Suppose \(A_1, A_2,\ldots,A_{30}\) are thirty sets each having 5 elements and \(B_1, B_2,\ldots,B_n\) are \(n\) sets each with 3 elements.
Let
\[ \bigcup_{i=1}^{30} A_i=\bigcup_{j=1}^{n} B_j=S \]
and each element of \(S\) belongs to exactly 10 of the \(A_i\)’s and exactly 9 of the \(B_j\)’s, then \(n\) is equal to
(a) 15
(b) 3
(c) 45
(d) 35
Solution
Total number of element occurrences in all \(A_i\)’s:
\[ 30\times5=150 \]
If \(S\) has \(k\) elements and each element occurs in 10 sets,
\[ 10k=150 \]
\[ k=15 \]
Now total number of element occurrences in all \(B_j\)’s:
\[ n\times3 \]
Each element of \(S\) occurs in 9 sets,
\[ 9\times15=n\times3 \]
\[ 135=3n \]
\[ n=45 \]
Answer
\[ \boxed{45} \]
Correct option: (c)