+355 In how many ways can you distribute 8 indistinguishable balls among 5 distinguishable boxes, if at least one of the boxes must be empty?
+964 There are two cases to consider:
Case 1: One box is empty
There are 5 ways to choose which box will be empty. The remaining 7 balls can be distributed among the remaining 4 boxes using stars and bars formula. Therefore, the total number of ways in this case is 5 * (7+4-1) choose (4-1) = 5 * 10 = 50 ways.
Case 2: Two boxes are empty
There are a total of 5 choose 2 = 10 ways to choose which 2 boxes will be empty. The remaining 6 balls can be distributed among the remaining 3 boxes using stars and bars formula. Therefore, the total number of ways in this case is 10 * (6+3-1) choose (3-1) = 10 * 8 = 80 ways.
Adding the two cases together, the total number of ways to distribute 8 indistinguishable balls among 5 distinguishable boxes if at least one of the boxes must be empty is 50 + 80 = 130 ways.
