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

An equivalence relation corresponds to a partition of the set; count partitions of $\{1,2,3\}$ in which $1$ and $2$ lie in the same block.

The set $A = \{1,2,3\}$ has exactly 5 possible equivalence relations (equal to the number of partitions of a 3-element set, the Bell number $B_3=5$):

1. $\{1\}\{2\}\{3\}$ — finest, only diagonal pairs
2. $\{1,2\}\{3\}$
3. $\{1,3\}\{2\}$
4. $\{2,3\}\{1\}$
5. $\{1,2,3\}$ — the universal relation $A\times A$ …

Equivalence relations containing $(1,2)$ correspond to partitions of $\{1,2,3\}$ that keep $1$ and $2$ in the same block.

An equivalence relation on $A=\{1,2,3\}$ corresponds to a partition of $A$. Since $(1,2)$ must belong to the relation, elements $1$ and $2$ must lie in the same block of the partition.

The possible partitions of $\{1,2,3\}$ with $1,2$ together are:

1. $\{\{1,2,3\}\}$ — the universal relation ($A\times A$).
2. $\{\{1,2\},\{3\}\}$ — $1,2$ related to each other and to themselves, $3$ related only to itself. …

A relation is any subset of $A\times B$; since $A\times B$ has $mn$ pairs, there are $2^{mn}$ relations, option (b).

Concept. A relation from $A$ to $B$ is any subset of the Cartesian product $A\times B$. Counting relations = counting subsets.

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Equivalence relations correspond to partitions of the set; count partitions of $\{4,5,6\}$ where 4 and 5 lie in the same block.

Equivalence relations on a set correspond one-to-one with partitions of that set. The partitions of $A=\{4,5,6\}$ are:

1. $\{4\},\{5\},\{6\}$ — does not contain $(4,5)$ since 4 and 5 are in different blocks.
2. $\{4,5\},\{6\}$ — contains $(4,5)$. ✓
3. $\{4,6\},\{5\}$ — does not contain $(4,5)$.
4. $\{5,6\},\{4\}$ — does not contain $(4,5)$. …

With $|A|=3<|B|=5$, no function from $A$ to $B$ can be onto (there aren't enough domain elements to cover all 5 codomain elements). The number that matches a given option is the count of one-one functions from $A$ to $B$.

Since $O(A)=3$ and $O(B)=5$: an onto function from $A$ to $B$ would need every element of $B$ to have a pre-image in $A$, which is impossible when $|A|<|B|$ — so the number of onto functions from $A$ to $B$ is exactly $0$, which is not among the choices $[30,60,10,45]$.

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The number of relations on a set of $n$ elements is $2^{n^2}$; for $n=3$ this is $2^9$.

A relation on a set $A$ is any subset of $A\times A$. If $|A|=n$, then $|A\times A|=n^2$, and the number of subsets is $2^{n^2}$.

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An equivalence relation must be reflexive, symmetric, and transitive. For $A = \{1,3,5\}$, forcing $(1,3)$ into the relation forces $(3,1)$ by symmetry, and then transitivity forces $1$ and $3$ to be in the same equivalence class. The only freedom is whether $5$ joins that class or stays alone, giving exactly 2 possible relations.

Only one reflexive-symmetric relation on $\{1,2,3\}$ containing $(1,2)$ and $(1,3)$ fails to be transitive — the answer is (a) 1.

Let $A=\{1,2,3\}$ and let $R$ be a relation on $A$ that is reflexive and symmetric and contains $(1,2)$ and $(1,3)$.

Step 1 — Reflexivity forces the diagonal.

Reflexivity requires $(1,1),(2,2),(3,3)\in R$.

Step 2 — Symmetry forces the reverse pairs.

Since $(1,2)\in R$, symmetry forces $(2,1)\in R$. Since $(1,3)\in R$, symmetry forces $(3,1)\in R$.

So every such $R$ must contain at least

$R_0=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}.$

Step 3 — Check transitivity of $R_0$.

$(2,1)\in R_0$ and $(1,3)\in R_0$ but $(2,3)\notin R_0$ — so $R_0$ is not transitive. This gives one valid relation.

Step 4 — Can we add more pairs and still avoid transitivity? …

There are exactly 2 equivalence relations on $A=\{1,2,3\}$ that contain $(1,2)$.

An equivalence relation partitions the set into disjoint blocks, and $(a,b)$ belongs to the relation exactly when $a,b$ lie in the same block. Since $(1,2)$ must be in the relation, elements $1$ and $2$ must lie in the same block.

The possible partitions of $\{1,2,3\}$ with $1$ and $2$ together are: …