Question

If ${}^n{C}_9={}^n{C}_8$, what is the value of ${}^n{C}_{17}$ ?

• A. 1
• B. 0
• C. 3
• D. 17

Answer 1

Answer: (A)

$\frac{n!}{9!\left(n-9\right)!}=\frac{n!}{8!\left(n-8\right)!}$
$\Rightarrow \frac{1!}{9\times 8!\left(n-9\right)!}=\frac{1!}{8!\left(n-8\right)\left(n-9\right)!}$
$\Rightarrow \frac{1}{9}=\frac{1}{\left(n-8\right)}\Rightarrow 9=n-8$
$\Rightarrow 9+8=n\Rightarrow n=17$
${\therefore }^n{C}_{17}{=}^{17}{C}_{17}=1$ $[\because {}^n{C}_n=1]$