Given 5 different balls and 5 different boxes, in how many ways can you put the balls in the boxes such that at most 3 boxes are empty?

Question

My answer: As, at least 2 boxes should be not empty, we need to have 2 boxes with one ball each. No of ways in which we can do that is $\binom52^22$ (ways of choosing 2 balls * ways of choosing 2 boxes * ways of placing the chosen balls in the box). We will have 3 balls remaining, which can be placed in $5^3$ ways. $$2\binom525^3= 8×5^5$$ But the given answer is $5^5 - 5$. Where did I go wrong?

Answer 1

Obviously the answer cannot be greater than $5^5$, the number of ways to put the balls in the boxes without restrictions. Now the only way to have more than 3 boxes empty is to have 4 empty and the remaining box having all the balls, which can occur in 5 ways, hence the given correct answer of $5^5-5$.