If \(A\) and \(B\) have \(n\) elements in common, then the number of elements common to \(A \times B\) and \(B \times A\) is

Let \(\quad \mathrm{C}=\mathrm{A} \cap \mathrm{B}\), then
\(C \times C=(A \cap B) \times(A \cap B)\)
\(=(A \cap B) \times(B \cap A)\)
\[
\begin{gathered}
=(A \times B) \cap(B \times A) \\
{[\because(A \cap C) \times(B \cap D)=(A \times B) \cap(C \times D)]}
\end{gathered}
\]
Since, \(A \cap B\) has \(\mathrm{n}\) elements, so \(C\) has n elements. Hence, \(\mathrm{C} \times \mathrm{C}\) has \(\mathrm{n}^{2}\) elements. \(\therefore \quad(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\) has \(\mathrm{n}^{2}\) elements.
Hence, \(\mathrm{A} \times \mathrm{B}\) and \(\mathrm{B} \times \mathrm{A}\) has \(\mathrm{n}^{2}\) elements in common.