Question

If two sets $A$ and $B$ have $99$ elements in common, then the number of elements common to each of the sets $A\times B$ and $B\times A$ are

• A. $2^{99}$
• B. ${99}^2$
• C. $100$
• D. $18$

Answer 1

The correct option is B ${99}^2$

To find $A\times B$ we take one element from set $A$ and one from set $B$.
Given that $99$ elements are common to both set $A$ and set $B$.
Suppose these common elements are $N_1,N_2,N_3,...N_{99}$.
Select an ordered pair for $A\times B$ such that both are selected out of these common elements. Examples: $(N_1,N_2),(N_3,N_5)$
All these will also be elements of $B\times A$. Hence number of elements common to $A\times B$ and $B\times A$ is $99\times 99={99}^2$ ( first element in ordered pair can be selected in 99 ways; second element can also be selected in 99 ways)
$n[(A\times B)\cap (B\times A)]=n[(A\cap B)\cap (B\cap A)]=\left(99\right)\left(99\right)={99}^2$