Q.Let $A = \{1, 2, 3\}$. Then number of relations containing $(1, 2)$ and $(1, 3)$ which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4

Concept: Relation Counting — We count relations by deciding which ordered pairs must be present, which must be absent, and which are optional, while satisfying given properties.

Step 1: Reflexive requirement

All $(a,a)$ for $a\in A$ must be in the relation. So $(1,1), (2,2), (3,3)$ are forced.

Step 2: Symmetric requirement

Given $(1,2)$ and $(1,3)$ are already included, symmetry forces $(2,1)$ and $(3,1)$ as well.

Step 3: Not transitive condition

The relation must fail transitivity. The only possible failure is with the pair $(2,3)$ (and its symmetric counterpart $(3,2)$). …

We count relations $R$ on $A = \{1,2,3\}$ that (i) contain $(1,2)$ and $(1,3)$, (ii) are reflexive, (iii) are symmetric, and (iv) are not transitive.

Which pairs are forced?

• Reflexive $\Rightarrow$ $R$ must contain $(1,1),\ (2,2),\ (3,3)$.
• Given pairs $(1,2)$ and $(1,3)$ must be in $R$.
• Symmetric $\Rightarrow$ their reverses $(2,1)$ and $(3,1)$ must be in $R$.

So every valid relation must contain these seven pairs:

The only pairs left to decide are $(2,3)$ and $(3,2)$. By symmetry they must be both in or both out — giving just two candidate relations.

Candidate 1: neither $(2,3)$ nor $(3,2)$ included

$R_1 = \{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}.$ …

Method: Counting relations with prescribed properties

Use this when asked "how many relations on a set satisfy conditions X, Y, Z?" — the trick is to separate the pairs you are forced to include from the pairs you are free to choose.

Steps

Step 1: Write down the forced pairs

• Reflexivity forces every diagonal pair $(a,a)$.
• Any pairs the question says the relation must contain are forced.
• Symmetry then forces the reverse of each forced non-diagonal pair.

Step 2: Identify the remaining "free" pairs

List the ordered pairs not yet decided. Symmetry usually links them in couples $\{(a,b),(b,a)\}$ that must be both-in or both-out, which cuts the number of independent choices. …

Common Mistakes

Mistake 1: Forgetting that symmetry forces the reverse of the given pairs

Why it's wrong: students count relations containing $(1,2)$ and $(1,3)$ but forget symmetry compels $(2,1)$ and $(3,1)$ too, so they overcount the "free" pairs. Correct approach: close the given pairs under symmetry and reflexivity first, then count only what is genuinely undecided.

Mistake 2: Overlooking that adding both $(2,3)$ and $(3,2)$ makes the relation transitive …