P white balls, Q black balls, N boxes First of all sorry if this has been asked before, I could find "similiar" questions which seem to be harder but not quite this specific question.   
You are given P white balls and Q black balls, how many ways can you put them into N different boxes?   
My idea was to put first the P white balls into the N different boxes which can be done in $\binom{P+N-1}{P}$ ways (right?) then for each of these you do the same with the black balls so overall the answer is $\binom{P+N-1}{P}\binom{Q+N-1}{Q}$  
Is this correct? If so is there a way to do it so you get a nicer form? 
 A: Your answer is indeed correct, assuming that your bins are distiguishable and the balls of same color are identical.  For those browsing, the OP arrived at his answer using Theorem 2 at the link below:
http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
Also, I'm fairly sure this is the nicest form you'll get.  If you attempt to expand this into factorials and try to get a single $\binom{n}{k}$ expression, I think you'll simply end up with a mess of factorials.
A: Putting first the P white balls in the boxes, then the black ball seems the right way to do it. The formula for this would rather be P(k,n) though: The number of partitions of the integer k into n parts.
Oh ! Are your bins distinguishable or not ? Here I assume the bins do all look the same...
So final formula would be P(P,N) * P(Q, N)
See here for an explanation about k balls in n boxes,
and here for the calculation of P(k,n)
A: You can use numeric representation. For each ball, you write the box that it is in.
For example, if $P = 2$, $Q = 3$ and $N = 3$ than for each ball, you have 3 different choices. You write all the numbers from $11111$ (all in box number 1) to $33333$ (all in box number 3). That is $3^5$ ($N^{(P+Q)}$.
If none of the boxes should be emtpty, you subtract the cases that two of the boxes are empty. That covers the cases where at least two of the numbers are different. For $N$ boxes, that is $\binom{P+Q}{N} \times \binom{P + Q - N}{N}$
Although, if the boxes are identical, you need to eliminate symmetrical solutions as well.
Your result is $(\frac{(P+Q)!}{P!Q!})$ $\times$ $(\frac{1}{K!(P+Q-K)!)})$ $+$ $(\frac{1}{(K-1)!(P+Q-K+1)!})$ $+$ $(\frac{1}{(K-2)!(P+Q-K+2)!})$ $...$ 
