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Suppose $$A_1,A_2,\ldots,A_{33}$$ be $$33$$ sets each containing $$6$$ elements and $$B_1,B_2,\ldots,B_n$$ be $$n$$ sets each with $$8$$ elements. If $$\bigcup_{i=1}^{33}A_i=\bigcup_{i=1}^{n}B_i=S$$ and each element of $$S$$ occurs exactly $$9$$ times in $$A_1,\ldots,A_{33}$$ and exactly $$4$$ times in $$B_1,\ldots,B_n$$, then $$n$$ is
Counting incidences in the first family gives $$33\times6=9|S|$$, so $$|S|=22$$. Counting incidences in the second family gives $$8n=4|S|=88$$. Hence $$n=11$$.
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