Let $\mathrm{A}=\{2,3,4\}$ and B={8,9,12}. Then the number of elements in the relation $\mathrm{R}=\left\{\left(\left({\mathrm{a}}_1,{\mathrm{b}}_1\right),\left({\mathrm{a}}_2,{\mathrm{b}}_2\right)\right)\in (\mathrm{A}\times \mathrm{B},\mathrm{A}\times \mathrm{B}):{\mathrm{a}}_1\right.$ divides b₂ and a₂ divides b₁} is

$$\begin{gathered} \begin{array}{l}{a}_1\in A,b_2\in B\\ a_1\text{ divides }{b}_2\\ \mathrm{Each} \mathrm{element} \mathrm{has} 2 \mathrm{choices}\\ \left(a_1,b_2\right)\in \left\{\right(2,4),(2,1 2),(3,9),(3,12),(4,8),(4,12\left)\right\}\end{array} \\ {a}_2\in A,b_1\in B and {\mathrm{a}}_2\text{ divides }{\mathrm{b}}_1 \end{gathered}$$

$\therefore The \mathrm{the} \mathrm{number} of \mathrm{elements} \mathrm{in} \mathrm{the} \mathrm{relation} R is 6\times 6=36$