Variation of distributing $k$ balls into $n$ distinguishable boxes I am interested in understanding a variation problem of distributing balls into boxes. It seems to be not any of the individual case mentioned in twelvefold way classification.
The problem is described as follows: 
In total there are $K$ balls, there are $I$ distinguishable groups by color, and in each color group the balls are indistinguishable with number to be $n_i$ where $i = 1, 2, \ldots, I$. Meanwhile, we have $N$ distinguishable boxes where $N$ is always bigger than any $n_i$. Therefore, how many ways are there to distribute these $K$ balls into $N$ boxes, when no box can contain more than one ball from same indistinguishable color group and no empty boxes?
Any suggestion will be appreciated.
 A: This is a straightforward application of the principle of inclusion-exclusion. First, count up all the ways to put all the balls into boxes, ignoring the condition that no box can be empty. This is simply
$$
\prod_{i=1}^I \binom{N}{n_i}
$$
Next, you have to subtract out the "bad" cases where some box is empty. For each of the $N$ boxes, there are $\prod_{i=1}^I \binom{N-1}{n_i}$ ways to place the balls where that box is empty, and you subtract this for each box, so we are left with
$$
\prod_{i=1}^I \binom{N}{n_i}-N\cdot \prod_{i=1}^I \binom{N-1}{n_i}
$$
However, distributions with two empty boxes were subtracted out twice by the above formula, so they must be added back in. The result is
$$
\prod_{i=1}^I \binom{N}{n_i}-N\cdot \prod_{i=1}^I \binom{N-1}{n_i}+\binom{N}2\prod_{i=1}^I \binom{N-2}{n_i}
$$
The summation continues developing in this way. The end result is
$$
\boxed{\sum_{j=0}^N (-1)^j\binom{N}j\prod_{i=1}^I \binom{N-j}{n_i}.}
$$
A: I got $$P = \prod_{i=1}^{I} {N\choose n_i} - \prod_{i=1}^{I} {{N-1}\choose n_i}$$
The first part is obviously the combinations of "no box contains more than one ball of the color".
The second part is "no box contains more than one ball of the color AND there is at least one empty box".
This is the Venn diagram I was going by:
The pink part is what we're looking for.
