# If $S = A_1 \cup A_2 \cup \cdots \cup A_6 = B_1 \cup \cdots \cup B_n$, find $n$

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 $n$.

Hint: We know that there are in total $6\cdot4=24$ (not necessarily different) elements in all A's together. (i.e. $\sum_{i=1}^6 |A_i|=24$) However, each element of $S$ appears in exactly four of the A's, so $S$ contains 6 elements. Now apply a similiar reasoning to the B's.
• Thus $n = 9$ right? – user19405892 Feb 15 '16 at 20:40
• How do we know there are $24$ elements in all A's together? its only the maximum value right and even if union of all A's together has $24$ elements can you explain how you concluded $S$ has $6$ elements – Umesh shankar Apr 27 '17 at 2:00
• @Umeshshankar I mean that there are $6 \cdot 4 = 24$ not-necessarily different values in all $A$s together, before taking the union. Since since each element of $S$ appears in exactly 4 $A$s (given), $S$ contains $\frac{24}{4} = 6$ elements. – wythagoras Apr 27 '17 at 7:02