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Exercise 245Combinatorics › Inclusion-exclusionHigh schoolAdded 3 Nov 2026

Proof of the two-set formula

#IE#two sets#proof

Statement

Prove that ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣|A\cup B|=|A|+|B|-|A\cap B|.

Hint

Partition A∪BA\cup B into three disjoint parts: A∖BA\setminus B, A∩BA\cap B, B∖AB\setminus A.

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Solution

Partition. A∪B=(A∖B)∪(A∩B)∪(B∖A)A\cup B=(A\setminus B)\cup(A\cap B)\cup(B\setminus A) (disjoint).

Sizes. ∣A∣=∣A∖B∣+∣A∩B∣|A|=|A\setminus B|+|A\cap B|, ∣B∣=∣B∖A∣+∣A∩B∣|B|=|B\setminus A|+|A\cap B|.

Sum. ∣A∣+∣B∣=∣A∖B∣+∣B∖A∣+2∣A∩B∣=∣A∪B∣+∣A∩B∣|A|+|B|=|A\setminus B|+|B\setminus A|+2|A\cap B|=|A\cup B|+|A\cap B|.

Conclusion. ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣|A\cup B|=|A|+|B|-|A\cap B|.