Exercise · ZW-INCLUSION-EXCLUSION-PRINCIPLE-006 · ★★☆ Intermediate

Suppose \(|A|=10\), \(|B|=8\), \(|C|=6\), \(|A\cap B|=3\), \(|A\cap C|=2\), \(|B\cap C|=1\) and \(|A\cap B\cap C|=1\). What is \(|A\cup B\cup C|\)?

Choose one option.

Show answer
  • C. \(19\)

Explanation

\(10+8+6-3-2-1+1=19\). The final \(+1\) restores what was subtracted too often in the three pairwise intersections.

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Review the concept: Inclusion-exclusion principle

It counts the elements of the union of finite sets \(A\), \(B\), \(C\), … by adding their sizes and correcting what was counted too often in the intersections.

\[\begin{aligned}|A\cup B|&=|A|+|B|-|A\cap B|\\|A\cup B\cup C|&=|A|+|B|+|C|\\&\quad-|A\cap B|-|A\cap C|-|B\cap C|\\&\quad+|A\cap B\cap C|\end{aligned}\]

More practice problems: inclusion-exclusion principle

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