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A set A has 3 elements and another set B has 6 elements. Then

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$3 \leq n (A \cup B) \leq 6$

$0 \leq n (A \cup B) \leq 9$

$3 \leq n (A \cup B) \leq 9$

$6 \leq n (A \cup B) \leq 9$

answer is C.

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Detailed Solution

We know , $min n (A \cup B) = max {n (A), n (B)} = max {3, 6} = 6$

$\begin{aligned} Max n (A \cup B) = n (A) + n (B) = 9 \\ ∴ 6 \leq n (A \cup B) \leq 9 \end{aligned}$

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