Let $A \times B$ denote the Cartesian product of sets $A$ and $B$. If the cardinality of $A \cap B$ is $10$, then the cardinality of $(A \times B) \cap (B \times A)$ is:

The standard property of set theory states that for any sets $A, B, C,$ and $D$:

$(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$

In this case, replacing $C$ with $B$ and $D$ with $A$:

$(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$

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The intersection of sets is commutative, meaning $A \cap B = B \cap A$.

Therefore:

$(A \times B) \cap (B \times A) = (A \cap B) \times (A \cap B)$

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The cardinality of a Cartesian product $X \times Y$ is the product of the cardinalities of the individual sets:

$n(X \times Y) = n(X) \times n(Y)$

Applying this to the simplified expression:

$n((A \times B) \cap (B \times A)) = n((A \cap B) \times (A \cap B))$

$n((A \times B) \cap (B \times A)) = n(A \cap B) \times n(A \cap B)$

$n((A \times B) \cap (B \times A)) = [n(A \cap B)]^2$

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Given that $n(A \cap B) = 10$:

$n((A \times B) \cap (B \times A)) = 10 \times 10$

$n((A \times B) \cap (B \times A)) = 100$

Therefore, the cardinality of $(A \times B) \cap (B \times A)$ is $100$.