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Henryk Flachowsky [1] ; Marko Riedel [1] ; Stefanie Reim [1] ; Magda-Viola Hanke [1] [1] Federal Centre for Breeding Research on Cultivated Plants Institute of Fruit Breeding Localización: Electronic Journal of Biotechnology, ISSN-e 0717-3458, Vol. 11, Nº. 1, 2008, págs.
HINT: Start with a box of $i$ balls numbered from $1$ through $i$, so that they can be individually identified. There are $\binom{i}k$ ways to choose $k$ of them to move to a second box. There are then $\binom{k}j$ ways to choose $j$ of the balls in the second box to move to a third box. That leaves $k-j$ balls in the second box; there are $2^{k-j}$ ways to pick a subset of them to move to a fourth box. Thus, for each possible choice of $k$ there are possible different outcomes.
For each $a \in \{1,2,\ldots,n\}$, there are $k+1$ possibilities: $a$ can be element of $S_1$, of $S_2$, ..., of $S_n$ or of none of the $S_i$. It is not possible that $a$ is an element of more than one of the $S_i$, since that would contradict the condition $S_i \cap S_j$ for $i \neq j$. Multiplying these $k+1$ possibilities for all $a \in \{1,2,\ldots,n\}$ yields the answer of $(k+1)^n$ possibilities.
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