If $$^{2n}C_3 : ^nC_3 = 10:1$$, then the ratio $$n^2+3n : n^2-3n+4$$ is

Just use basic PnC formula:-

$$^nC_r=\frac{n!}{r!\ \left(n-r\right)!}$$

$$\frac{\frac{2n!}{\left(2n-3\right)!3!}}{\frac{n!}{\left(n-3\right)!3!}}=\frac{10}{1}$$

$$\frac{2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{\left(2n-3\right)!\ 3!}=\frac{10\ n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{1\ \left(n-3\right)!\ 3!}$$

$$2n\left(2n-1\right)\left(2n-2\right)=10n\left(n-1\right)\left(n-2\right)$$

Solve it and get the value of n.

You will get n=1,8

1 will be rejected because n>=3 is allowed only.

Putting the n = 8 in the given equation you will get 2:1 as answer.