m+n)!/ m!n!

In the situation given above, if m = n, that is, the groups have the same size, then the total number of ways of breaking down 2n distinct items into two equal groups is expressed as 2nCn/2!.This can be written as (2n)!/n!n!2!.

- The total number of ways of dividing (m + n + p) different items into three unequal groups m, n, p is given by (m + n + p)!/ m!n!p!.
- In the case given above, if m = n = p, the total number of ways reduces to (3n)!/(n!)3 3!
- The number of ways in which ‘l’ groups of n distinct objects could be made in a way that ‘p’ groups are of object n1, q groups of object n2 are given by n!/ (n1)!p(n2)!q(p!)(q!)
- If (a + b + c) distinct items have to be divided into 3 groups and distributed between three people, then the number of ways of doing this is given by (a + b + c)!. 3!/ a!b!c!

**Number of divisors**

Suppose N = p1α1 p2α2 ….. pkαk, then,

Number of divisors = Number of ways of choosing zero or more objects from the group that has identical objects (α1+1)( α2+1)…(αk+1)

This also includes 1 and N.

All of the divisors excluding 1 and N are known as Proper divisors.

**Sum of divisors**-

If N = p1α1 p2α2 ….. pkαk, then the sum of divisors of N is

(1+ p1 + p12 +…+ p1α1) × (1 + p2 + p22 +…+ p2α2) ….. (1 + pk + pk2 +…+ pkαk)

=

**The number of ways of using N as the product of two natural numbers is -**

When n is not a perfect square = ½ (a1 + 1)(a2 + 1) ….. (ak +1)

When n is a perfect square = ½ [(a1 + 1)(a2 + 1) ….. (ak +1) + 1].

When n things are arranged in a row, the number of ways through which they could be deranged such that r things take wrong places while (n-r) things take their original places is given by -

= nCn-r Dr, where

Dr =

When n things are arranged in a row, the number of ways of deranging them such that not one of them takes its original place is given by -

= nC0 Dn

=

When there are n points in plane from which m (< n) are collinear, the results given below would hold good -

The total number of distinct straight lines found by joining these n points is nC2 – mC2 +1

- The total number of different triangles made by connecting these n points is nC3 – mC3
- The number of diagonals in a polygon with n sides is nC2 – n
- When m parallel lines in a plane are intersected by a family of other n parallel lines, the total number of parallelograms thus formed is given by mC2 × nC2
- The number of triangles created by connecting the vertices of a convex polygon of n sides is nC3 of which
- The number of triangles that have exactly 2 sides common to the polygon = n
- The number of triangles that have exactly 1 side common to the polygon = n(n-4)
- The number of triangles that have no side common to the polygon =

**Q1. 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.

**Q2. Suppose a class consists of n students, form a group of the students including a minimum of two students and also excluding a minimum of two students. The number of ways in which the group can be formed is**

(a) 2n − 2n

(b) 2n − 2n − 2

(c) 2n − 2n − 4

(d) 2n − 2n − 6

**Solution: **

The required number of ways

Therefore, the answer is (b) 2n − 2n − 2.

**Q3. Assume that Tn is the number of all possible triangles that can be formed by joining the vertices of an n-sided regular polygon. Tn+1 − Tn = 10. Then, the value of n would be -**

(a) 7

(b) 5

(c) 10

(d) 8

**Solution: **

We have,

Therefore, the answer is (b) 5.

**Q4. The total number of ways in which 5 balls of distinct colors can be distributed between 3 people such that every person receives a minimum of one ball is -**

(a) 75

(b) 150

(c) 210

(d) 243

**Solution:**

Given, 5 different balls are to be distributed between 3 persons such that every person gets a minimum of one ball.

Thus, we have the possible combinations -

1, 2, 2 or 1, 3, 1

The number of ways to distribute the 5 balls are -

Therefore, the answer is (b) 150.

**Q5. When the number of ways of choosing n cards out of an unlimited number of cards containing the numbers 0, 2, 3, such that it cannot be utilized to write the number 203 is 93, then n equals to -**

(a) 3

(b) 4

(c) 5

(d) 6

**Solution: **

203 cannot be written if in the choice of n cards we get either (2 or 3), (3 or 0), ( 0 or 2), (only 0), (only 2) or (only 3).

Therefore, the answer is (c) 5.

Attend the classes of the coaching institute you have enrolled in. This will help you grasp the basic concepts of Permutation and Combination.

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.

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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