Question

how many ways are there to put $4$ balls into $3$ boxes, given that the balls are not distinguished and neither are the boxes?

Answer 1

Explanation:

think that there are 15 ways ... call the boxes A, B, and C.

You could put all four b***s into either box A, or box B, or box C ---> 3 ways.

You could put three ball into one box and the other ball into another box:

-- 3 into A, 1 into B, 0 into C; or 3 into A, 0 into B, 1 into C; or 1 into A, 3 into B, 0 into C; or 1 into A, 0 into B, 3 into C; or 0 into A, 3 into B, 1 into C; or 0 into A, 1 into B, 3 into C ---> 6 ways

You could put two balls into one box and the other two balls into another box:

-- 2 into A, 2 into B, 0 into C; or 2 into A, 0 into B, 2 into C; or 0 into A, 2 into B, 2 into C ---> 3 ways.

You could put two b***s into one box and one ball into each of the other two boxes:

-- There are three boxes that could get the two b***s ---> 3 ways

Total: 3 + 6 + 3 + 3 = 15 ways.