Question

If ${}^{2n}{C}_3:{}^n{C}_3=11:1,$ find n.

Answer 1

We have

${}^{2n}{C}_3=\frac{2n(2n-1)(2n-2)}{3!}=\frac{2n(n-1)(2n-1)}{3}$

and ${}^n{C}_3=\frac{n(n-1)(n-2)}{3!}=\frac{n(n-1)(n-2)}{6}.$

$\therefore \frac{{}^{2n}{C}_3}{{}^n{C}_3}=\frac{2n(n-1)(2n-1)}{3}\times \frac{6}{n(n-1)(n-2)}=\frac{4(2n-1)}{(n-2)}.$

Now, $\frac{{}^{2n}{C}_3}{{}^n{C}_3}=\frac{11}{1}\Rightarrow \frac{4(2n-1)}{(n-2)}=\frac{11}{1}$

$\Rightarrow 8n-4=11n-22\Rightarrow 3n=18\Rightarrow n=6.$

Hence, $n=6.$