Permutations and Combinations JEE Main PYQs (2020–2026)

Solved Permutations and Combinations PYQs

3 recent questions with answers. The full set of 171 is free to practise.

1. JEE Main 2026MediumFundamental Principles of Counting
   Q1

   A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the $10^{\text{th }}$ floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to :

   • A2184
   • B3064
   • C7056
   • D11340

   Show answer & solution

   Correct answer: (C)

   First, decide which $4$ out of the $9$ persons will get down together. The remaining $5$ persons are then fixed automatically.
   
   Number of ways to form the two groups of sizes $4$ and $5$ is $\frac{9!}{4!5!}=126$
   
   Now, the lift does not stop at the $1^{\text{st}}$ and $2^{\text{nd}}$ floors. So people can get down only on floors $3$ to $10$ , i.e., $8$ possible floors.
   
   Choose any $2$ different floors out of these $8$ floors for the two groups to exit: ${}^8{C}_2=28$
   
   After choosing the two floors, we must decide which floor is for the group of $4$ and which floor is for the group of $5$ . This can be done in $2$ ways.
   
   So, total number of ways is $126\times 28\times 2=7056$
2. JEE Main 2026MediumLinear Permutations & Arrangement of Objects
   Q2

   The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, is:

   • A2670
   • B2840
   • C2920
   • D3600

   Show answer & solution

   Correct answer: (D)

   From the word INCONSEQUENTIAL, let us first identify the vowels and consonants available.
   
   The letters are:
   
   $$I,N,C,O,N,S,E,Q,U,E,N,T,I,A,L$$
   
   We have to form $4$ -letter words containing:
   
   exactly $2$ vowels
   
   exactly $2$ consonants
   
   no letter repeated
   
   Since letters of the word may repeat in the original word, we count only distinct available letters for selection.
   
   Step 1: Distinct vowels
   
   The vowels present are:
   
   $$A,E,I,O,U$$
   
   So, number of distinct vowels $=5$ .
   
   Step 2: Distinct consonants
   
   The consonants present are:
   
   $$N,C,S,Q,T,L$$
   
   So, number of distinct consonants $=6$ .
   
   Step 3: Choose 2 vowels and 2 consonants
   
   Number of ways to choose $2$ vowels from $5$ :
   
   $${}^5{C}_2=10$$
   
   Number of ways to choose $2$ consonants from $6$ :
   
   $${}^6{C}_2=15$$
   
   So, total selections of letters:
   
   $${}^5{C}_2{\times }^6{C}_2=10\times 15=150$$
   
   Step 4: Arrange the 4 chosen letters
   
   All 4 letters are distinct, so number of arrangements:
   
   $$4!=24$$
   
   Hence, total number of such 4-letter words is:
   
   $$150\times 24=3600$$
   
   Therefore, the correct answer is:
   
   $$\boxed{3600}$$
   
   So, Option D is correct.
3. JEE Main 2026MediumNumber of Integral Solutions & Geometric Applications
   Q3

   Let $A=\{(a,b,c):a,b,c$ are non-negative integers and $a+b+2c=22\}$ .
   
   Then $n(A)$ is equal to :

   • A121
   • B124
   • C144
   • D169

   Show answer & solution

   Correct answer: (C)

   We need to count the number of ordered triples $(a,b,c)$ of non-negative integers satisfying
   
   $$a+b+2c=22$$
   
   Since $a,b,c\ge 0$ , let us fix $c$ first.
   
   For a fixed value of $c$ , the equation becomes
   
   $$a+b=22-2c$$
   
   Now, the number of non-negative integer solutions of
   
   $$a+b=n$$
   
   is
   
   $$n+1$$
   
   So, for fixed $c$ , number of solutions is
   
   $$(22-2c)+1=23-2c$$
   
   Now find all possible values of $c$ .
   
   Since $2c\le 22$ ,
   
   $$c=0,1,2,\dots ,11$$
   
   Hence total number of solutions is
   
   $$\sum _{c=0}^{11}(23-2c)$$
   
   This is an arithmetic series:
   
   first term $=23$
   
   last term $=23-2(11)=1$
   
   number of terms $=12$
   
   So,
   
   $$n(A)=\frac{12}{2}(23+1)$$
   
   $$n(A)=6\cdot 24=144$$
   
   Therefore,
   
   $$n(A)=144$$
   
   So the correct option is:
   
   $$\boxed{144}$$
   
   Hence, $\boxed{\text{Option C}}$ .

Topic-wise question count

Where JEE Main has asked from Permutations and Combinations, most frequent first.

• Combinations & Selection of Objects40
• Fundamental Principles of Counting31
• Linear Permutations & Arrangement of Objects25
• Permutations & Combinations with Repetition24
• Number of Integral Solutions & Geometric Applications22
• Dearrangement & Rank of a Word in Dictionary17
• Division and Distribution of Objects11
• Circular Permutations1

Year-wise Permutations and Combinations questions in JEE Main

Number of questions from this chapter in each year's paper(s).