Question

If $\frac{\left(2n\right)!}{3!(2n-3)!}$ and $\frac{n!}{2!(n-2)!}$ are in the ratio $44:3$ find $n$.

Answer 1

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

$\frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}$

$\frac{\left(2n\right)!}{3!(2n-3)!}:\frac{n!}{2!(n-2)!}=44:3$

$\frac{4}{3}(2n-1)=\frac{44}{3}$

$⟹n=6$