Question

If $2\times$ ${}^n{C}_5=9\times$ ${}^{n-2}{C}_5$, then the value of n will be:

• A. 7
• B. 10
• C. 9
• D. 5

Answer 1

The correct option is B 10

Given, $2\left({}^n{C}_5\right)=9\left({}^{n-2}{C}_5\right)$
$2.\frac{n!}{5!.(n-5)!}=9.\frac{(n-2)!}{(n-7)!.5!}$
$\frac{2(n-1)\left(n\right)}{(n-6)(n-5)}=9$
$2n^2-2n=9(n^2-11n+30)$
$7n^2-97n+270=0$
Hence $n=10$ and $n=\frac{27}{7}$
Since $nϵN$.
Hence $n=10$.