Competency-Based
Assessment Made Simple.

Step-by-Step Solution

1. First, list all possible pairs of elements from set A and set B: $A = \{2, 3, 6, 7\}$ and $B = \{4, 5, 6, 8\}$.

2. We are given the relation $R$ defined on $A \times B$ such that $(a_1, b_1) R (a_2, b_2)$ if and only if $a_1 + a_2 = b_1 + b_2$. We need to find the number of elements in $R$.

3. Let's consider all possible pairs $(a_1, b_1)$ and $(a_2, b_2)$ from $A \times B$ and check the condition $a_1 + a_2 = b_1 + b_2$. Since we are looking for the number of elements in $R$, we need to find pairs of pairs that satisfy the given condition.

4. Instead of checking all possible pairs of pairs, let's analyze the possible values of $a_1 + a_2$ and $b_1 + b_2$.

   The possible values of $a_1 + a_2$ are: $2+2=4, 2+3=5, 2+6=8, 2+7=9, 3+2=5, 3+3=6, 3+6=9, 3+7=10, 6+2=8, 6+3=9, 6+6=12, 6+7=13, 7+2=9, 7+3=10, 7+6=13, 7+7=14$.

   The possible values of $b_1 + b_2$ are: $4+4=8, 4+5=9, 4+6=10, 4+8=12, 5+4=9, 5+5=10, 5+6=11, 5+8=13, 6+4=10, 6+5=11, 6+6=12, 6+8=14, 8+4=12, 8+5=13, 8+6=14, 8+8=16$.

5. Now, we need to find the number of pairs $(a_1, b_1), (a_2, b_2)$ such that $a_1 + a_2 = b_1 + b_2$.

   Let's consider all pairs $(a, b) \in A \times B$. $A \times B = \{(2,4), (2,5), (2,6), (2,8), (3,4), (3,5), (3,6), (3,8), (6,4), (6,5), (6,6), (6,8), (7,4), (7,5), (7,6), (7,8)\}$. There are $4 \times 4 = 16$ such pairs.

6. We need to find how many pairs of pairs satisfy the condition. This is equivalent to counting the number of pairs $(a_1, b_1), (a_2, b_2)$ such that $a_1 + a_2 = b_1 + b_2$.

7. Instead, let's consider the possible values of $S = a_1 + a_2 = b_1 + b_2$.

   The minimum value of $a_1 + a_2$ is $2+2=4$ and the maximum value is $7+7=14$. The minimum value of $b_1 + b_2$ is $4+4=8$ and the maximum value is $8+8=16$.

   So, we are looking for values of $S$ that are possible for both $a_1 + a_2$ and $b_1 + b_2$. The possible values of $S$ are $8, 9, 10, 12, 13, 14$.

8. Now, let's count the number of ways to obtain each value of $S$:

   • $S=8$: $a_1+a_2=8 \implies (2,6), (6,2)$. $b_1+b_2=8 \implies (4,4)$. Number of pairs: $2 \times 1 = 2$
   • $S=9$: $a_1+a_2=9 \implies (2,7), (3,6), (6,3), (7,2)$. $b_1+b_2=9 \implies (4,5), (5,4)$. Number of pairs: $4 \times 2 = 8$
   • $S=10$: $a_1+a_2=10 \implies (3,7), (7,3)$. $b_1+b_2=10 \implies (4,6), (6,4), (5,5)$. Number of pairs: $2 \times 3 = 6$
   • $S=12$: $a_1+a_2=12 \implies (6,6)$. $b_1+b_2=12 \implies (4,8), (6,6), (8,4)$. Number of pairs: $1 \times 3 = 3$
   • $S=13$: $a_1+a_2=13 \implies (6,7), (7,6)$. $b_1+b_2=13 \implies (5,8), (8,5)$. Number of pairs: $2 \times 2 = 4$
   • $S=14$: $a_1+a_2=14 \implies (7,7)$. $b_1+b_2=14 \implies (6,8), (8,6)$. Number of pairs: $1 \times 2 = 2$

9. Total number of elements in $R$ is $2 + 8 + 6 + 3 + 4 + 2 = 25$.