Question

In how many ways can $4$ balls be placed in $3$ boxes if the balls are distinguishable, and the boxes are indistinguishable, and no box can be empty?

Answer 1

First, we need to consider that each box must have at least one ball. This means that there are three possibilities for each ball:

1. The first ball can be placed in any of the three boxes.
2. The second ball can be placed in any of the three boxes.
3. The third ball can be placed in any of the three boxes.
4. The fourth ball can be placed in any of the three boxes.

Therefore, there are $3 \times 3 \times 3 \times 3 = 3^4 = 81$ ways to place $4$ distinguishable balls in $3$ indistinguishable boxes such that no box is empty.

Answer 2

no, you have to consider the case where only three balls are placed.
There are 3^3 = 27 of those

Answer 3

You are correct, I missed the case where only three balls are placed in the boxes.

So, there are 27 ways to place 3 distinguishable balls in 3 indistinguishable boxes and 81 ways to place 4 distinguishable balls in 3 indistinguishable boxes such that no box is empty.

Thus, the total number of ways to place 4 distinguishable balls in 3 indistinguishable boxes, without any box being empty, is 81 (for 4 balls) + 27 (for 3 balls) = 108 ways.