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If $A$ and $B$ have $n$ elements in common, then the number of elements common to $A \times B$ and $B \times A$ is
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Verified Answer
The correct answer is:
$\mathrm{n}^{2}$
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.
$C \times C=(A \cap B) \times(A \cap B)$
$=(A \cap B) \times(B \cap A)$
$$
\begin{gathered}
=(A \times B) \cap(B \times A) \\
{[\because(A \cap C) \times(B \cap D)=(A \times B) \cap(C \times D)]}
\end{gathered}
$$
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.
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