Permutations and Combinations: Previous Year Questions with Solutions

1.0Introduction

Permutations and Combinations Previous Year Questions typically cover topics like fundamental counting principle, factorial notation, permutations of distinct and identical objects, circular permutations, and combinations with or without restrictions. Examples include arranging letters of a word, selecting teams or committees, finding the number of ways to distribute items, and solving problems involving both selection and arrangement under specific conditions. Solutions involve applying standard formulas such as $n{P}_r=\frac{n!}{(n-r)!}andn{C}_r=\frac{n!}{r!(n-r)!}$, along with logical reasoning and constraints handling. Practicing these questions helps in building a strong foundation in counting techniques and enhances problem-solving speed and accuracy for competitive exams.

2.0Permutations and Combinations Previous Year Questions for JEE with Solutions

JEE questions in Permutations and Combinations often test concepts related to the fundamental principles of counting, arrangements, selections, and problems involving constraints. Some common types of problems include:

• Basic Counting and Factorials: Questions involving application of the fundamental principle of counting and factorial concepts to calculate total possible outcomes.
• Permutations: Problems on arranging objects (distinct or identical), circular permutations, and arrangements under restrictions (e.g., certain people always or never together).
• Combinations: Questions on selecting items or forming groups/teams from a given set, sometimes with additional constraints such as at least/at most selections or mutually exclusive choices.
• Mixed Problems: Problems that combine both permutations and combinations, including word formation, distribution of objects, and selection followed by arrangement.

These questions are aimed at testing logical reasoning, understanding of counting techniques, and the ability to apply formulas under different scenarios.
Note: In the JEE Main Mathematics exam, you can generally expect 2 to 3 questions from the Permutations and Combinations chapter.

Related Video:

3.0Key Concepts to Remember – Permutation

1. Definition

• Permutation refers to arrangement of objects. Order matters.

2. Basic Formula

$n{P}_r=\frac{n!}{(n-r)!}$

(Number of ways to arrange r objects out of n)

• With Repetition:

$n^r$

(Each position can be filled in n ways, total r positions)

3. Permutations with Identical Items

• When some objects are identical:

$\frac{n!}{p!q!r!},$

(e.g., for the word “BALLOON” with repeating letters)

4. Circular Permutations

• Without Reflection: (n – 1)!
• With Reflection (like a necklace):

$\frac{(n-1)!}{2}$

5. Common Applications

• Arranging people or objects in a line or around a circle
• Arranging digits/letters with or without restrictions
• Placing people with conditions (e.g., certain people together or apart)

6. JEE Tips for Permutations

• Read whether the arrangement is linear or circular
• Watch for words like "together", "not together", "starting with", "ending with", etc.
• Use the difference between identical and distinct objects wisely

4.0Key Concepts to Remember – Combination

1. Definition

• Combination refers to selection of objects. Order does not matter.

2. Basic Formula

$n{C}_r=\frac{n!}{r!(n-r)!}$

3. Important Properties

• Symmetry:

${}^n{C}_r={}^n{C}_{n-r}$

• Pascal’s Identity:   

${}^r{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$

4. Common Applications

• Selecting members for teams or committees
• Choosing questions from a paper
• Problems with at least / at most conditions
• Distributing items among people or groups (when only selection is involved)

5. JEE Tips for Combinations

• Focus on whether the problem involves selection only
• In mixed problems (selection + arrangement), use combination first, then multiply by permutation
• Carefully handle restrictions like minimum or maximum selections

5.0JEE Mains Past Year Questions with Solutions on Coordinate Geometry

1. There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is …..

Ans. (5626)

Sol.

2. There are 5 points P₁, P₂, P₃, P₄, P₅ on the side AB, excluding A and B, of a triangle ABC. Similarly, there are 6 points P₆, P₇, …, P₁₁ on the side BC and 7 points P₁₂, P₁₃, …, P₁₈ on the side CA of the triangle. The number of triangles that can be formed using the points P₁, P₂, …, P₁₈ as vertices is :

(1) 776

(2) 751

(3) 796

(4) 771

Ans. (2)

Sol. ¹⁸C₃ – ⁵C₃ – ⁶C₃ – ⁷C₃

= 751

3. 60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the 50^th word is :

(1) OBBHJ

(2) HBBJO

(3) OBBJH

(4) JBBOH

Ans. (3)

