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Question

Let A={x:(x1)(x210+21)=0} and B={y:y is a prime factor of 42}, then the number of common elements between A×B and B×A is

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Solution

A={x:(x1)(x210+21)=0}
(x1)(x210+21)=0
(x1)(x27x3x+21)=0
(x1)(x7)(x3)=0
x=1,3,7

B={y:y is a prime factor of 42}
Prime factors of 42 is 2,3,7

Common elements between A and B is 2.
The number of common elements between A×B and B×A is 22=4

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