If AAA and BBB are disjoint, n(AcupB)=n(A \\cup B) = n(AcupB)=
n(A)+n(B)n(A) + n(B)n(A)+n(B)
n(A)+n(B)−1n(A) + n(B) - 1n(A)+n(B)−1
n(A)timesn(B)n(A) \\times n(B)n(A)timesn(B)
0