Question

If the ratio ${}^{2n}{C}_3{:}^n{C}_3$ is equal to $11:1$ , find $n$

Answer 1

We have ${}^{2n}{C}_3{:}^n{C}_3=11:1$

$\Rightarrow \frac{{}^{2}n{C}_3}{{}^n{C}_3}=\frac{11}{1}$

$\Rightarrow \frac{\frac{\left(2n\right)!}{(2n-3)!3!}}{\frac{n!}{(n-3)!(3!)}}=\frac{11}{1}$

$\Rightarrow \frac{\left(2n\right)!}{(2n-3)!}\times \frac{(n-3)!}{n!}=\frac{11}{1}$

$\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)(2n-3)!}{(2n-3)!}\times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!}=\frac{11}{1}$

$\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)}{n(n-1)(n-2)}=\frac{11}{1}$

$\Rightarrow \frac{4(2n-1)}{n-2}=\frac{11}{1}$

$\Rightarrow 8n-4=11n-22$

$\Rightarrow 3n=18$

$\Rightarrow n=6.$