If $C (n, 9) = C (n, 8)$ , find $C (18, n)$

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Solution

As $^{n} C_{9} =^{n} C_{8}$

$\frac{n!}{9! (n - 9)!} = \frac{n!}{8! (n - 8)!}$

$\frac{n!}{9 \times 8! (n - 9)!} = \frac{n!}{8! (n - 8) (n - 9)!}$

$n - 8 = 9$

$n = 17$

Then the value of $^{18} C_{n}$ is,

$^{18} C_{17} = \frac{18!}{17! (18 - 17)!}$

$= \frac{18 \times 17!}{17!}$

$= 18$

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