Class 11 Maths Chapter 6: Permutations and Combinations worksheet with answer key
A 20-mark worksheet on Permutations and Combinations for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
Show answer
Answer: (c) 3
ⁿP₀ = 1, ⁿCn = 1 and 0! = 1, so the sum is 1 + 1 + 1 = 3.
Show answer
Answer: (a) 105
¹⁵C₁₃ = ¹⁵C₂ = 15 × 142 = 105. 210 is ¹⁵P₂, the count without dividing by 2!.
Show answer
Answer: (d) 360
By Theorem 3, n!/p! = 6!/2! = 720/2 = 360.
Show answer
Answer: (a) 210
The medals are different, so order matters: ⁷P₃ = 7 × 6 × 5 = 210. ⁷C₃ = 35 ignores which medal goes to whom; 73 lets one runner win twice.
Show answer
Answer: (c) 243
Each of the 5 letters has 3 choices of box, and a box may take any number of letters, so the count is 3 × 3 × 3 × 3 × 3 = 35 = 243. 53 gives each box a choice of letter, which is the wrong way round.
Show answer
10! = (1 × 3 × 5 × 7 × 9) × (2 × 4 × 6 × 8 × 10), and 2 × 4 × 6 × 8 × 10 = 25 × (1 × 2 × 3 × 4 × 5) = 25 × 5!. So 1 × 3 × 5 × 7 × 9 = 10!25 × 5! = 3628800/3840 = 945.
- splits 10! into the odd and the even factors (1)
- writes the even product as 25 × 5! and gives 10!/(25 × 5!) (1)
Show answer
The pair of spinners has no order: picking X and then Y gives the same pair as Y and then X. So the count is ⁶C₂ = 6 × 52 = 15.
- states that order does not matter in a selection (1)
- ⁶C₂ = 15 (1)
Show answer
(n + 1)! = (n + 1) × n × (n − 1)!. Dividing by (n − 1)!: n(n + 1) = 12, so n2 + n − 12 = 0, (n + 4)(n − 3) = 0. As n is a natural number, n = 3. Check: 4! = 24 = 12 × 2!.
- writes (n + 1)! = (n + 1)n(n − 1)! and gets n(n + 1) = 12 (1)
- solves and rejects n = −4 to get n = 3 (1)
Show answer
(i) Each place has 10 choices: 104 = 10000. (ii) No repetition: ¹⁰P₄ = 10 × 9 × 8 × 7 = 5040. (iii) At least one repeat is the complement of (ii): 10000 − 5040 = 4960.
- (i) 104 = 10000 (1)
- (ii) ¹⁰P₄ = 5040 (1)
- (iii) 10000 − 5040 = 4960 (1)
Show answer
Method 1: choose the team, ⁷C₄ = 35 ways, then its leader, 4 ways: 35 × 4 = 140. Method 2: choose the leader first, 7 ways, then the other 3 members from the remaining 6, ⁶C₃ = 20 ways: 7 × 20 = 140. Both give 140.
- method 1: ⁷C₄ × 4 = 140 (1)
- method 2: 7 × ⁶C₃ = 140 (1)
- uses the multiplication principle and states they agree (1)
Show answer
(i) Pairs of points: ¹⁰C₂ = 45. The 4 collinear points give ⁴C₂ = 6 pairs but only 1 line, so lines = 45 − 6 + 1 = 40. (ii) Sets of 3 points: ¹⁰C₃ = 120; the ⁴C₃ = 4 sets of collinear points make no triangle, so triangles = 120 − 4 = 116.
- ⁴C₂ pairs on one line counted once (1)
- lines = 45 − 6 + 1 = 40 (1)
- triangles = 120 − 4 = 116 (1)
← Chapter 5: Linear InequalitiesChapter 7: Binomial Theorem →