Suppose $A_1, \ldots , A_6$ are six sets each with four elements and $B_1, \ldots, B_n$ are $n$ sets each with two elements. Let $S = A_1 \cup A_2 \cup \cdots \cup A_6 = B_1 \cup \cdots \cup B_n$. Given that each of the elements of $S$ belongs to exactly four of the $A$’s and to exactly three of the $B$’s, find n.
The first thing I tried is looking at cardinality because we are trying to prove equality by the axiom of extension. We have $\left | A_1 \cup A_2 \cup \cdots \cup A_6 \right| \leq 24$ and $\left |B_1 \cup \cdots \cup B_n \right | \leq 2n$. So let's suppose $\left |S \right| = m$ where $m \leq 2n$ and $m \leq 12$. I am trying to figure out how to utilize the last sentence . It seems if we can find $m$ we will be able to find $n$.