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.
 A: 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 $5!$
A: There won't be more than 5 groups. We can consider three cases: that there are 3, 4 or 5 groups respectively. Let's consider the case where number of groups is equal to 4. Then there are additional 2 cases: there is one group of size 5 and one of 4 or three of four (the size of rest is 3).
To summarize there are:
$$\frac{15!}{5!\cdot 5! \cdot5! \cdot 3!}+\frac{15!}{3!\cdot 3! \cdot 4! \cdot 5! \cdot 2!}+\frac{15!}{3!\cdot 4!\cdot 4! \cdot 4! \cdot 3!}+\frac{15!}{3!\cdot 3!\cdot 3! \cdot 3! \cdot 3! \cdot 5!}$$
ways to solve this problem.
