JEE Main 2025 Apr 7 Shift 1, Mathematics Q22: Relations

Setting up the problem. The relation $R$ must be reflexive, so it must contain $(1,1),(2,2),(3,3)$. It must also contain $(1,2)$ (Given). That fixes 4 pairs. Since $|R|\le 6$, we can add at most 2 more pairs, chosen from the remaining 5 candidates:

$\text{Candidates}=\{(2,1),(1,3),(3,1),(2,3),(3,2)\}$

Two shortcuts to speed up checking:

• Non-symmetric: Since $(1,2)\in R$, the relation is automatically non-symmetric as long as we do not add $(2,1)$. If we do add $(2,1)$, we must double-check.
• Transitivity: For each pair $(a,b),(b,c)\in R$, we need $(a,c)\in R$ too. If a chain forces a missing pair, the relation fails.

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Case 1 — Add 0 extras (relation size = 4).

Only pair we need to check is $(1,2)\circ (2,2)\Rightarrow (1,2)$ ✓. Non-symmetric since $(2,1)\notin R$.

✅ 1 valid relation.

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Case 2 — Add exactly 1 extra (relation size = 5). Check each candidate:

Extra added	Transitivity check	Symmetry check	Valid?
$(2,1)$	All required pairs present for transitivity	Now $(1,2)$ and $(2,1)$ both present $\Rightarrow$ symmetric	❌
$(1,3)$	All required pairs present for transitivity	Non-symmetric ✓	✅
$(3,1)$	$(3,1)\circ (1,2)\Rightarrow (3,2)$ needed, but missing	—	❌
$(2,3)$	$(1,2)\circ (2,3)\Rightarrow (1,3)$ needed, but missing	—	❌
$(3,2)$	All required pairs present for transitivity	Non-symmetric ✓	✅

✅ 2 valid relations.

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Case 3 — Add exactly 2 extras (relation size = 6). There are $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ pairs of candidates. Check each:

Extras added	Transitivity check	Symmetry check	Valid?
$\{(2,1),(1,3)\}$	$(2,1)(1,3)\Rightarrow (2,3)$ missing	—	❌
$\{(2,1),(3,1)\}$	$(3,1)(1,2)\Rightarrow (3,2)$ missing	—	❌
$\{(2,1),(2,3)\}$	$(1,2)(2,3)\Rightarrow (1,3)$ missing	—	❌
$\{(2,1),(3,2)\}$	$(3,2)(2,1)\Rightarrow (3,1)$ missing	—	❌
$\{(1,3),(3,1)\}$	$(3,1)(1,2)\Rightarrow (3,2)$ missing	—	❌
$\{(1,3),(2,3)\}$	$(1,2)(2,3)\Rightarrow (1,3)$ ✓ present. All required pairs present for transitivity.	Non-symmetric ✓	✅
$\{(1,3),(3,2)\}$	$(1,3)(3,2)\Rightarrow (1,2)$ ✓ present. All required pairs present for transitivity.	Non-symmetric ✓	✅
$\{(3,1),(2,3)\}$	$(2,3)(3,1)\Rightarrow (2,1)$ missing	—	❌
$\{(3,1),(3,2)\}$	$(3,1)(1,2)\Rightarrow (3,2)$ ✓ present. All required pairs present for transitivity.	Non-symmetric ✓	✅
$\{(2,3),(3,2)\}$	$(2,3)(3,2)\Rightarrow (2,2)$ ✓, but $(1,2)(2,3)\Rightarrow (1,3)$ missing	—	❌

✅ 3 valid relations.

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Grand total:

Case	Count
Size 4 (no extras)	$1$
Size 5 (one extra)	$2$
Size 6 (two extras)	$3$
Total	$6$

$\therefore$ Answer = $\boxed{6}$.