If ${ }^{2 n} C_{3}:{ }^{n} C_{3}=10: 1$, then the ratio $\left(n^{2}+3 n\right):\left(n^{2}-3 n+4\right)$ is :

$$ \begin{aligned} & \text {We have, }{ }^{2 n} C_3:{ }^n C_3=10: 1 \\\\ & \Rightarrow \frac{{ }^{2 n} C_3}{{ }^n C_3}=\frac{10}{1} \\\\ & \Rightarrow \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=\frac{10}{1} \\\\ & \Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{(n)(n-1)(n-2)}=\frac{10}{1} \\\\ & \Rightarrow 4 n^2-6 n+2=5\left(n^2-3 n+2\right) \\\\ & \Rightarrow n^2-9 n+8=0 \\\\ & \Rightarrow n^2-8 n-n+8=0 \\\\ & \Rightarrow n(n-8)-1(n-8)=0 \\\\ & \Rightarrow (n-8)(n-1)=0 \\\\ & \Rightarrow n=8(n=1 \text { not valid }) \end{aligned} $$ <br/><br/>$\therefore \frac{n^2+3 n}{n^2-3 n+4}=\frac{88}{44}=\frac{2}{1}=2: 1$