Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is :
(1) 15400
(2) 17800
(3) 16800
(4) 29400
Show answer and solution
Answer: Option (3)
Key idea: Permutations with repetition - distributing objects with constraints
We have 5 distinct digits {1, 2, 3, 5, 7} and need to form 7-digit numbers with each digit used at least once. Since $5 + 2 = 7$, we need 2 extra digits.
**Case 1:** Two digits appear twice each, three appear once. Choose 2 digits to repeat: $\binom{5}{2} = 10$ ways.
Arrange 7 digits with two pairs: $\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260$. Case 1 total: $10 \times 1260 = 12600$.
**Case 2:** One digit appears three times, four appear once. Choose 1 digit to repeat thrice: $\binom{5}{1} = 5$ ways.
Arrange 7 digits with one triplet: $\frac{7!}{3!} = \frac{5040}{6} = 840$. Case 2 total: $5 \times 840 = 4200$.
Total number of 7-digit numbers: $12600 + 4200 = 16800$.
Let $p_n$ denote the total number of triangles formed by joining the vertices of an n-side regular polygon. If $P_{n+1} - P_n = 66$, then the sum of all distinct prime divisors of n is :
(1) 7
(2) 8
(3) 5
(4) 6
Show answer and solution
Answer: Option (3)
Key idea: Combinations and Prime Factorization
Step 1: Number of triangles from $n$ vertices is $P_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6}$
Two players A and B play a series of games of badminton. The player, who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player A wins the series is______.
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Answer: 126
Key idea: Combinations with constraint - counting sequences where a specific event occurs at a fixed position
Player A wins the series when A gets 5 wins first. The last game must be won by A (to clinch the series).
Before the final winning game, A must have exactly 4 wins. B can have 0, 1, 2, 3, or 4 wins.
For each case where B has $k$ wins $(k = 0, 1, 2, 3, 4)$: Total games = $5 + k$, and we choose positions for A's 4 wins in the first $(4 + k)$ games.
Number of ways = $\binom{4}{4} + \binom{5}{4} + \binom{6}{4} + \binom{7}{4} + \binom{8}{4}$
A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is:
(1) 4100
(2) 4140
(3) 4230
(4) 4290
Show answer and solution
Answer: Option (1)
Key idea: Combinations with constraints - distributing objects with minimum requirements
Identify constraints: Select 8 balls with at least 2 of each color from 5 blue, 6 yellow, 4 red balls
Set up: Let $(b, y, r)$ be the number of blue, yellow, red balls. Need $b+y+r=8$ with $b \geq 2, y \geq 2, r \geq 2$ and $b \leq 5, y \leq 6, r \leq 4$
Find all valid distributions: $(2,2,4), (2,3,3), (2,4,2), (3,2,3), (3,3,2), (4,2,2)$
Calculate ways for each case using $\binom{n}{r}$: $150 + 800 + 900 + 600 + 1200 + 450$
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:
(1) 2670
(2) 2840
(3) 2920
(4) 3600
Show answer and solution
Answer: Option (4)
Key idea: Combinations and Permutations - Selecting and arranging distinct objects
Identify distinct letters in INCONSEQUENTIAL: Vowels are I, O, E, U, A (5 distinct) and Consonants are N, C, S, Q, T, L (6 distinct)
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 $10^{th}$ 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 :
(1) $2184$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is:
(1) 18
(2) 36
(3) 39
(4) 72
Show answer and solution
Answer: Option (2)
Key idea: Distribution of distinct objects into distinct boxes with restriction (Stirling numbers of second kind)
Since 4 books go into 3 bags with no bag empty, the distribution must be $(2,1,1)$ - one bag gets 2 books, other two get 1 each
Choose which 2 books go together: $\binom{4}{2} = 6$ ways
Now we have 3 distinct groups: one pair of books and two single books
Assign these 3 groups to 3 different bags: $3! = 6$ ways
Total number of ways = $\binom{4}{2} \times 3! = 6 \times 6 = 36$
FAQs
Frequently asked questions
How many questions come from Permutations and Combinations in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 8 questions on it, about 0.9 per shift.
Where can I practise Permutations and Combinations PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
