How many ways are there to put 4 balls in 3 boxes if the balls are not distinguishable and the boxes are not distinguishable?
With empty boxes allowed, you have: 4 0 0 = 3 permutations 3 1 0 = 6 p 2 2 0 = 3 p 2 1 1 = 3 p
Total = 3 + 6 + 3 + 3 = 15 ways
