If ${{}^{2n}C}_3:{{}^{n}C}_2=44:3$, find $n$.

${{}^{2n}C}_3:{{}^{n}C}_2=44:3$

$⟹\frac{\left(2n\right)!}{(2n-3)!\times 3!}\times \frac{(n-2)!\times 2!}{n!}=\frac{44}{3}$
$⟹\frac{2n(2n-1)(2n-2)}{n(n-1)\times 3}=\frac{44}{3}$
$⟹\frac{2(2n-1)(n-1)}{n-1}=22$

$⟹2n^2-3n+1=11n-11$
$⟹2n^2-14n+12=0$
$⟹(2n-2)(n-6)=0$
$⟹n=1,6$
Since ${}^1{C}_2$ is not possible
$⟹\therefore n=6$