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 $b_2$ and $a_2$ divides $b_1}$ is :
Correct Answer:
C
Explanation
Given sets are $A = {2, 3, 4}$ and $B = {8, 9, 12}$. We want to find the number of elements of the form $((a_1, b_1), (a_2, b_2))$ such that $a_1$ divides $b_2$ and $a_2$ divides $b_1$.
For the first condition, $a_1$ divides $b_2$, with $a_1 \in A$ and $b_2 \in B$, we can list the pairs: $(a_1, b_2) \in {(2, 8), (2, 12), (3, 9), (3, 12), (4, 8), (4, 12)}$. This gives $6$ pairs.
For the second condition, $a_2$ divides $b_1$, we can similarly list the pairs. This again has $6$ valid pairs.
Now, for every pair from the first condition, we can have any pair from the second condition. This leads to $6 \times 6 = 36$ relations.