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.

Choose which two balls are to be paired, $\binom{5}{2}=10$, and then arrange the four objects, $4!=24$, for a total of $240$.
