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n(A ∪ B) =

An(A) + n(B)
Bn(A) + n(B) - n(A ∩ B) (inclusion-exclusion)
C|A| × |B|
Dn(A ∩ B)
Answer & Solution
Correct answer: B. n(A) + n(B) - n(A ∩ B) (inclusion-exclusion)
**Inclusion-exclusion:** union = sum minus intersection (avoid double counting). **For 3 sets:** n(A∪B∪C) = sum singles - sum pairs + n(A∩B∩C).
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