Sol. B  B  H  J  O

$$\begin{gathered} \text{B}\Rightarrow 4!=24 \\ \text{H}\Rightarrow \frac{4!}{2!}=12 \\ \text{J}\Rightarrow \frac{4!}{2!}=12 \end{gathered}$$

O  B  B  H  J

O  B  B  J  H → 50^th rank

4. Let the set S = {2, 4, 8, 16, ....., 512} be partitioned into 3 sets A, B, C with equal number of elements such that A ∪ B ∪ C = S and A ∩ B = B ∩ C = A ∩ C = φ. The maximum number of such possible partitions of S is equal to :

(1) 1680

(2) 1520

(3) 1710

(4) 1640

Ans. (1)

Sol.

$\frac{9!}{(3!3!3!)}\times 3!$

5. The number of ways of getting a sum 16 on throwing a dice four times is __________.

Ans. (125)

Sol. (x¹ + x² ....+ x⁶)⁴

$x^4{\left(\frac{1-x^6}{1-x}\right)}^4$

x⁴·(1–x⁶)⁴·(1–x)^–4

x⁴[1–4x⁶ + 6x¹²....] [(1–x)^–4]

(x⁴ – 4x¹⁰ + 6x¹⁶....) (1–x)^–4

(x⁴ – 4x¹⁰ + 6x¹⁶) (1 + ¹⁵C₁₂ x¹² + ⁹C₆x⁶ ....)

(¹⁵C₁₂ – 4⋅⁹C₆ + 6)x¹⁶

(¹⁵C₃ – 4⋅⁹C₆ + 6)

= 35 × 13 – 6 × 8 × 7 + 6

= 455 – 336 + 6

= 125

7. The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is 

(1) 24

(2) 56

(3) 16

(4) 48

Ans. (3)

Sol. no. of triangles having no side common with a n sided polygon =

$\frac{{}^n{C}_1\cdot {}^{n-4}{C}_2}{3}=\frac{{}^8{C}_1\cdot {}^4{C}_2}{3}=16$

8. The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to :

(1) 175

(2) 181

(3) 177

(4) 179

Ans. (4)

Sol. AA, MM, TT, H, I, C, S, E

(1) All distinct

⁸C₅ → 56

(2) 2 same, 3 different

³C₁ × ⁷C₃ → 105

(3) 2 same I^st kind, 2 same 2^nd kind, 1 different

³C₂ × ⁶C₁ → 18

Total → 179

9. The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to ______.

Ans. (36)

Sol. 2, 3, 4, 5, 7

total number of three digit numbers not divisible by 3 will be formed by using the digits 

(4, 5, 7)

(3, 4, 7)

(2, 5, 7)

(2, 4, 7)

(2, 4, 5)

(2, 3, 5)

number of ways = 6 × 3! = 36

11. If n is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to:

(1) 47

(2) 53

(3) 51

(4) 43

Ans. (3)

Sol.

Total ways to partition 5 into 4 parts are :

$$\begin{gathered} 5,0,0,0\Rightarrow 1\text{ way} \\ 4,1,0,0\Rightarrow \frac{5!}{4!}=5\text{ ways} \\ 3,2,0,0\Rightarrow \frac{5!}{3!2!}=10\text{ ways} \\ 2,2,0,1\Rightarrow \frac{5!}{2!2!1!}=15\text{ ways} \\ 2,1,1,1\Rightarrow \frac{5!}{2(1!{)}^3}=10\text{ ways} \\ 3,1,1,0\Rightarrow \frac{5!}{3!2!}=10\text{ ways} \\ Total1+5+10+15+10+10=51ways \end{gathered}$$

12. The number of elements in the set

S = {(x, y, z) : x, y,z ∈ Z, x + 2y + 3z = 42, x, y, z ≥ 0} equals __________.

