To find the number of elements in the relation $$R$$ defined on $$A \times B$$, we need to determine all pairs $$((a_1, b_1), (a_2, b_2))$$ such that $$a_1 + a_2 = b_1 + b_2$$, where $$a_1, a_2 \in A$$ and $$b_1, b_2 \in B$$.

First, consider all possible sums of pairs from set $$A$$ and set $$B$$.

Possible sums from set $$A = \{2, 3, 6, 7\}$$:

• $$2 + 2 = 4$$
• $$2 + 3 = 5$$
• $$2 + 6 = 8$$
• $$2 + 7 = 9$$
• $$3 + 2 = 5$$
• $$3 + 3 = 6$$
• $$3 + 6 = 9$$
• $$3 + 7 = 10$$
• $$6 + 2 = 8$$
• $$6 + 3 = 9$$
• $$6 + 6 = 12$$
• $$6 + 7 = 13$$
• $$7 + 2 = 9$$
• $$7 + 3 = 10$$
• $$7 + 6 = 13$$
• $$7 + 7 = 14$$

Possible sums from set $$B = \{4, 5, 6, 8\}$$:

• $$4 + 4 = 8$$
• $$4 + 5 = 9$$
• $$4 + 6 = 10$$
• $$4 + 8 = 12$$
• $$5 + 4 = 9$$
• $$5 + 5 = 10$$
• $$5 + 6 = 11$$
• $$5 + 8 = 13$$
• $$6 + 4 = 10$$
• $$6 + 5 = 11$$
• $$6 + 6 = 12$$
• $$6 + 8 = 14$$
• $$8 + 4 = 12$$
• $$8 + 5 = 13$$
• $$8 + 6 = 14$$
• $$8 + 8 = 16$$

Now, identify the common sums from both sets:

Common sums: $$8, 9, 10, 12, 13, 14$$

For each common sum, count the pairs from set $$A$$ and set $$B$$ that produce these sums:

• Sum = 8: From $$A$$: {(2,6), (6,2)} - 2 pairs; From $$B$$: {(4,4)} - 1 pair; Hence, 2 * 1 = 2 pairs
• Sum = 9: From $$A$$: {(2,7), (3,6), (6,3), (7,2)} - 4 pairs; From $$B$$: {(4,5), (5,4)} - 2 pairs; Hence, 4 * 2 = 8 pairs
• Sum = 10: From $$A$$: {(3,7), (7,3)} - 2 pairs; From $$B$$: {(4,6), (5,5), (6,4)} - 3 pairs; Hence, 2 * 3 = 6 pairs
• Sum = 12: From $$A$$: {(6,6)} - 1 pair; From $$B$$: {(4,8), (6,6), (8,4)} - 3 pairs; Hence, 1 * 3 = 3 pairs
• Sum = 13: From $$A$$: {(6,7), (7,6)} - 2 pairs; From $$B$$: {(5,8), (8,5)} - 2 pairs; Hence, 2 * 2 = 4 pairs
• Sum = 14: From $$A$$: {(7,7)} - 1 pair; From $$B$$: {(6,8), (8,6)} - 2 pairs; Hence, 1 * 2 = 2 pairs

Adding all these, we get the number of elements in the relation $$R$$:

2 + 8 + 6 + 3 + 4 + 2 = 25

Thus, the number of elements in $$R$$ is 25.