This problem can be solved using stars and bars method. Since no box can be empty, we must distribute all $4$ balls among the $3$ boxes.
We can think of this problem as distributing $4$ identical balls into $3$ distinct boxes. Using stars and bars method, we can insert $2$ dividers (bars) among the $4$ balls to separate them into $3$ groups representing the $3$ boxes.
For example, if the $4$ balls were represented by o's and the dividers were represented by |'s, we could have an arrangement like:
oo|oo||
This represents $2$ balls in the first box, $2$ balls in the second box, and $0$ balls in the third box.
The total number of ways to arrange the $4$ balls and $2$ dividers is given by the formula $\binom{4+3-1}{3-1} = \binom{6}{2} = 15$.
Therefore, there are $15$ ways to place $4$ balls in $3$ boxes if neither the balls nor the boxes are distinguishable.
In how many ways can $4$ balls be placed in $3$ boxes if neither the balls nor the boxes are distinguishable, and no box can be empty?
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