Ans. (169)

Sol. x + 2y + 3z = 42, x, y, z ≥ 0

z = 0; x + 2y = 42 ⇒ 22

z = 1; x + 2y = 39 ⇒ 20

z = 2; x + 2y = 36 ⇒ 19

z = 3; x + 2y = 33 ⇒ 17

z = 4; x + 2y = 30 ⇒ 16

z = 5; x + 2y = 27 ⇒ 14

z = 6; x + 2y = 24 ⇒ 13

z = 7; x + 2y = 21 ⇒ 11

z = 8; x + 2y = 18 ⇒ 10

z = 9; x + 2y = 15 ⇒ 8

z = 10; x + 2y = 12 ⇒ 7

z = 11; x + 2y = 9 ⇒ 5

z = 12; x + 2y = 6 ⇒ 4

z = 13; x + 2y = 3 ⇒ 2

z = 14 ;         x + 2y = 0 ⇒ 1

Total: 169

13. Let and $\alpha =\frac{(4!)!}{(4!{)}^3!},and\beta =\frac{(5!)!}{(5!{)}^4!}.$

Then:

$$\begin{gathered} (1)\alpha \in \mathbb{N}and\beta \notin \mathbb{N} \\ (2)\alpha \notin \mathbb{N}and\beta \in \mathbb{N} \\ (3)\alpha \in \mathbb{N}and\beta \in \mathbb{N} \\ (4)\alpha \notin \mathbb{N}and\beta \notin \mathbb{N} \end{gathered}$$

Ans. (3)

Sol.

$$\begin{gathered} \alpha =\frac{(4!)!}{(4!{)}^3\cdot 3!},\beta =\frac{(5!)!}{(5!{)}^4\cdot 4!} \\ \alpha =\frac{24!}{(4!{)}^6\cdot 6!},\beta =\frac{120!}{(5!{)}^{24}\cdot 24!} \end{gathered}$$

Let 24 distinct objects are divided into 6 groups of 4 objects in each group.

No. of ways of formation of group = $\frac{24!}{(4!{)}^6\cdot 6!}\in \mathbb{N}$

Similarly,

Let 120 distinct objects are divided into 24 groups of 5 objects in each group.

No. of ways of formation of groups 

$\frac{120!}{(5!{)}^{24}\cdot 24!}\in \mathbb{N}$

16. In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections : A, B and C . A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is _______ .

Ans. (11376)

Sol. If 4 questions from each section are selected

Remaining 3 questions can be selected either in (1, 1, 1) or (3, 0, 0) or (2, 1, 0)

$\text{Total ways}={}^8{C}_5\cdot {}^6{C}_5\cdot {}^6{C}_5+{}^8{C}_6\cdot {}^6{C}_5\cdot {}^6{C}_4\cdot 2+{}^8{C}_5\cdot {}^6{C}_6\cdot {}^6{C}_4\cdot 2+{}^8{C}_4\cdot {}^6{C}_6\cdot {}^6{C}_5\cdot 2$

= 56 . 6 . 6 + 28 . 6 . 15 . 2 + 56 . 15 . 2 + 70 . 6 . 2 + 8 . 15 . 15

= 2016 + 5040 + 1680 + 840 + 1800 = 11376

17. The total number of words (with or without meaning) that can be formed out of the letters of the word ‘DISTRIBUTION’ taken four at a time, is equal to _____

Ans. (3734)

Sol. We have III, TT, D, S, R, B, U, O, N

Number of words with selection (a, a, a, b)

$={}^8{C}_1\times \frac{4!}{3!}=32$

Number of words with selection (a, a, b, b)

$=\frac{4!}{2!\cdot 2!}=6$

Number of words with selection (a, a, b, c)

$={}^2{C}_1\times {}^8{C}_2\times \frac{4!}{2!}=672$

Number of words with selection (a, b, c, d)

$={}^9{C}_4\times 4!=3024$

total = 3024 + 672 + 6 + 32

= 3734

20. The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is ______.

Ans. (70)

Sol. N = a b c

(i) All distinct digits

a + b + c = 14

a ≥ 1

b, c ∈ {0 to 9}

by hit & trial : 8 cases

(6, 5, 3) (8, 6, 0) (9, 4, 1)

(7, 6, 1) (8, 5, 1) (9, 3, 2)

(7, 5, 2) (8, 4, 2)

(7, 4, 3) (9, 5, 0)

(ii) 2 same, 1 diff a = b ; c

2a + c = 14

by values : 

$$\begin{gathered} (3,8),(4,6),(5,4),(6,2),(7,0) \\ \text{For each, total permutations =}\frac{3!}{2!}=3,\text{and total 5 such combinations:} \end{gathered}$$

= 14 cases

(iii) all same :

3a = 14

a = $\frac{14}{3}$ × rejected

0 cases

Hence, Total cases :

8 × 3! + 2 × (4) + 14

= 48 + 22

= 70 

Also Read:-