Question

Let $A$ and $B$ be two sets containing $4$ and $7$ elements respectively. If the minimum and maximum number of elements in $A\cup B$ are $m$ and $n$ respectively, then $m+n$ is

• A. $18$
• B. $19$
• C. $20$
• D. $21$

Answer 1

The correct option is A $18$

$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$
For maximum value of $n(A\cup B)$, $n(A\cap B)$ should be minimum, i.e., $n(A\cap B)=0$
$\Rightarrow max\left(n\right(A\cup B\left)\right)=n\left(A\right)+n\left(B\right)\therefore n=4+7=11$

For minimum value of $n(A\cup B)$, $n(A\cap B)$ should be maximum, i.e., $n(A\cap B)=4$
$\Rightarrow min\left(n\right(A\cup B\left)\right)=n\left(A\right)+n\left(B\right)-4\therefore m=4+7-4=7$

Hence, $m+n=7+11=18$