Permutations and Combinations — Class 11 MCQs with Answers
Class 11 CBSE Mathematics · Chapter 7
90 practice questions · 30 Easy · 30 Medium · 30 Hard · Updated
Practise the most important Class 11 CBSE Mathematics questions from Chapter 7, "Permutations and Combinations". You get 9 timed quizzes made from 90 NCERT-based MCQs, with answers and explanations. The questions are split into 30 Easy, 30 Medium and 30 Hard. Warm up on the basics, then move on to the exam-level questions that set top scorers in CBSE & Maharashtra HSC Board exams, JEE Main, MHT-CET and JEE Advanced apart.
To score well in "Permutations and Combinations", focus on fast problem-solving, formula recall and step-by-step working. Each MCQ here is timed and uses exam-style marking (+4 correct, −1 wrong, 0 skipped). This trains you to stay accurate under time pressure, as real papers need. Every question has a short explanation, so a wrong answer becomes a quick lesson. It is the fastest way to fix gaps before a test.
Use this chapter for focused revision. Start with the Easy set to check your basics on Permutations and Combinations, then move to Medium and Hard to practise applying them. Your accuracy, streaks and XP save automatically. This chapter also adds to your overall Class 11 Mathematics mastery score. 14 sample questions are solved in full below, with the answer and a worked explanation. Sign in free to start practising.
Key concepts: Permutations and Combinations (Class 11 Mathematics)
Counting problems ask how many ways an outcome can happen without listing them all. This chapter builds the fundamental principle of counting, then permutations for ordered arrangements and combinations for unordered selections, developing nPr and nCr, the identities linking them, and methods for repetition and restrictions.
- Fundamental principle of counting
- If one stage happens in m ways and the next independently in n ways, together they happen in m × n ways; multiply the choices at every stage.
- Addition principle
- When outcomes are mutually exclusive alternatives, add their counts rather than multiply; the guide is OR → add, AND → multiply the possibilities.
- Factorial notation
- n! = n × (n−1) × … × 2 × 1 with 0! defined as 1; it counts the arrangements of n distinct objects placed in a row.
- Permutations (nPr)
- An ordered arrangement of r objects chosen from n distinct objects; nPr = n!/(n−r)!, since order matters and each position is filled distinctly.
Permutations and Combinations — important questions & MCQs with answers (Class 11 Mathematics)
14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.
- Q1Easy
Value of 5! is:
A.60B.120✓ CorrectC.25D.100Answer: B. 120
Explanation: 5! means 5×4×3×2×1 = 120, the product of all positive integers up to 5 — not 5×5 = 25, and not a partial product like 4×3×2×1 = 24.
- Q2Easy
C(n,r) =
A.r!/n!B.n!/r!C.n!D.n!/(r!(n−r)!)✓ CorrectAnswer: D. n!/(r!(n−r)!)
Explanation: ⁿCᵣ = n!/(r!(n−r)!) — selections where order doesn't matter. Skipping the r! divisor gives ⁿPᵣ instead, which overcounts each selection r! times.
- Q3Easy
If event A can occur in m ways and event B in n ways, both can occur in:
A.m × n ways (multiplication principle)✓ CorrectB.m + n waysC.m − n waysD.m^n waysAnswer: A. m × n ways (multiplication principle)
Explanation: The fundamental counting principle multiplies independent choices: m ways for one event and n for another give m×n combined outcomes, not m+n, which only suits mutually exclusive alternatives.
- Q4Easy
n! equals:
A.n × (n−1)B.1 × 2 × 3 × ... × n✓ CorrectC.2ⁿD.n²Answer: B. 1 × 2 × 3 × ... × n
Explanation: n! is the product of every positive integer from 1 up to n, e.g. 4! = 1×2×3×4 = 24 — not n×(n−1), which only captures the last two factors.
- Q5Easy
Number of permutations of n distinct objects taken r at a time, P(n, r) =
A.r!/n!B.n!C.n·rD.n!/(n − r)!✓ CorrectAnswer: D. n!/(n − r)!
Explanation: ⁿPᵣ = n!/(n−r)! counts ordered arrangements. Dropping the (n−r)! divisor and writing just n! wrongly includes the (n−r) objects that weren't even chosen.
- Q6Easy
Number of combinations of n distinct objects taken r at a time, C(n, r) =
A.n!/r!B.n!/(n − r)!C.n!/(r!(n − r)!)✓ CorrectD.n·rAnswer: C. n!/(r!(n − r)!)
Explanation: ⁿCᵣ = n!/(r!(n−r)!) counts selections, ignoring order. Skipping the r! in the denominator turns this into ⁿPᵣ, which counts every selection r! times over.
- Q7Medium
P(5,2) =
A.60B.10C.20✓ CorrectD.5Answer: C. 20
Explanation: ⁿPᵣ = n!/(n−r)! counts ordered arrangements: P(5,2) = 5×4 = 20. Dividing by 2! as for combinations gives 10, wrong here since order matters.
