Library / /491 of 1394

, 2022

Combinatorics Difficulty 5.1 AIME, harder Prove it United States

Problem:

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 $|A \cap B \cap C|$ .

Solution

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.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.