There are two sections on this homework exercise I'm stuck on. I'll appreciate some help!
We have 8 different boxes and 15 different glasses, divided between them. 3. What is the probability exactly two boxes will remain empty? 4. What is the probability the first 4 boxes would have an equal number of glasses?
For 3., I figured you have $C(8,6)$ ways to choose the non-empty boxes, and then you can multiply that by the number of ways to divide 15 different glasses in 6 boxes such that no box is empty. However, I don't know how many ways there are to do that (I only know the formula when the glasses are equivalent)
EDIT regarding 3.: Earlier today I had asked this question Ways to put numbered balls in boxes no box being empty, which has since gotten an answer. Using that I can solve 3. now. I'm still interested if there's an easier calculation, though!
For 4., I'm quite lost. I tried a number of methods, none of them seem to work. Only progress I really made is, you need to divide the question into cases: 1 glass in each of the first 4 boxes, 2 in each, and 3 in each.
I'd appreciate any help I can get! Thanks a lot.