Question

If two sets $A$and $B$ are having $99$ 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. $18$
• E. $9$

Answer 1

The correct option is B

${99}^2$

Explanation for the correct answer:

Finding the number of elements common to each of the sets $A\times B$and $B\times A$ :

Given,

$\begin{array}{rcl}n(A\cap B)& =& 99\end{array}$

We know,

$\begin{array}{rcl}(A\times B)\cap (C\times D)& =& (A\cap C)\times (B\cap D)\\ (A\times B)\cap (B\times A)& =& (A\cap B)\times (B\cap A)\\ & =& 99*99\\ & =& {99}^2\end{array}$

Hence, the correct option is (B) .