Let A, B, C be three sets such that n(A cap B | vedclass

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(D) We know that for any sets $A, B, C$,the intersection of Cartesian products is given by the formula: $(A 	imes B) \cap (B 	imes C) = (A \cap B) \

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$4$

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(D) We know that for any sets $A, B, C$,the intersection of Cartesian products is given by the formula: $(A 	imes B) \cap (B 	imes C) = (A \cap B) 	imes (B \cap C)$.Given that $n(A \cap B) = 2$ and $n(B \cap C) = 2$.Therefore,the number of elements in the intersection is $n((A \cap B) 	imes (B \cap C)) = n(A \cap B) 	imes n(B \cap C)$.Substituting the given values,we get $2 	imes 2 = 4$.Thus,$n((A 	imes B) \cap (B 	imes C)) = 4$.

Let $A, B, C$ be three sets such that $n(A \cap B) = n(B \cap C) = n(C \cap A) = n(A \cap B \cap C) = 2$. Then $n((A \times B) \cap (B \times C))$ is equal to -

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