 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-2/3f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Every allocation is determined by the set of

n

children on the red bus, so there are

(n 2 n)

equally likely allocations. Exactly one places every boy on red and every girl on green. Hence

P (A_{n}) = (n 2 n)^{- 1} = \frac{( n ! ) ^{2}}{( 2 n )!} .

(n 2 n) \sim \frac{4 ^{n}}{πn},

and therefore

P (A_{n}) \sim \frac{πn}{4 ^{n}} .

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