Question

If $\frac{{}^{2n}{C}_3}{{}^n{C}_3}=11$, then the value of $n$ is

Answer 1

$\frac{{}^{2n}{C}_3}{{}^n{C}_3}=11\Rightarrow \frac{2n!(n-3)!}{n!(2n-3)!}=11\Rightarrow \frac{2n\times (2n-1)\times (2n-2)}{n\times (n-1)(n-2)}=11\Rightarrow \frac{4(2n-1)}{n-2}=11\Rightarrow 8n-4=11n-22\therefore n=6$