1. We need to count the number of ordered pairs $((a_1,b_1),(a_2,b_2)) \in (A\times B)\times (A\times B)$ for which $a_1+a_2=b_1+b_2.$

Equivalently, $a_1-b_1 = b_2-a_2 = -(a_2-b_2),$ so it is easiest to group elements of $A\times B$ by the value of $a-b$.

2. First list all pairs in $A\times B$ and compute $a-b$.

Given $A=\{2,3,6,7\},\qquad B=\{4,5,6,8\}.$

All $16$ pairs $(a,b)$ are:

• For $a=2$: $(2,4): -2,\ (2,5): -3,\ (2,6): -4,\ (2,8): -6$
• For $a=3$: $(3,4): -1,\ (3,5): -2,\ (3,6): -3,\ (3,8): -5$
• For $a=6$: $(6,4): 2,\ (6,5): 1,\ (6,6): 0,\ (6,8): -2$
• For $a=7$: $(7,4): 3,\ (7,5): 2,\ (7,6): 1,\ (7,8): -1$

3. Count frequency of each value of $a-b$.

From the above table:

4. Condition for relation.

For two elements $(a_1,b_1)$ and $(a_2,b_2)$, $a_1+a_2=b_1+b_2$ iff $a_1-b_1 = -(a_2-b_2).$

So if a value $d=a-b$ occurs $f(d)$ times, then it contributes $f(d)\,f(-d)$ ordered pairs to the relation.

Thus total number of elements in $R$ is $\sum_d f(d)f(-d).$

5. Compute the total.

Using the table:

• $d=-6$: $f(-6)f(6)=1\cdot 0=0$
• $d=-5$: $1\cdot 0=0$
• $d=-4$: $1\cdot 0=0$
• $d=-3$: $2\cdot 1=2$
• $d=-2$: $3\cdot 2=6$
• $d=-1$: $2\cdot 2=4$
• $d=0$: $1\cdot 1=1$
• $d=1$: $2\cdot 2=4$
• $d=2$: $2\cdot 3=6$
• $d=3$: $1\cdot 2=2$

Adding, $0+0+0+2+6+4+1+4+6+2=25.$

Hence, $|R|=25.$

6. Comparison with stored answer.

Stored correct answer = $25$. Our derived answer also equals $25$, so they agree.