How many ways can you arrange 7 distinct balls into 5 distinct boxes such that at least two of the boxes are empty?
Here is my solution, which seems too big an answer:
I'm not sure if this is valid but here is my attempt:
I will do "TOTAL - (no box empty + one box empty)"
TOTAL: Each ball has $5$ possible choices (boxes) they can go in. Thus the total is simply $5^7$
NO EMPTY: Pair each box with a ball. There are $5!$ ways of doing this (first ball has 5 options, next 4, next 3 etc..). That's 5 balls dealt with after matching them up with boxes, so two left. We deal with these by $5^2$ since they can go in any box (condition has already been satisfied by previous). Thus the total is $5! \cdot 5^2$
ONE EMPTY: First choose a box to be empty $\binom{5}{1}$. The other boxes must be filled - so "matching up balls with boxes" again: $4!$. Then the other 3 balls $5^3$. Thus the total is $\binom{5}{1}\cdot 4! \cdot 5^3$.
My answer is $5^7 - \left(5! \cdot 5^2 + \binom{5}{1}\cdot 4! \cdot 4^3\right) = 60125$