 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/11i/b/iii/solution

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

For column

j

, let

N_{j}

of the

M

codewords have a one. A pair contributes a one in that column exactly when one word is chosen from those

N_{j}

words and the other from the remaining

M - N_{j}

. Hence column

j

contains

N_{j} (M - N_{j})

ones among all pairwise sums. Part (i), summed over the

n

columns, gives the upper bound

#1 (A) \leq {n M^{2} /4, n (M^{2} - 1) /4, M even, M odd .

There are

(2 M)

rows, one for each unordered pair of codewords. By part (ii), every row has weight equal to that pair's distance, which is at least

d

. Therefore

#1 (A) \geq d (2 M) = \frac{d M ( M - 1 )}{2} .

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