JEE Advanced Previous Year Questions (2018 - 2025): Permutations and Combinations

2024

Q1: Let S = {1, 2, 3, 4, 5, 6} and X be the set of all relations R from S to S that satisfy both the following properties:
i. R has exactly 6 elements.
ii. For each (a, b) ∈ R, we have |a - b| ≥ 2.
Let Y = {R ∈ X : The range of R has exactly one element} and
Z = {R ∈ X : R is a function from S to S}.
Let n(A) denote the number of elements in a set A.
If n(X) = mC₆, then the value of m is _____.      [JEE Advanced 2024 Paper 2]
Ans: 20
|a - b| ≥ 2 or |b - a| = 2
Total:
a = 1, b = 3, 4, 5, 6 → 8
a = 2, b = 4, 5, 6 → 6
a = 3, b = 5, 6 → 4
a = 4, b = 6 → 2
Sum = 20
n(X) = ²⁰C₆ = ᵐC₆
m = 20

Q2: Let S = {1, 2, 3, 4, 5, 6} and X be the set of all relations R from S to S that satisfy both the following properties:
i. R has exactly 6 elements.
ii. For each (a, b) ∈ R, we have |a - b| ≥ 2.
Let Y = {R ∈ X : The range of R has exactly one element} and
Z = {R ∈ X : R is a function from S to S}.
Let n(A) denote the number of elements in a set A.
If the value of n(Y) + n(Z) is k², then |k| is _______.    [JEE Advanced 2024 Paper 2]
Ans: 36
Given |a - b| ≥ 2, so if
a = 1, b = 3, 4, 5, 6 → 4 × 2 = 8
a = 1, b = 4, 5, 6 → 3 × 2 = 6
a = 1, b = 5, 6 → 2 × 2 = 4
a = 1, b = 6 → 2 × 1 = 2
Total = 20
i.e. Total elements in X is ²⁰C₆
Now for n(Y), the range of R has exactly one element, i.e., second elements must be constant in R. Since R must have 6 elements, it is not possible to satisfy both conditions, so n(Y) = 0.
For n(Z):
1 → 3, 4, 5, 6
2 → 4, 5, 6
3 → 1, 5, 6
4 → 1, 2, 6
5 → 1, 2, 3
6 → 1, 2, 3, 4
Number of relations that are functions will be:
= ⁴C₁ × ³C₁ × ³C₁ × ³C₁ × ³C₁ × ⁴C₁
= (4 × 3 × 3)² = k²
i.e., k = 36

Q3: A group of 9 students, s₁, s₂, ..., s₉, is to be divided to form three teams X, Y, and Z of sizes 2, 3, and 4, respectively. Suppose that s₁ cannot be selected for the team X, and s₂ cannot be selected for the team Y. Then the number of ways to form such teams is __________.    [JEE Advanced 2024 Paper 1]
Ans: 665

Given Data:

C-i) When x does not contain S₁, but contains S₂

7C₁ × 7!3!4! = 245

C-ii) When x does not contain S₁, S₂ and y does not contain S₂

i.e. 7C₂ × 6!3!3! = 420

So total number of ways = 665.

2022

Q1: Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen ?
(a) 21816
(b) 85536
(c) 12096
(d) 156816               [JEE Advanced 2022 Paper 2]
Ans: (a)

Case 1 :
Among four bags from one bag 4 balls are taken. Number of ways of choosing one bag from 4 bags = 4C₁
From this bag number of ways of taking 4 balls are
(a) 3 Red and 1 Blue balls, which can be chosen in   ways.
(b) 2 Red and 2 Blue balls, which can be chosen in .
∴ Total number of ways of choosing this 4 balls from this bag = ⁴C₁
Now, two balls are taken from remaining three bags.
From each bag two balls can be taken in ³C₁ × ²C₁ ways.So, for three bags two balls can be taken in (³C₁ × ²C₁)³ ways.
∴ Total number of ways of chosing 10 balls from these four bags
=

Case 2 :
Among four bags from two bags 3 balls are taken. Number of ways of chosing two bags from 4 bags = ⁴C₂
From each bag, number of ways of taking 3 balls are
(a) 2 Red and 1 Blue balls, which can be chosen in ³C₂ × ²C₁ ways.
(b) 1 Red and 2 Blue balls, which can be chosen in ³C₁ × ²C₂ ways.
So, for two bags three balls can be taken in (³C₂ × ²C₁ + ³C₁ × ²C₂)² ways.
∴ Total number of ways of choosing this 3 balls from this two bags = ⁴C₂(³C₂ × ²C₁ + ³C₁ × ²C₂)²
Now, two balls are taken from remaining two bags.
From each bag two balls can be taken in ³C₁ × ²C₁ ways.
So, for two bags two balls can be taken in (³C₁ × ²C₁)2 ways.
∴ Total number of ways of chosing 10 balls from these four bags
= From Case 1 and Case 2, total number of ways of chosing 10 balls from these 4 boxes so that from each box at least one red ball and one blue ball are chosen

