Question

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

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

Answer 1

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

Given
$n(A\cap B)=99n\left(\right(A\times B)\cap (B\times A\left)\right)=n\left(\right(A\cap B)\times (B\cap A\left)\right)=n(A\cap B)\times n(B\cap A)={99}^2$