I managed to figure this out for n = 1.
$ \frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} $
I thought it would work by putting this to the power of n for all n.
But I can't manage to figure it out for n = 2, more importantly for all n.
I saw solutions online using matrices, however I have no experience with matrices yet so I was wondering if this was possible without it.
Thanks in advance for answers. Also, I know what the answers should be since I ran a computer simulation with pseudo-random numbers, and I'm definitely not getting the right answer.
For $n=2$ the number possible results from twelve rolls is $6^{12}$.
If out of $12$ you want two rolls to be equal to $1$, two of the rest to be equal to $2$ etc the number of successful rolls is $$\binom {12}2\binom {10}2\binom 82\binom 62 \binom 42 \binom 22$$
Note that for six rolls you had $$\binom 61\binom 51\binom 41\binom 31\binom 21\binom 11$$ out of $6^6$ possibilities in total.
I am sure you can generalise (and maybe simplify) from that.