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. $18$

Answer 1

Answer: (B)

Here, given that $n(A\cap B)=99$
Now, $(A\times B)\cap (B\times A)=n(A\cap B)\times (B\cap A)$
$=n(A\cap B)\cdot n(B\cap A)$
$=n(A\cap B)\cdot n(A\cap B)$
$=99\cdot 99=99^2$