Previous year questions: Permutations and Combinations

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Six items are real JEE Main 2026 questions from the official National Testing Agency papers and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the source and the marks. The stems were checked against the official PDF pages and the values against the official key. In every question, decide first whether order matters, then choose between the permutation and combination formulas.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Seven digit numbers are formed from the digits 1, 2, 3, 5 and 7 so that every one of the five digits appears at least once. Find how many numbers can be formed.

Answer: Seven places are filled from five distinct digits with all five used at least once, so the multiplicities must split as either 3, 1, 1, 1, 1 or 2, 2, 1, 1, 1. In the first case choose the digit that appears three times in 5 ways and arrange the multiset in 7 factorial by 3 factorial ways, giving 5 times 840 equal to 4200. In the second case choose the two digits that appear twice in 5C2 equal to 10 ways and arrange in 7 factorial by 2 factorial by 2 factorial ways, giving 10 times 1260 equal to 12600. Adding the two cases gives 16800 numbers, which is the official answer.

Q2 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

Let pn be the number of triangles formed by joining the vertices of a regular polygon with n sides. If pn plus 1 minus pn equals 66, find the sum of all distinct prime divisors of n.

Answer: A triangle is fixed by choosing three vertices and order does not matter, so pn equals nC3. The difference between consecutive values is (n plus 1)C3 minus nC3, which by the Pascal rule equals nC2. Setting nC2 equal to 66 gives n(n minus 1) by 2 equal to 66, so n squared minus n minus 132 equals 0. The positive root is 12, since 12 times 11 by 2 is 66. The distinct prime divisors of 12 are 2 and 3, and their sum is 5, which is the official answer.

Q3 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Four boys and three girls stand in a line. In how many of these lines are the three girls not all next to each other?

Answer: Count the total queues first, which is 7 factorial equal to 5040. Now count the queues where all three girls are together: treat the three girls as one block, so the objects are the block and the four boys, five objects in all, arranged in 5 factorial ways, and the three girls can be ordered inside the block in 3 factorial ways. The together count is 120 times 6 equal to 720. The required number is 5040 minus 720 equal to 4320, which is the official answer. The subtraction works because every queue either has the girls together or has at least one boy between two of them, and the two groups do not overlap.

Q4 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Two players A and B play a series of badminton games. The first player to win 5 games wins the series, and no game is drawn. Find the number of ways in which player A can win the series.

Answer: For A to win the series, the last game must be A's fifth win. In the games before that A has exactly 4 wins and B has k wins, where k ranges from 0 to 4, and the order of those games can be any arrangement with 4 A wins among 4 plus k places. The count for a fixed k is (4 plus k)C4. Adding for k equal to 0, 1, 2, 3 and 4 gives 1 plus 5 plus 15 plus 35 plus 70, which is 126. The telescoping identity (4 plus k)C4 summed over these values is 9C5, also 126, matching the official answer.

Q5 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

Let A be the set of ordered triples (a, b, c) of non-negative integers with a plus b plus 2c equal to 22. Find the number of elements in A.

Answer: Fix c first. For a fixed c the equation becomes a plus b equal to 22 minus 2c, and the number of non-negative solutions is one more than the right side, that is 23 minus 2c. The value c can range from 0 up to 11, because 2c cannot exceed 22. Summing 23 minus 2c for c equal to 0 through 11 gives 12 times 23 minus 2 times the sum of 0 to 11. The sum of 0 to 11 is 66, so the count is 276 minus 132 equal to 144, which is the official answer.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

A person has three different bags and four different books. Find the number of ways to put the books into the bags so that no bag is empty.

Answer: Each book independently chooses one of 3 bags, giving 3 to the fourth power equal to 81 assignments in total. Remove the assignments that leave at least one bag empty using inclusion and exclusion. Assignments missing a specified bag number 2 to the fourth power, and there are three such bags, contributing 48 to subtract; assignments missing two specified bags number 1, and there are three such pairs, which were subtracted twice and must be added back once, giving 3. The count is 81 minus 48 plus 3 equal to 36, which is the official answer.

Q7 (NCERT Exemplar, 3 marks)

Eight chairs are numbered 1 to 8. Two women and three men each occupy one chair. The women choose first from chairs 1 to 4, then the men choose from the remaining chairs. Find the total number of arrangements.

Answer: The two women occupy two of the four chairs numbered 1 to 4, and order matters because the women are distinct, so they have 4P2 equal to 12 ways. Six chairs remain, including chairs 5 to 8 and the two unused chairs from 1 to 4, and the three men choose three of these in 6P3 equal to 120 ways. Multiplying gives 12 times 120 equal to 1440 arrangements. Source: NCERT Exemplar Class 11 Mathematics, Chapter 7, Exercise 7.3, Short Answer Type, Q1.

Q8 (NCERT Exemplar, 3 marks)

A bag contains 5 black balls and 6 red balls. Find the number of ways to select 2 black balls and 3 red balls from the bag.

Answer: Choose the black balls in 5C2 equal to 10 ways and the red balls in 6C3 equal to 20 ways. The two selections are independent and order inside a colour group does not matter, so multiply: 10 times 20 equal to 200 ways. Source: NCERT Exemplar Class 11 Mathematics, Chapter 7, Exercise 7.3, Short Answer Type, Q8.

Q9 (NCERT Exemplar, 3 marks)

There are 10 lamps in a hall and each can be switched on or off independently. Find the number of ways in which the hall can be illuminated.

Answer: Each lamp has 2 states, so the 10 lamps have 2 to the tenth power equal to 1024 switch settings. One of these settings has every lamp off, which does not illuminate the hall. The number of illuminating settings is therefore 1024 minus 1 equal to 1023. Source: NCERT Exemplar Class 11 Mathematics, Chapter 7, Exercise 7.3, Short Answer Type, Q13.

Q10 (NCERT Exemplar, 3 marks)

A convex polygon has 44 diagonals. Find the number of its sides.

Answer: A polygon with n vertices has nC2 segments joining pairs of vertices, and n of these are the sides, so the number of diagonals is nC2 minus n. Set n(n minus 1) by 2 minus n equal to 44. Multiplying by 2 gives n squared minus n minus 2n equal to 88, that is n squared minus 3n minus 88 equal to 0. The positive root is 11, since 11 times 8 is 88 and 11 minus 8 is 3. The polygon has 11 sides. Source: NCERT Exemplar Class 11 Mathematics, Chapter 7, Exercise 7.3, Short Answer Type, Q20.