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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-2/10f/d/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 2 / 10F / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

View a restricted-growth string as a set partition. Reverse the underlying order and relabel blocks by first appearance. This is a bijection of

A_{n}

which sends the size of the block containing the first position to the size of the block containing the last position. Part (c) shows that it preserves probability, so the two probabilities in part (b) are equal. Replacing

n

there by

n + 1

gives

E M_{n} = 2 (\frac{\prod _{i = 1}^{n} ( i + 1/2 )}{n !} - 1) .

(1)

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