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If two sets $A$ and $B$ have 99 elements in common, then the number of elements common to the sets $A \times B$ and $B \times A$ is
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Verified Answer
The correct answer is:
$99^{2}$
Since, $n(A \cap B)=99$
$n((A \times B) \cap(B \times A))$
$$
\begin{aligned}
&=n((A \cap B) \times(B \cap A)) \\
&=n(A \cap B) \cdot n(B \cap A) \\
&=n(A \cap B) \cdot n(A \cap B)=99 \cdot 99=(99)^{2}
\end{aligned}
$$
$n((A \times B) \cap(B \times A))$
$$
\begin{aligned}
&=n((A \cap B) \times(B \cap A)) \\
&=n(A \cap B) \cdot n(B \cap A) \\
&=n(A \cap B) \cdot n(A \cap B)=99 \cdot 99=(99)^{2}
\end{aligned}
$$
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