JEE Main 2024 — Sets And Relations Question with Solution

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.