Suppose A₁, A₂, ... ₃₀ are thirty sets, each with five elements and B₁, B₂, ...., B_n are n sets each with three elements. Let $\bigcup_{i=1}^{30} A_i= \bigcup_{j=1}^{n} Bj= S$. If each element of S belongs to exactly ten of the Ai' s and exactly nine of the Bj' 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 \(n\times 3=135 \Rightarrow n=\dfrac{135}{3}=\boxed{45}.\)