If $n(A) = 12$, $n(B) = 8$, and $A$ and $B$ are disjoint, then $n(A \cup B)$ is
If $n(A) = 12$, $n(B) = 8$, and $A$ and $B$ are disjoint, then $n(A \cup B)$ is
- $20$
- $0$
- $4$
- $16$
Solution
For disjoint sets $n(A \cap B) = 0$, so $n(A \cup B) = 12 + 8 = 20$.
Asked in: MH-SSC-9
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