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

Answer 1

To solve this problem, we can use stars and bars method with a twist. We can first distribute one ball to each box to ensure that no box is empty. This reduces the problem to distributing the remaining ball among the boxes.

With the remaining ball, we can think of distributing it as distributing two balls among three boxes (a total of 2 balls). Using stars and bars, we can represent this as placing two dividers among the balls and boxes. There are two indistinguishable dividers and three indistinguishable boxes, so the number of ways to distribute the remaining ball is $\binom{2 + 3 -1}{3-1} = \binom{4}{2} = 6$.

Therefore, the total number of ways to place $4$ distinguishable balls in $3$ distinguishable boxes, with no box empty, is $3 \times 6 = \boxed{18}$.