1.If AAA and BBB are not disjoint, then n(A∪B)n(A \cup B)n(A∪B) is equal to —a.n(A)+n(B)n(A) + n(B)n(A)+n(B)b.n(A)+n(B)−n(A∩B)n(A) + n(B) - n(A \cap B)n(A)+n(B)−n(A∩B)c.n(A)+n(B)+n(A∩B)n(A) + n(B) + n(A \cap B)n(A)+n(B)+n(A∩B)d.n(A)⋅n(B)n(A) \cdot n(B)n(A)⋅n(B)Login to continueOnly logged in users canattempt or see the solution.
