A combinatorial problem about distributing different kinds of balls to different people Assume there are k kinds of balls. The number of the $i$-th kind of ball is $a_i$, thus there are $\sum_{i=1}^k a_i$ balls in total. The same kind of balls are identical. There are m different men. The $j$-th man takes $b_j$($b_j\neq0$) balls and $\sum_{i=1}^k a_i=\sum_{j=1}^m b_j$, so all balls are taken. Then in how many ways can these balls be distributed?
 A: Fixed $i$, we want to know the number of possible distributions of the $a_i$ identical "$i$-balls" among the $m$ "men" (???), which I assume to be enumerated: man 1, man 2,$\ldots$, man $m$. If the $r$-th man receives $c_{ri}$ $i$-balls, then you have $c_{1i}+c_{2i}+\cdots+c_{mi}=a_i$. The number of nonnegative (i.e., perhaps some men get no $i$-th ball) solutions of this equation is well-known to be $\binom{a_i+m-1}{m-1}$.
Obviously the distribution of the different $i$-balls are independent, so the total number of distributions is equal to $$\prod_{i=1}^k\binom{a_i+m-1}{m-1}\ldots$$ 
but beware! you are including the case in that some unlucky man get no balls at all (what a pity!). How to count these unfortunate cases? If man $i_1,\ldots,$ man $i_t$ (with $1\leq t\leq m-1$) are unlucky, then you can repeat the argument above, with the $m-t$ remaining men, so the number of ways to distribute all the balls among these guys is 
$$\prod_{i=1}^k\binom{a_i+m-t-1}{m-t-1}\,.$$
Note that among these distributions there are additional unlucky men, but this does not matter. Now apply inclusion-exclusion to determine the number of ways in that some guy is inlucky.
P.S. I find my solution rather ugly, I will try hard to find a clever argument.
