In how many ways can I distribute 5 distinguishable balls into 4 distinguishable boxes such that no box is left empty. So what I think is the way to solve this question is to first count the total number of ways of putting all the balls into boxes such that the restriction isn't satisfied (i.e. the total number of ways of putting the balls into the boxes). Then using the inclusion-exclusion principle for situations where the boxes are empty.
I think the number of total ways that you can distribute them would be $4^5$ but I'm unsure how to do the second part of the question.

One box will have two balls. Choose which box that will be, choose two balls to put in it, then put one ball in each of the remaining boxes.
$$\binom{4}{1}\binom{5}{2}\times 3!=4\times 10\times 6=240$$
