If two sets A and B have 99 elements in commo | vedclass

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(B) The number of elements common to $A 	imes B$ and $B 	imes A$ is given by $n((A 	imes B) \cap (B 	imes A))$.Using the property of Cartesian pro

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$99^2$

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(B) The number of elements common to $A 	imes B$ and $B 	imes A$ is given by $n((A 	imes B) \cap (B 	imes A))$.Using the property of Cartesian products,$(A 	imes B) \cap (B 	imes A) = (A \cap B) 	imes (B \cap A)$.Therefore,$n((A 	imes B) \cap (B 	imes A)) = n((A \cap B) 	imes (A \cap B)) = n(A \cap B) 	imes n(A \cap B)$.Given that $n(A \cap B) = 99$,we have $99 	imes 99 = 99^2$.

If two sets $A$ and $B$ have $99$ elements in common,then the number of elements common to each of the sets $A \times B$ and $B \times A$ is

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