0606/23 May/Jun 2025, Question 11: Permutations: Counting Arrangements
Cambridge IGCSE Additional Mathematics 0606 · Paper 23 (Calculator) · Calculator · 4 marks
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The question, in short
Solve the permutation equation with ratio ⁿ⁺¹P₅ ÷ 437 = ⁿP₃ algebraically to determine n.
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Solution · 4 marks
- 1(n + 1)!/(437(n − 4)!) = n!/(n − 3)!
- 2(n + 1)!/n! = n + 1 and (n − 3)!/(n − 4)! = n − 3
- 3(n + 1)(n − 3) = 437
- 4n² − 2n − 440 = 0, (n − 22)(n + 20) = 0
- 5n = 22 (n must be positive)
Answern = 22
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Also searched as: 0606/23/M/J/25 · 0606_s25_qp_23 · 0606 s25 23 · question 11.
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