Probability of selecting jellybeans 8 red and 9 blue jellybeans are distributed randomly to 4 students. What is the probability that each student got at least one jellybean of each color?
I am getting $\binom{7}{3} \binom{8}{3} / \binom{23}{3}$
Is this correct or is my calculation off
 A: The reasoning that led to the answer is not described. Presumably it is some variant of a "Stars and Bars" approach. In that case, the answer obtained should not be right.
We need to make it clear how the distribution takes place. My interpretation is that for every jellybean, the giver decides "at random" who will get it. If so, our probability is the product of the probability that everyone gets at least one red and the probability that everyone gets at least one blue.
Let us start to calculate the probability that everyone gets at least one red. Imagine the red jellybeans have labels. There are $4^8$ equally likely ways to distribute them. Now we could quote a result about Stirling numbers, or do an explicit Inclusion/Exclusion argument. Call our people A, B, C, D. There are $3^8$ ways to distribute the beans so that A gets none. Our first estimate of the number of ways to distribute so that at least one person gets none is $4\cdot 3^8$. However, for each pair of people, we have double-counted the ways in which these two people each get none. So we should subtract $6\cdot 2^8$. But we have subtracted too many times the cases in which $3$ people get none. So we should add back $4\cdot 1^8$.
After a while, we find that the probability that everyone gets at least one red is
$$1-\frac{4\cdot 3^8-6\cdot 2^8+4\cdot 1^8}{4^8}.$$
