Question

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

Answer 1

$\frac{\frac{2n!}{3!(2n-3)!}}{\frac{n!}{2!(n-2)!}}=\frac{44}{3}\Rightarrow \frac{2n!2!(n-2)!}{3!(2n-3)!n!}=\frac{44}{3}\Rightarrow \frac{2n!2!(n-2)}{3\times 2\times (2n-3)!\times n(n-1)(n-2)!}=\frac{44}{3}\Rightarrow \frac{2n!}{3n(n-1)(2n-3)!}=\frac{44}{3}\Rightarrow 2n!=44\left(n\right)(n-1)(2n-3)!\Rightarrow 2n(2n-1)(2n-2)=44n(n-1)\Rightarrow (2n-1)(n-1)=11(n-1)Qn=6\therefore n=6$