- Q8Medium
C(7,3) =
A.21B.35✓ CorrectC.42D.7Answer: B. 35
Explanation: ⁿCᵣ = n!/(r!(n−r)!): C(7,3) = (7×6×5)/(3×2×1) = 35. Using the ordered count 7×6×5 = 210 forgets to remove the 3! order-doesn't-matter overcounting.
- Q9Medium
Number of arrangements of letters of the word INDIA:
A.60✓ CorrectB.120C.20D.240Answer: A. 60
Explanation: INDIA has 5 letters with I repeated twice: arrangements = 5!/2! = 120/2 = 60. Treating both I's as distinguishable and using plain 5! = 120 double-counts every arrangement.
- Q10Medium
Number of ways to select 3 students from 10:
A.100B.720C.30D.120✓ CorrectAnswer: D. 120
Explanation: Selecting 3 students from 10, order irrelevant: C(10,3) = 10!/(3!7!) = 120. Using the ordered count P(10,3) = 720 instead treats different orderings of the same trio as separate teams.
- Q11Medium
Number of arrangements of MATHEMATICS:
A.11!B.4989600✓ CorrectC.11!/2!D.11!/24Answer: B. 4989600
Explanation: MATHEMATICS has 11 letters with M, A, T each repeated twice: arrangements = 11!/(2!2!2!) = 39916800/8 = 4989600. Dividing by only one pair's 2! misses the other two repeats.
- Q12Hard
Number of arrangements of n distinct objects in a circle:
A.n!B.(n−1)!✓ CorrectC.n!/2D.n²Answer: B. (n−1)!
Explanation: Circular arrangements of n distinct objects give (n−1)!, not n!, because fixing one object's seat removes the rotational duplicates a linear count would include.
- Q13Hard
C(n,r) + C(n,r−1) =
A.C(n−1,r)B.C(n,r+1)C.C(n+1,r)✓ CorrectD.C(2n,r)Answer: C. C(n+1,r)
Explanation: Pascal's identity: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ — adding one more object either excludes it (ⁿCᵣ) or includes it (ⁿCᵣ₋₁). The upper index bumps up, not the lower.
- Q14Hard
Number of 4-letter words using letters from MATHS that contain only consonants:
A.120B.60C.24✓ CorrectD.6Answer: C. 24
Explanation: MATHS has 4 consonants (M,T,H,S); a 4-letter word using only consonants must use all of them in some order: P(4,4) = 4! = 24. No extra selecting step since letters used equal letters available.
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Start this chapter free →Permutations and Combinations — FAQs
What are the key concepts in Class 11 Mathematics Permutations and Combinations?+
Counting problems ask how many ways an outcome can happen without listing them all. This chapter builds the fundamental principle of counting, then permutations for ordered arrangements and combinations for unordered selections, developing nPr and nCr, the identities linking them, and methods for repetition and restrictions. Key ideas include Fundamental principle of counting, Addition principle, Factorial notation, Permutations (nPr).
What does Class 11 Mathematics Chapter 7 (Permutations and Combinations) cover on XamBaaz?+
It has 90 NCERT-based MCQs on "Permutations and Combinations": 30 Easy, 30 Medium and 30 Hard. Together they make 9 timed quizzes, and you never get the same set twice. Every question has an instant explanation. They help you prepare for CBSE & Maharashtra HSC Board exams, JEE Main, MHT-CET and JEE Advanced.
Are these "Permutations and Combinations" questions free to practise?+
Yes. Sign in with Google to practise "Permutations and Combinations" free. Full unlimited access is ₹999/year on a launch offer until 1 December 2026. No chapter is charged separately.
How should I revise "Permutations and Combinations" for the exam?+
Start with the Easy quiz to check your basics, then try Medium and Hard to practise applying them. There are 9 timed quizzes on this chapter, so you can come back for a fresh set instead of one you have seen. Read each explanation, retry the questions you miss, and track your accuracy until it stays high.
Are these "Permutations and Combinations" MCQs available with answers?+
Yes. 14 sample questions are shown here in full, each with the correct option and a step-by-step "Why" explanation. Sign in free with Google to start practising, with instant scoring.
Is there negative marking in the "Permutations and Combinations" quizzes?+
Yes. The timed quizzes use exam-style marking: +4 for a right answer, −1 for a wrong one and 0 for a skip, the same negative marking as JEE Main and JEE Advanced. MHT-CET and CBSE board papers have no negative marking. Our mocks for those are scored their way.
What are the important questions from Permutations and Combinations (Class 11 Mathematics)?+
The questions that matter most test Fundamental principle of counting, Addition principle, Factorial notation, Permutations (nPr). This page shows 14 solved important MCQs with answers and explanations; all 90 questions on the chapter are available as timed quizzes once you sign in.
Is there an online quiz for Permutations and Combinations?+
Yes — Class 11 Mathematics Permutations and Combinations has timed online quizzes at Easy, Medium and Hard levels, with instant scoring and a worked explanation on every question. The first quiz on the chapter is free.