If $5$ dice are rolled, $3$ 6-sided die and $2$ 8-sided die, how do I come up with the chances that the sum will be $12$?
I've figured that there are $13824$ total combinations, but can't figure out the number of combinations that equal $12$.
What formula or method should I use?
This is rather straightforward to do using inclusion-exclusion, since only one of the constraints can be violated at a time.
There are $\binom{12-1}{5-1}=330$ ways to distribute $12$ balls to $5$ non-empty bins, of which $3\cdot\binom{12-6-1}{5-1}=15$ have more than $6$ balls in one of the $3$ bins with capacity $6$. All other constraint violations and combinations thereof are impossible, so the result is $330-15=315$, in agreement with the result of the generating function solution given by lulu in a comment.