Q.Let $A = \{1, 2, 3\}$. Then number of equivalence relations containing $(1, 2)$ is (A) 1 (B) 2 (C) 3 (D) 4

Concept: Relation Counting — For equivalence relations, containing $(1,2)$ forces $1$ and $2$ to be in the same equivalence class. We count all possible partitions of $\{1,2,3\}$ that keep $1$ and $2$ together.

Step 1: Since $(1,2)$ is in the relation, $1$ and $2$ must belong to the same equivalence class. The possible classes containing both are $\{1,2\}$ or $\{1,2,3\}$.

Step 2:

• If the class is $\{1,2\}$, the remaining element $3$ can either form its own class $\{3\}$ (giving partition $\{\{1,2\},\{3\}\}$) or join the same class (giving $\{\{1,2,3\}\}$).
• If the class is $\{1,2,3\}$, that's the same as the second case above — no new partition. …

We are counting equivalence relations on a 3-element set that contain the ordered pair $(1,2)$. An equivalence relation is the same as a partition of the set into disjoint, non-empty subsets (the equivalence classes). The pair $(1,2)$ being in the relation means $1$ and $2$ belong to the same class. So the problem reduces to: How many partitions of $\{1,2,3\}$ have $1$ and $2$ in the same block?

Let’s reason step by step.

1. Reflexivity is forced. Every equivalence relation on $A$ must contain $(1,1)$, $(2,2)$, and $(3,3)$. These are automatic and don’t affect the count — they are always present.

2. Symmetry is forced for the given pair. Since $(1,2)$ is in the relation, symmetry demands $(2,1)$ must also be present. So the pair $(1,2)$ and its symmetric counterpart $(2,1)$ are locked in.

3. Transitivity will now decide the rest. With $1$ and $2$ in the same class, the only question is: where does $3$ go?

   • Case 1: $3$ is in its own separate class. Then the partition is $\{\{1,2\}, \{3\}\}$. This is a valid equivalence relation.
   • Case 2: $3$ joins the class containing $1$ and $2$. Then the partition is $\{\{1,2,3\}\}$ — a single class containing all three elements. This is also valid.

   Are there any other possibilities? Could $3$ be in a class with only one of $1$ or $2$? No — because if $3$ were in the same class as $1$ but not $2$, then transitivity would force $2$ and $3$ to be related (since $1 \sim 2$ and $1 \sim 3$ implies $2 \sim 3$), collapsing the classes. So the only two partitions are the ones listed. …

Method: Counting equivalence relations via partitions

Use this whenever you must count equivalence relations on a finite set, possibly subject to "must contain the pair $(a,b)$".

Steps

Step 1: Translate equivalence relations into partitions

Every equivalence relation on a set corresponds to exactly one partition of that set into disjoint non-empty blocks (its equivalence classes), and vice versa. So counting equivalence relations = counting partitions.

Step 2: Apply the constraint to the blocks

If the relation must contain $(a,b)$, then $a$ and $b$ must lie in the same block. This removes some partitions from consideration — only those keeping $a,b$ together survive. …

Common Mistakes

Mistake 1: Trying to count by choosing individual pairs instead of partitions

Why it's wrong: building equivalence relations pair-by-pair almost always misses the transitivity closure and produces sets that aren't actually transitive. Correct approach: use the fact that equivalence relations correspond exactly to partitions, and count partitions instead — transitivity is then automatic.

Mistake 2: Forgetting that $(1,2)$ forces $1$ and $2$ into the same block …