 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-4/6d/a/solution

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

For

0 \leq k \leq n

(k n) = \frac{n !}{k ! ( n - k )!} .

When

1 \leq k \leq n - 1

(k n - 1) + (k - 1 n - 1) = \frac{( n - 1 )!}{k ! ( n - 1 - k )!} + \frac{( n - 1 )!}{( k - 1 )! ( n - k )!} = \frac{( n - 1 )! [( n - k ) + k ]}{k ! ( n - k )!} = (k n) .

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