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

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Question

(B) The number of elements common to the Cartesian products $A 	imes B$ and $B 	imes A$ is given by the intersection of the two sets:$n((A 	imes B)

Vedclass · Accepted Answer

$99^2$

Vedclass · Answer

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

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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