Question

Find the value of $n$ if ${}^{2n}{C}_3:{}^n{C}_3=12:1$.

Answer 1

$\frac{{{}^{2n}C}_3}{{{}^{n}C}_3}=\frac{12}{1}$

$\Rightarrow \frac{\frac{2n!}{(2n-3)!3!}}{\frac{n!}{(n-3)!3!}}=12$

$\Rightarrow \frac{(n-3)!}{n!}\times \frac{2n!}{(2n-3)!}=12$

$\Rightarrow \frac{(2n-2)}{(n-2)(n-1)\left(n\right)}(2n-1)\left(2n\right)=12$

$\Rightarrow \frac{2\times 2\times (2n-1)}{n-2}=12$

$\Rightarrow 8n-4=12n-24$

$\Rightarrow 4n=20$

$\Rightarrow n=5.$

Hence, the answer is $5.$