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 $n=$

Let \(|S|=m\).

Count incidences via the \(A_i\): There are 30 sets each of size 5, so total memberships \(=30\times 5=150\). Each element of \(S\) lies in exactly 10 of the \(A_i\), so also \(= m\times 10\). Hence \(m=\dfrac{150}{10}=15\).

Count incidences via the \(B_j\): There are \(n\) sets each of size 3, so total memberships \(=n\times 3\). Each element of \(S\) lies in exactly 9 of the \(B_j\), so also \(= m\times 9 = 15\times 9=135\).

Thus \(n\times 3=135 \Rightarrow n=\dfrac{135}{3}=\boxed{45}.\)