Let $\mathrm{A}=\{2,3,4\}$ and $\mathrm{B}=\{8,9,12\}$. Then the number of elements in the relation $$\mathrm{R}=\left\{\left(\left(a_{1}, \mathrm{~b}_{1}\right),\left(a_{2}, \mathrm{~b}_{2}\right)\right) \in(A \times B, A \times B): a_{1}\right.$$ divides $\mathrm{b}_{2}$ and $\mathrm{a}_{2}$ divides $\left.\mathrm{b}_{1}\right\}$ is :

<p>Given sets : <br/>$ A = {2,3,4} $ <br/>$ B = {8,9,12} $</p> <p>We want to find the number of elements of the form $( (a_1, b_1), (a_2, b_2) )$ such that :</p> <ol> <li>$ a_1 $ divides $ b_2 $</li> <li>$ a_2 $ divides $ b_1 $</li> </ol> <p>For the first condition : <br/>$ a_1 $ divides $ b_2 $ <br/>Given $ a_1 \in A $ and $ b_2 \in B $, we can list the pairs: <br/>$ (a_1, b_2) \in {(2,8),(2,12),(3,9),(3,12),(4,8),(4,12)} $ <br/>This gives 6 pairs.</p> <p>For the second condition, the pairs are the same, because it&#39;s just the reversed relation. So : <br/>$ a_2 $ divides $ b_1 $ <br/>Again has 6 valid pairs.</p> <p>Now, for every pair from the first condition, we can have any pair from the second condition. This leads to : <br/>$ 6 \times 6 = 36 $ relations.</p>