Counting Problem: How to distribute 20 donuts to 6 kids where you have 4 types of donuts (No restriction on type or number of donuts) The difficulty in this question is the fact that there are 4 types of donuts. If there was just one type of donut then the question is relatively simple using either bars and stars or generating functions.
But when you introduce types of donuts, it gets a bit confusing (for me at least):


*

*To pick 20 donuts of which there are 4 types is equivalent to the number of integer solutions to
$$n_a+n_b+n_c+n_d = 20, n_i \geq 0$$

*For each selection, we now need to distribute the 4 types of donuts to every kid (clearly each kid is distinct). So this is equivalent to the number of integer solutions of
$$k_1+k_2+k_3+k_4+k_5+k_6 = n, n\in \{n_a,n_b,n_c,n_d\}$$
Obviously, solving this exhaustively isn't feasible (i.e. for every solution of 1. get number of ways of doing 2. and sum). Is there any combinatorics trick I'm not seeing. Studying for an exam.

Let $n_{xy}$ be the number of donuts of type $y$ given to kid $x$, where $1\leq x\leq6$ and $1\leq y\leq4$.  Then we want the number of solutions, in nonnegative integers to $$\sum_{x=1}^6\sum_{y=1}^4a_{xy}=20$$
Stars and bars is indicated. 
Wait.  This assumes that you many not have any of a particular type of donut.  Is that legitimate, or did you want at least one of each type?
