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

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

A
$3 \leq n (A \cup B) \leq 6$
B
$3 \leq n (A \cup B) \leq 9$
C
$6 \leq n (A \cup B) \leq 9$
D
$0 \leq n (Λ \cup B) \leq 9$

Solution:

$n (A \cup B)$ cannot exceed $n (A) + n (B)$ and cannot

be less than $n (B)$ .

$∴ 6 \leq n (A \cup B) \leq 3 + 6 i.e., 6 \leq n (A \cup B) \leq 9$

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