Question

Solve for n if $\frac{\left(2n\right)!}{3!(2n-3)!}:\frac{n!}{2!(n-2)!}=12:1$

Answer 1

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

$\Rightarrow \frac{\frac{\left(2n\right)!\times 2(n-1)(2n-1)2n}{3!\left(2n\right)!}}{\frac{n!(n-1)n}{2!n!}}=\frac{4(2n-1)}{6}\times 2$

$\frac{4(2n-1)}{3}=\frac{12}{1}\Rightarrow 8n-4=36$

$8n=40\Rightarrow n=5$