Question

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

Answer 1

We have,

${}^n{C}_9{=}^n{C}_8{\therefore }^n{C}_r=\frac{n!}{r!(n-r)!}$

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

$\frac{1}{9\times 8!(n-9)!}=\frac{1}{8!(n-8)!}$

$\frac{1}{9(n-9)!}=\frac{1}{(n-8)!}$

$9(n-9)!=(n-8)!$

$9(n-9)!=(n-8)(n-8-1)!$

$9=n-8$

$n=17$

Value of ${}^n{C}_{17}$

${}^{17}{C}_{17}=1$ ${\therefore }^n{C}_n=1$

Hence, this is the answer.