If A∩B=∅A \cap B = \emptysetA∩B=∅, then n(A∪B)n(A \cup B)n(A∪B) equals:
n(A)+n(B)n(A) + n(B)n(A)+n(B)
n(A)−n(B)n(A) - n(B)n(A)−n(B)
n(A)×n(B)n(A) \times n(B)n(A)×n(B)
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