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

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

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

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

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

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Solution

The correct option is C $6 \leq n (A \cup B) \leq 9$
$n (A) = m_{1}$
n(B)= $m_{2}$
we know that
n(A $\cup$ B)=n(A)+n(B)-n(A $\cap$ B)
$∴$ the greatest value of n(A $\cup$ B) is when
$n (A \cap B) = 0$
$∴$ n(A $\cup$ B)= $m_{1}$ + $m_{2}$
$∴$ least value of $n (A \cup B)$ = min(A,B)
$∴$ least value of n(A $\cup$ B) = $m_{2}$ ....when $n (A \cap B)$ = m $_{1}$
=m $_{1}$ ....when n $n (A \cap B)$ = $m_{2}$

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