Question #265892

Given three sets A, B, and C. Suppose we know that the union of the three sets has cardinality 182.

Further, |A| = 92, |B| = 41, |C| = 118. Also, |A ∩ B| = 15, |A ∩ C| = 42, and |A ∩ B ∩ C| = 10. Find

|B ∩ C|.


Expert's answer

Let us use the inclusion–exclusion principle:

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


It follows that

∣B∩C∣=∣A∣+∣B∣+∣C∣−∣A∩B∣−∣A∩C∣+∣A∩B∩C∣−∣A∪B∪C∣=92+41+118−15−42+10−182=22|B\cap C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|+|A\cap B\cap C|-|A\cup B\cup C|\\ =92+41+118-15-42+10-182\\=22



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