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B b blb c>. F q N b q U X d }. N(AuBuC) = n(A)n(B) n(C)n(AnB)n(Anc)n(BnC)n(AnBnC) As we know n(AuB)=n(A) n(B)n(AnB) Here when we add A and B we are adding (AnB) two times , as it is present in both A and B So we subtract (AnB) one time from A and B Which means we need.

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