How to solve combinatorics with variable set sizes? The question: How many ways to put $15$ students into groups, such that each group has $3$~$5$ students?
If the group was $3$ equal groups of $5$, then the answer would be $\cfrac{15!}{5! \cdot 5! \cdot 5! \cdot 3!}$, and if it was $5$ equal groups of $3$, then the answer would be $\cfrac{15}{3! \cdot 3! \cdot 3! \cdot 3! \cdot 3! \cdot 5!}$. How do you deal with various sized groups? 
Thanks.

Since it is unclear whether the groups are labelled or unlabelled,
I'll treat them as labelled, and indicate the correction needed if they are unlabelled.
Groups can be$\;$ 5-5-5, $\;$ 5-4-3-3, $\;$ 4-4-4-3, $\;$ or$\;$ 3-3-3-3-3 , so # of ways for labelled groups is:
$$\frac{15!}{5!\cdot 5! \cdot5!}+\frac{15!}{5!\cdot 4!\cdot 3! \cdot 3! }+\frac{15!}{4! \cdot 4! \cdot 4! \cdot 3!}+\frac{15!}{3!\cdot 3!\cdot 3! \cdot 3! \cdot 3!}$$
If the groups are unlabelled, divide respectively by $3!,\;2!,\;3!\;$ and $5!$
