Number of Ordered Pairs in Relation with Divisibility Condition

To find the number of elements in the relation $R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B) \times (A \times B) : a_1+b_2 \text{ divides } a_2+b_1\}$ where $A = \{1, 4, 7\}$ and $B = \{2, 3, 8\}$, we need to find all ordered pairs $((a_1, b_1), (a_2, b_2))$ such that: $(a_1 + b_2) \mid (a_2 + b_1)$

Here, $a_1, a_2 \in \{1, 4, 7\}$ and $b_1, b_2 \in \{2, 3, 8\}$.

We analyze the condition $(a_1 + b_2) \mid (a_2 + b_1)$ by considering all possible pairs $(a_1, a_2)$:

Case 1: $a_1 = a_2$

The divisibility condition becomes $(a_1 + b_2) \mid (a_1 + b_1)$.

1. Diagonal pairs where $b_1 = b_2$: Since $a_1 + b_1 = a_1 + b_1$, it holds trivially for all $3 \times 3 = 9$ combinations of $(a_1, b_1) = (a_2, b_2)$.

   • Count = 9

2. Off-diagonal pairs where $b_1 \neq b_2$:

   • For $a_1 = 1$: Possible values of $1 + b$ are $\{3, 4, 9\}$. We need $1 + b_2 \mid 1 + b_1$. The only non-trivial divisor pair is $3 \mid 9$, which corresponds to $b_2 = 2$ and $b_1 = 8$.
     • Pair: $((1, 8), (1, 2))$ $\implies (1+2) \mid (1+8)$ i.e., $3 \mid 9$.
     • Count = 1
   • For $a_1 = 4$: Possible values of $4 + b$ are $\{6, 7, 12\}$. The only divisor pair is $6 \mid 12$, corresponding to $b_2 = 2$ and $b_1 = 8$.
     • Pair: $((4, 8), (4, 2))$ $\implies (4+2) \mid (4+8)$ i.e., $6 \mid 12$.
     • Count = 1
   • For $a_1 = 7$: Possible values of $7 + b$ are $\{9, 10, 15\}$. No distinct pair satisfies $7 + b_2 \mid 7 + b_1$.
     • Count = 0

Total for Case 1 = $9 + 1 + 1 = 11$ pairs.

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Case 2: $a_1 \neq a_2$

1. For $(a_1, a_2) = (1, 4)$: We need $(1 + b_2) \mid (4 + b_1)$. Possible values: $1 + b_2 \in \{3, 4, 9\}$ and $4 + b_1 \in \{6, 7, 12\}$.

   • $1 + b_2 = 3 \implies b_2 = 2$: Divides $6$ ($b_1 = 2$) and $12$ ($b_1 = 8$).
     • Pairs: $((1, 2), (4, 2))$ and $((1, 8), (4, 2))$.
   • $1 + b_2 = 4 \implies b_2 = 3$: Divides $12$ ($b_1 = 8$).
     • Pair: $((1, 8), (4, 3))$.
   • Count = 3

2. For $(a_1, a_2) = (4, 1)$: We need $(4 + b_2) \mid (1 + b_1)$. Possible values: $4 + b_2 \in \{6, 7, 12\}$ and $1 + b_1 \in \{3, 4, 9\}$.

   • No value in $\{6, 7, 12\}$ divides any value in $\{3, 4, 9\}$.
   • Count = 0

3. For $(a_1, a_2) = (1, 7)$: We need $(1 + b_2) \mid (7 + b_1)$. Possible values: $1 + b_2 \in \{3, 4, 9\}$ and $7 + b_1 \in \{9, 10, 15\}$.

   • $1 + b_2 = 3 \implies b_2 = 2$: Divides $9$ ($b_1 = 2$) and $15$ ($b_1 = 8$).
     • Pairs: $((1, 2), (7, 2))$ and $((1, 8), (7, 2))$.
   • $1 + b_2 = 9 \implies b_2 = 8$: Divides $9$ ($b_1 = 2$).
     • Pair: $((1, 2), (7, 8))$.
   • Count = 3

4. For $(a_1, a_2) = (7, 1)$: We need $(7 + b_2) \mid (1 + b_1)$. Possible values: $7 + b_2 \in \{9, 10, 15\}$ and $1 + b_1 \in \{3, 4, 9\}$.

   • $7 + b_2 = 9 \implies b_2 = 2$: Divides $9$ ($b_1 = 8$).
     • Pair: $((7, 8), (1, 2))$.
   • Count = 1

5. For $(a_1, a_2) = (4, 7)$: We need $(4 + b_2) \mid (7 + b_1)$. Possible values: $4 + b_2 \in \{6, 7, 12\}$ and $7 + b_1 \in \{9, 10, 15\}$.

   • No divisor relationship exists among these sets.
   • Count = 0

6. For $(a_1, a_2) = (7, 4)$: We need $(7 + b_2) \mid (4 + b_1)$. Possible values: $7 + b_2 \in \{9, 10, 15\}$ and $4 + b_1 \in \{6, 7, 12\}$.

   • No divisor relationship exists among these sets.
   • Count = 0

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Total Number of Elements in $R$:

$\text{Total} = 11 + 3 + 0 + 3 + 1 + 0 + 0 = 18$