Without any restriction we have a permutation of $4n$ objects where $2n$ each are identical. So, ${}^{4n}C_{2n}$.
Now we have to subtract the cases where $n$ white balls are consecutive. We can choose the starting position for these $n$ balls in $3n+1$ ways. Remaining $3n$ balls would give ${}^{3n}C_{n}$. So, our answer should be
$${}^{4n}C_{2n} - (3n+1){}^{3n}C_{n}.$$
But this counts repeatedly the $n$ consecutive case- waiting for correct solution.