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

Answer 1

We can use stars and bars to solve this problem. Since no box can be empty, each box must contain at least one ball. Let's first distribute one ball to each box, leaving us with $1$ ball to distribute among $3$ boxes.

Using stars and bars, we can represent this as placing $2$ dividers among $1$ ball to divide the balls into $3$ groups. There are $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$ ways to do this.

Therefore, there are $6$ ways to place $4$ indistinguishable balls in $3$ distinguishable boxes such that no box is empty.