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0606/11 Oct/Nov 2025, Question 11: Permutations: Counting Arrangements

Cambridge IGCSE Additional Mathematics 0606 · Paper 11 (Non-calculator) · Non-calculator · 6 marks

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The question, in short

Relate the count of ordered arrangements of n objects chosen four at a time to one sixth of the ordered arrangements of 2n objects chosen three at a time, then work with n.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 4 marks

Derive the quadratic 3n² − 19n + 20 = 0 from the given relationship between the two permutation counts.

  1. 1ⁿP₄ = n!/(n − 4)!, ²ⁿP₃ = (2n)!/(2n − 3)!
  2. 2n(n − 1)(n − 2)(n − 3) = (1/6) × 2n(2n − 1)(2n − 2)
  3. 3Divide by n(n − 1): (n − 2)(n − 3) = (2/3)(2n − 1)
  4. 43(n² − 5n + 6) = 4n − 2
  5. 53n² − 19n + 20 = 0

Answer3n² − 19n + 20 = 0 shown

Part (b) · 2 marks

Use your equation from part a to determine the numerical value of n.

  1. 1(3n − 4)(n − 5) = 0
  2. 2n = 4/3 or n = 5
  3. 3n must be a whole number (at least 4), so n = 5

Answern = 5

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