# Twenty apples are placed randomly into five boxes. What is the expected number of boxes that contain exactly four apples?

I understand the given solution for this problem but am having issues understanding why the way I did it won't work. Using 'stars and bars' there are $${24 \choose 4}$$ possible arrangements for this scenario. If we let $${A_k}$$ be the event that 4 balls end up in the $$k^{th}$$ box then I figured that since there are $${19 \choose 3}$$ ways of ordering the remaining 16 balls (again from 'stars and bars') then it should be that: $$P(A_k) = \frac{19 \choose 3}{24 \choose 4}$$

If we then let $$I_k = I(A_k)$$ where $$I$$ is the indicator function we should have $$X = \sum_{k = 1}^{5} {I_k}$$, where $$X$$ is the random variable denoting the number of boxes with 4 balls. Therefore $$E(X) = E(\sum_{k = 1}^{5} {I_k})$$ which is simply $$\frac{5 {19 \choose 3}}{24 \choose 4}$$. This isn't correct however. I am assuming I have a misunderstanding to do with the selection of $${19 \choose 3}$$ but I can't quite get my head around it. Any help appreciated.

Your $$24 \choose 4$$ possible arrangements aren't all equally likely. A simpler example: suppose you have 2 apples placed randomly into 2 boxes. Stars and bars tells you that there are 3 possible arrangements: one in each box, two in the first box, or two in the second box. But the first of these options is twice as likely as each of the others.