Question

If ${}^n{C}_9={}^n{C}_8,$ find ${}^n{C}_{17}$

Answer 1

Finding the value of $n$
Given, ${}^n{C}_9={}^n{C}_8$
$\Rightarrow \frac{n!}{9!(n-9)!}=\frac{n!}{8!(n-8)!}$
$\Rightarrow \frac{1}{9\times 8!(n-9)!}=\frac{1}{8!(n-8)(n-9)!}$
$\Rightarrow \frac{1}{9}=\frac{1}{n-8}$
$\Rightarrow n-8=9$
$\Rightarrow n=17$

$\therefore {}^n{C}_{17}={}^{17}{C}_{17}=\frac{17!}{17!(17-17)!}=1$