We could state:

\(\displaystyle (1+x)^n=\sum_{k=0}^n{n \choose k}x^{k}\)

So, we require:

\(\displaystyle {n \choose 3}=4{n \choose 2}\)

\(\displaystyle \frac{n!}{3!(n-3)!}=4\frac{n!}{2!(n-2)!}\)

Can you finish, since it is for an exam, which hopefully, you are allowed outside help?