Suppose \(A_1,A_2,\ldots,A_{30}\) are 30 sets each with five elements and \(B_1,B_2,B_3,\ldots,B_n\) are n sets (each with three elements) such that \(\bigcup ^{30}_{i=1}{{A}}_i={{\bigcup }}^n_{j=1}{{B}}_i=S\, \) and each element of S belongs to exactly ten of the \(A_i\)'s and exactly 9 of the \(B^{\prime}_j\)'s. Then \(n=\)
Since each \( A_i \) has 5 elements, then: \[ \sum_{i=1}^{30} A_i = 30 \times 5 = 150 \] Let set \( S \) consist of \( m \) elements, and given that each element in \( S \) belongs to exactly 10 of \( A_i \)'s, we have: \[ 10m = 150 \Rightarrow m = 15 \] Now, since each \( B_i \)'s has 3 elements and each element in \( S \) belongs to exactly 9 of \( B_i \)'s, we have: \[ \Rightarrow \sum_{i=1}^n B_i = 3 \times n = 3n = 9m = 135 \] \[ \Rightarrow n = 45 \]
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