Question

Suppose $A_1,A_2,A_{30}$ are thirty sets each with five elements and $B_1,B_2,.,B_n$ are in sets each with three elements. Let $\stackrel{30}{\underset{i=1}{\cup }}Ai=\stackrel{n}{\underset{j=1}{\cup }}{B}_j=S$. Assume the each element of $S$ belongs to exactly ten of the $A_i$ and exactly $9$ of $B_{j}s$. Find n.

Answer 1

Since each $A{A}_i$ has $5$ elements, we have $\stackrel{30}{\underset{i=1}{\Sigma }}n\left(A_i\right)=5\times 30=150$ ......(1)

Let S consist of m distinct elements, Since each element of S belongs to exactly $10$ of the $A_i^{\prime }s,$ we also have $\stackrel{30}{\underset{i=1}{\Sigma }}n\left(A_i\right)=10m$ ....(2)

Hence from (1) and (2), $10m=150$ or $m=15.$

Again since each $B_i$ has $3$ elements and each element of S belongs to exactly $9$ of the $B_i^{\prime }S,$ we have $\stackrel{n}{\underset{j=1}{\Sigma }}n\left(B_i\right)=3n$ and $\stackrel{n}{\underset{j=1}{\Sigma }}n\left(B_j\right)=9m$

It follows that $3n=9\times 15[\therefore m=15]$

This gives $n=45$