Let A = {1, 2, 3, 4, 5} and B = {2, 3, 6, 7}. | vedclass

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(C) The Cartesian product of two sets $A$ and $B$ is defined as $A 	imes B = \{(a, b) : a \in A, b \in B\}$.For the intersection $(A 	imes B) \cap (

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$4$

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(C) The Cartesian product of two sets $A$ and $B$ is defined as $A 	imes B = \{(a, b) : a \in A, b \in B\}$.For the intersection $(A 	imes B) \cap (B 	imes A)$,an element $(x, y)$ must satisfy $(x, y) \in A 	imes B$ and $(x, y) \in B 	imes A$.This implies $x \in A, y \in B$ and $x \in B, y \in A$.Therefore,$x \in (A \cap B)$ and $y \in (A \cap B)$.Given $A = \{1, 2, 3, 4, 5\}$ and $B = \{2, 3, 6, 7\}$,the intersection is $A \cap B = \{2, 3\}$.The number of elements in $A \cap B$ is $n(A \cap B) = 2$.The number of elements in $(A 	imes B) \cap (B 	imes A)$ is given by $n(A \cap B) 	imes n(A \cap B) = 2 	imes 2 = 4$.

Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{2, 3, 6, 7\}$. Then the number of elements in $(A \times B) \cap (B \times A)$ is

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