 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/12i/iii/solution

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

Adding the all-one codeword is a bijection from

C

to itself, and it replaces every bit by its complement. It therefore sends weight

j

n - j

, proving

ν (n - j) = ν (j) .

The code has one word of weight zero and, since its minimum Hamming distance is three, no words of weight one or two. Combining this with part (i) gives

ν (n - 1) = 0, ν (n - 2) = 0, ν (n - 3) = \frac{n ( n - 1 )}{6} .

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