Q.Show that the number of equivalence relation in the set $\{1, 2, 3\}$ containing $(1, 2)$ and $(2, 1)$ is two.

Concept: Bell Numbers — the number of equivalence relations on an $n$-element set is the $n$th Bell number $B_n$. Here we count only those that contain the given ordered pairs.

Step 1: An equivalence relation containing $(1,2)$ and $(2,1)$ forces $1$ and $2$ to be in the same equivalence class. So we must have $\{1,2\}$ together.

Step 2: The element $3$ can either join the class $\{1,2\}$ or form its own separate class. These are the only possibilities.

Step 3: …

Equivalence relations on a set correspond exactly to partitions of that set. Each equivalence class is a block of the partition. So instead of counting relations directly, we count partitions — that’s the clean way.

The given condition says $(1,2)$ and $(2,1)$ are in the relation. For an equivalence relation, this means $1$ and $2$ are related to each other. Since equivalence is transitive, any element related to $1$ or $2$ must also be in the same block. So $1$ and $2$ are forced into the same equivalence class.

Now the only question is: where does $3$ go?

1. Case 1: $3$ is in the same class as $1$ and $2$.

   Then the partition is $\{\{1,2,3\}\}$ — a single block. This gives exactly one equivalence relation (the universal relation). It certainly contains $(1,2)$ and $(2,1)$.

2. Case 2: $3$ is in its own separate class.

   Then the partition is $\{\{1,2\}, \{3\}\}$ — two blocks. This also gives an equivalence relation, and it contains $(1,2)$ and $(2,1)$ because $1$ and $2$ are together.

No other partition is possible: $1$ and $2$ cannot be separated, and $3$ either joins them or doesn’t. So there are exactly two partitions, hence exactly two equivalence relations. …

Method: Counting Equivalence Relations via Partitions

The cleanest way to count equivalence relations on a finite set is to count partitions instead, because every equivalence relation corresponds to exactly one way of splitting the set into disjoint classes (and vice versa).

Steps

Step 1: Convert the relation condition into a class condition.

A pair $(a,b)$ in an equivalence relation means $a$ and $b$ share a class. Required pairs therefore merge elements: containing $(1,2)$ and $(2,1)$ forces $1$ and $2$ into the same block.

Step 2: Fix what the constraints force in the partition.

Group together every element that is forced to be related. Here $\{1,2\}$ must sit in one block.

Step 3: Enumerate the free choices for the remaining elements. …

Common Mistakes

Mistake 1: Thinking $(1,2)$ and $(2,1)$ only add symmetry, not merge classes.

Why it's wrong: in an equivalence relation those pairs mean $1$ and $2$ are related, which forces them into the same equivalence class — the real constraint is on the partition, not just symmetry. Correct approach: treat the given pairs as merging $1$ and $2$ into one block.

Mistake 2: Counting all 5 equivalence relations on $\{1,2,3\}$. …