+6 In how many ways can you distribute 8 indistinguishable balls among 5 distinguishable boxes, if at least one of the boxes must be empty?
There are two cases to consider:
Case 1: One of the boxes is empty. In this case, there are 4 ways to choose which box is empty, and then 7 ways to distribute the remaining 8 balls among the 4 boxes. So there are 4×7=28 ways in this case.
Case 2: No boxes are empty. In this case, there are 3 ways to choose which box has 1 ball, 2 ways to choose which box has 2 balls, and 1 way to choose which box has 3 balls. So there are 3×2×1=6 ways in this case.
Therefore, the total number of ways to distribute the balls is 28+6=34.
