Class 11 Maths Chapter 6 Permutations and Combinations – Extra Questions with Answers

Extra practice questions for Class 11 Maths Chapter 6 (Permutations and Combinations), beyond the textbook. These Class 11 Maths Chapter 6 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. Evaluate 5!.
Ans: 5×4×3×2×1 = 120.

Q2. Find ⁵P₂.
Ans: 5!/(5−2)! = 5!/3! = 120/6 = 20.

Q3. Find ⁵C₂.
Ans: 5!/[2!3!] = 120/[2×6] = 10.

Q4. In how many ways can 3 books be arranged on a shelf?
Ans: 3! = 6.

Q5. Should choosing a committee of people use permutations or combinations?
Ans: Combinations, since order of selection doesn’t matter for a committee.

Short Answer Questions (2–3 marks)

Q6. In how many ways can a President and a Vice-President be chosen from a group of 8 people (2 distinct roles)?
Ans: Since roles are distinct (order matters), use permutations: ⁸P₂ = 8!/6! = 8×7 = 56.

Q7. From a group of 10 students, how many ways can a team of 4 be selected (no distinct roles)?
Ans: Since no roles distinguish the 4 members, use combinations: ¹⁰C₄ = 10!/[4!6!] = 210.

Q8. How many different 4-letter arrangements (with no repetition) can be made using the letters of the word “MATH”?
Ans: All 4 letters are distinct, so this is a straightforward permutation of all 4: 4! = 24.

Higher-Order Thinking / Application Questions

Q9. A password must be exactly 4 digits long, using digits 0-9, with no digit repeated. Using the multiplication principle, find the total number of possible passwords, and explain why this matches the permutation formula.
Ans: Using the multiplication principle: first digit has 10 choices, second has 9 (one used), third has 8, fourth has 7: 10×9×8×7 = 5040. This matches ¹⁰P₄ = 10!/6! = 5040, because a permutation formula essentially counts exactly this kind of “choose without replacement, order matters” scenario.

Q10. Explain, using the relationship ⁿP_r = ⁿC_r × r!, why choosing a 3-person committee from 6 people gives fewer possibilities than arranging 3 people (from the same 6) in 3 distinct chairs, and calculate both values to confirm.
Ans: ⁶C₃ (committee, order doesn’t matter) = 6!/[3!3!] = 20. ⁶P₃ (arrangement in distinct chairs, order matters) = 6!/3! = 120. The committee count is smaller because each unique group of 3 people can be arranged in 3! = 6 different orders as permutations, but all 6 of those orderings count as just ONE combination (one committee) — confirmed since 20 × 6 = 120, matching ⁿP_r = ⁿC_r × r!.

Written by Satish

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