Let A={2,3,4} and B={8,9,12}. Then the number of elements in the relation R=a1,b1,a2,b2∈(A×B,A×B):a1 divides b2 and a2 divides b1} is

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Let $A = {2, 3, 4}$ and B={8,9,12}. Then the number of elements in the relation $R = \{((a_{1}, b_{1}), (a_{2}, b_{2})) \in (A \times B, A \times B) : a_{1}$ divides b₂ and a₂ divides b₁} is

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Detailed Solution

$\begin{array}{l} a_{1} \in A, b_{2} \in B \\ a_{1} divides b_{2} \\ Each element has 2 choices \\ (a_{1}, b_{2}) \in {(2, 4), (2, 1 2), (3, 9), (3, 12), (4, 8), (4, 12)} \end{array} a_{2} \in A, b_{1} \in B a n d a_{2} divides b_{1}$