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Step-by-Step Solution
Step 1: Understand the Problem
We have 30 sets, denoted by $A_{1}, A_{2}, A_{3}, \dots, A_{30}$, each containing 5 elements. We also have $n$ sets, denoted by $B_{1}, B_{2}, \dots, B_{n}$, each containing 3 elements. Let
$$
\bigcup_{i=1}^{30} A_i \;=\;\bigcup_{j=1}^{n} B_j \;=\;S.
$$
It is given that every element of $S$ belongs to exactly 10 of the $A_i$ and exactly 9 of the $B_j$. We want to find the value of $n$.
Step 2: Sum of the Sizes of All Ai
Each $A_i$ has 5 elements and there are 30 such sets. Therefore,
$$
\sum_{i=1}^{30} |A_i| = 30 \times 5 = 150.
$$
Step 3: Relate to Each Element’s Membership in Ai
Let $k$ be the total number of distinct elements in $S$. Each element belongs to exactly 10 of the $A_i$ sets. Thus, from the viewpoint of each element's membership:
$$
\sum_{i=1}^{30} |A_i| = 10\,k.
$$
Step 4: Determine k
From Step 2, we have
$$
\sum_{i=1}^{30} |A_i| = 150.
$$
Hence,
$$
10\,k = 150 \quad \Rightarrow \quad k = 15.
$$
So there are 15 distinct elements in the set $S$.
Step 5: Sum of the Sizes of All Bj
There are $n$ sets $B_{j}$, each with 3 elements. Therefore,
$$
\sum_{j=1}^{n} |B_j| = 3\,n.
$$
However, each element of $S$ is in exactly 9 of the $B_j$ sets. From the element viewpoint:
$$
\sum_{j=1}^{n} |B_j| = 9\,k.
$$
Step 6: Solve for n
Equating the two expressions for the sum of the sizes of all $B_j$:
$$
3\,n = 9\,k.
$$
Since we found $k = 15$, substituting gives:
$$
3\,n = 9 \times 15 \quad \Rightarrow \quad 3\,n = 135 \quad \Rightarrow \quad n = 45.
$$
Final Answer
$n = 45$.
