Let A = {2, 3, 4} and B = {8, 9, 12}. Then th | vedclass

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(A) We are given $A = \{2, 3, 4\}$ and $B = \{8, 9, 12\}$.We need to find the number of pairs $((a_1, b_1), (a_2, b_2))$ such that $a_1$ divides $b_2$

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$36$

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(A) We are given $A = \{2, 3, 4\}$ and $B = \{8, 9, 12\}$.We need to find the number of pairs $((a_1, b_1), (a_2, b_2))$ such that $a_1$ divides $b_2$ and $a_2$ divides $b_1$,where $a_1, a_2 \in A$ and $b_1, b_2 \in B$.Let $S$ be the set of pairs $(a, b) \in A 	imes B$ such that $a$ divides $b$.For $a = 2$,$b \in \{8, 12\}$ ($2$ pairs: $(2, 8), (2, 12)$).For $a = 3$,$b \in \{9, 12\}$ ($2$ pairs: $(3, 9), (3, 12)$).For $a = 4$,$b \in \{8, 12\}$ ($2$ pairs: $(4, 8), (4, 12)$).Thus,there are $2 + 2 + 2 = 6$ such pairs in $S$.The relation $R$ is defined as the set of pairs $((a_1, b_1), (a_2, b_2))$ such that $(a_1, b_2) \in S$ and $(a_2, b_1) \in S$.Since there are $6$ choices for the pair $(a_1, b_2)$ and $6$ choices for the pair $(a_2, b_1)$,the total number of elements in $R$ is $6 	imes 6 = 36$.

Let $A = \{2, 3, 4\}$ and $B = \{8, 9, 12\}$. Then the number of elements in the relation $R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B) \times (A \times B) : a_1 \text{ divides } b_2 \text{ and } a_2 \text{ divides } b_1\}$ is:

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