1. Sets $A, B$, and $C$ satisfy $|A|=92,|B|=35,|C|=63,|A \cap B|=16,|A \cap C|=51,|B \cap C|=19$. Compute the number of possible values of $|A \cap B \cap C|$.
Answer:
10
Solution: Suppose $|A \cap B \cap C|=n$. Then there are $16-n$ elements in $A$ and $B$ but not $C, 51-n$ in $A$ and $C$ but not $B$, and $19-n$ in $B$ and $C$ but not $A$. Furthermore, there are $25+n$ elements that are only in $A, n$ only in $B$, and $n-7$ that are only in $C$. Therefore, $7 \leq n \leq 16$, so there are 10 possible values.