Q2: The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is _________. [JEE Advanced 2022 Paper 1]
Ans: 569

This is a 4-digit integer whose 0th digit called D, 10th digit called C, 100th digit called B and 1000th digit called A.
Given range for possible number is 2022 to 4482.
So, position A can have digits 2, 3 or 4.
0, 6 and 7 can't put in A as number starts with 0, 6 or 7 don't fall in the range between 2022 to 4482.
So, Position A can be filled with 2, 3 or 4 in ³C₁=3 ways.
Position B can be filled by one of 0, 2, 3, 4, 6, 7
So, total possible ways for position B = ⁶C₁=6 ways
Similarly, Position C can be filled in ⁶C₁=6 ways
Position D can be filled in ⁶C₁=6 ways
∴ Total possible integer numbers starting with 2 or 3 or 4
= 3 × 6 × 6 × 6 = 3 × 216 = 648
Now lets find those numbers which starts with 2 and 4 but don't fall in the range between 2022 to 4482.
Situation 1:

∴ Total possible numbers starting with 200 are = 6

Situation 2:

∴  Total possible number in this case = 1
Situation 3: 

Total numbers in this case = 2 × 6 × 6 = 72
∴ Total number that don't fall in the range 2022 to 4482 which starts with 2 or 4 are
= 6 + 1 + 72=79
∴ Total numbers falls in the range 2022 to 4482
=648 - 79
=569

2020

Q1: An engineer is required to visit a factory for exactly four days during the first 15 days of every month and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1 - 15 June 2021 is ...........     [JEE Advanced 2020 Paper 2]
Ans: 495
Let the engineer visits the factory first time after x₁ days to 1 June, second time after x₂ days to first visit and so on.
∴ x₁ + x₂ + x₃ + x₄ + x₅ = 11
where x₁, x₅ ≥ 0 and x₂, x₃, x₄ ≥ 1 according to the requirement of the question.
Now, let x₂ = a + 1, x₃ = b + 1 and x₄ = c + 1 where a, b, c ≥ 0
∴ New equation will be
x₁ + a + b + c + x₅ = 8
Now, the number of all possible ways in which the engineer can made visits is equals to the non-negative integral solution of equation
x₁ + a + b + c + x₅ = 8, and it is equal to

=495

Q2: In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is .......... [JEE Advanced 2020 Paper 2]
Ans: 1080
The groups of persons can be made only in 2, 2, 1, 1
∴ So the number of required ways is equal to number of ways to distribute the 6 distinct objects in group sizes 1, 1, 2 and 2
=
= 360 x 3 = 1080

2018

Q1: In a high school, a committee has to be formed from a group of 6 boys M₁, M₂, M₃, M₄, M₅, M₆ and 5 girls G₁, G₂, G₃, G₄, G₅.
(i) Let α₁ be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls.
(ii) Let α₂ be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls.
(iii) Let α₃ be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls.
(iv) Let α₄ be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls such that both M₁ and G₁ are NOT in the committee together.

(a) P → 4; Q → 6; R → 2; S → 1

(b) P → 1; Q → 4; R → 2; S → 3
(c) P → 4; Q → 6; R → 5; S → 2
(d) P → 4; Q → 2; R → 3; S → 1                 [JEE Advanced 2018 Paper 2]
Ans: (c)
Given 6 boys M₁, M₂, M₃, M₄, M₅, M₆ and 5 girls G₁, G₂, G₃, G₄, G₅
(i) α₁ → Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls.
i.e., 
∴ α₁ = 200
(ii) α₂ → Total number of ways selecting at least 2 member and having equal number of boys and girls i.e.,  
= 30 + 150 + 200 + 75 + 6 = 461
α₂ = 461
(iii) α₃ → Total number of ways of selecting 5 members in which at least 2 of them girls
i.e., 
= 200 + 150 + 30 + 1 = 381
α₃ = 381
(iv) α₄ → Total number of ways of selecting 4 members in which at least two girls such that M₁ and G₁ are not included together.
G₁ is included →

40 + 30 + 4 = 74
M₁ is included →
 = 30 + 4 = 34
G₁ and M₁ both are not included
= 1 + 20 + 60 = 81
∴ Total number = 74 + 34 + 81 = 189
α₄ = 189
Now, P → 4 ; Q → 6 ; R → 5 ; S → 2
Hence, option (C) is correct.