Question

If $A$ and $B$ be two sets containing $3$ and $6$ elements respectively, what can be the minimum number of elements in $A\cup B$? Find also, the maximum number of elements in $(A\cup B)$?

• A. $6$ and $9$
• B. $3$ and $9$
• C. $5$ and $9$
• D. none

Answer 1

The correct option is A $6$ and $9$

$\Rightarrow$ $n\left(A\right)=3$ [ Given ]

$\Rightarrow$ $n\left(B\right)=6$ [ Given ]

$\Rightarrow$ $n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$

Now we have already given $n\left(A\right)=3$ and $n\left(B\right)=6$

So $n(A\cup B)$ can be minimum when $n(A\cap B)$ is maximum or these two sets have maximum overlapping or in other words maximum elements in common. $A$ and $B$ can have atmost $3$ elements common.

Then minimum no. of elements of $A\cup B$ can be achieved only when $A\subset B$.

$\Rightarrow$ $min\left(n\right(A\cup B\left)\right)=n\left(A\right)+n\left(B\right)-max\left(n\right(A\cap B\left)\right)=6+3-3=6$

Now max value can be achieved when $n(A\cap B)$ is minimum or $n(A\cap B)=0$

$\Rightarrow$ $max\left(n\right(A\cup B\left)\right)=n\left(A\right)+n\left(B\right)-min\left(n\right(A\cap B\left)\right)=6+3-0=9$