Balls in five non-empty boxes

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In how many ways can $20$ identical balls be distributed into $5$ distinct boxes, if no box may stay empty?

Solution

Since each box must contain at least one ball, we immediately place one ball in each of the $5$ boxes: $20-5=15$ remain to be distributed freely. More directly, we use the formula for distributions of $k$ identical objects into $n$ boxes with no empties, which amounts to choosing $n-1$ separators among the $k-1$ gaps: $N = \binom{k-1}{n-1} = \binom{20-1}{5-1} = \binom{19}{4} = \frac{19\cdot 18\cdot 17\cdot 16}{24} = \boxed{3876}.$