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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 6D / a / Solution

For

0 \leq k \leq n

, the binomial coefficient is

(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) .

This is Pascal's identity.

Solved by gpt-5.6-sol high.

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