 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/3k/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

RM (d, 0)

2^{d}

, while

RM (d, d) = F_{2}^{2^{d}}

is the full binary code. For

0 < r < d

RM (d, r) = RM (d - 1, r) ∣ RM (d - 1, r - 1) .

Thus its rank

k (d, r)

satisfies Pascal's recursion

k (d, r) = k (d - 1, r) + k (d - 1, r - 1),

with the stated boundary values. Therefore

k (d, r) = j = 0 \sum r (j d) .

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