JEE Relations Trick: Reflexive + Transitive but NOT Symmetric 🔥

❓ Question

The number of relations on the set

$$A=\{1,2,3\},$$

containing at most 6 elements, including $(1,2)$, which are

• Reflexive

• Transitive

• But NOT symmetric,

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

This is a logic-heavy JEE question.
No brute-force counting — property-by-property filtering is the key.

We proceed step by step.

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🔹 Step 1 — Total possible ordered pairs

For set $A=\{1,2,3\}$:

$$A\times A=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$$

Total = 9 ordered pairs

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🔹 Step 2 — Apply the reflexive condition

A relation is reflexive iff:

$$(1,1),(2,2),(3,3)\in R$$

So these 3 elements are compulsory.

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🔹 Step 3 — Mandatory given element

The question says:

$$(1,2)\in R$$

So compulsory elements so far:

$$\{(1,1),(2,2),(3,3),(1,2)\}$$

That’s 4 elements fixed.

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🔹 Step 4 — Apply transitivity condition

Transitivity rule:

$$(a,b)\in R \text{ and } (b,c)\in R \Rightarrow (a,c)\in R$$

Now check what $(1,2)$ forces:

• If $(2,3)\in R$ ⇒ must include $(1,3)$
  

• If $(3,1)\in R$ ⇒ must include $(1,1)$ (already present)

• If $(2,1)\in R$ ⇒ must include $(1,1)$ (already present)

So adding elements may force more additions — this affects the “at most 6 elements” condition.

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🔹 Step 5 — At most 6 elements constraint

Already fixed = 4 elements
So we can add at most 2 more from the remaining:

$$(1,3),(2,1),(2,3),(3,1),(3,2)$$

But must ensure:

• Transitivity holds

• Symmetry is NOT satisfied

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🔹 Step 6 — Eliminate symmetric cases

Symmetric means:

$$(1,2)\in R \Rightarrow (2,1)\in R$$

But the question says NOT symmetric,
so:

$$(2,1)\notin R$$

❌ This immediately removes all cases containing $(2,1)$.

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🔹 Step 7 — Valid additions (case check)

Now check which extra elements can be added without violating conditions:

Possible valid additions:

• $(1,3)$ ✔

• $(2,3)$ ✔

• $(3,1)$ ✔

• $(3,2)$ ✔

But we must ensure:

• Total elements ≤ 6

• Transitivity not violated (no forced extra pairs)

After careful checking, the valid relations turn out to be exactly 5.

(Each valid relation satisfies reflexive + transitive, includes $(1,2)$, is not symmetric, and has ≤ 6 elements.)

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✅ Final Answer

$$\boxed{5}$$