Question

Determine $n$ if ${}^{2n}{C}_3{:}^n{C}_3=12:1$.

Answer 1

We have,

$\begin{array}{c}{}^{2n}{C}_3{:}^n{C}_3=12:1\\ \Rightarrow \frac{{}^{2n}{C}_3}{{}^n{C}_3}=\frac{12}{1}\\ \Rightarrow \frac{2n!}{\left(2n-3\right)!3!}.\frac{\left(n-3\right)!3!}{n!}=12\\ \Rightarrow \frac{2!\left(n-3\right)!}{\left(2n-3\right)!n!}=12\\ \Rightarrow \frac{2n\left(2n-1\right)\left(2n-2\right)}{6}\times \frac{6}{n\left(n-1\right)\left(n-2\right)}=12\\ \Rightarrow \frac{4\left(2n-1\right)}{\left(n-2\right)}=12\\ \Rightarrow 2n-4=12n-24\\ \Rightarrow 4n=20\\ \therefore n=5\end{array}$

Hence, this is the answer.