If  $2n_{C_3}:{}^n{C}_3=10:1$, then  $n+3:n^2-n$

$$\begin{gathered} { }^{2n}{C}_3{:}^n{C}_3=10 \\ \Rightarrow n=8 \\ \therefore Ans=\frac{11}{56} \end{gathered}$$

$\begin{array}{r}\frac{2n!}{3!(2n-3)!}\times \frac{3!(n-3)!}{n!}=10\\ \Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)}{n(n-1)(n-2)}=10\\ \Rightarrow 4(2n-1)=10n-20\\ \Rightarrow 2n=16\Rightarrow n=8\\ \therefore \frac{n+3}{n^2-n}=\frac{8+3}{64-8}=\frac{11}{56}\end{array}$