If $n(A \cup B) = 18$, $n(A) = 10$, $n(B) = 12$, then $n(A \cap B)$ is

If $n(A \cup B) = 18$, $n(A) = 10$, $n(B) = 12$, then $n(A \cap B)$ is
  1. $4$
  2. $2$
  3. $22$
  4. $8$

Solution

$n(A \cap B) = n(A) + n(B) - n(A \cup B) = 10 + 12 - 18 = 4$.

Asked in: MH-SSC-9

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