In how many ways can $12$ different balls go into $3$ different boxes so that in every box there are $4$ balls? I need help with this question:
In how many ways can $12$ different balls go into $3$ different boxes so that in every box there are $4$ balls?
The answer should be $34650$.
Thank you.

The number of ways to permute $12$ different balls is $12!$.
If we draw an example we get something like:
$$6312|A4B9|5C87$$
But each box has $4!=24$ combinations, so we need to divide by $4!^3$.
So the final answer is $\dfrac{12!}{4!^3}=34650$.
