Let A = {1, 2, 3}. The number of relations on | vedclass

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(A) Let $R$ be a relation on $A = \{1, 2, 3\}$.Since $R$ is reflexive,$(1, 1), (2, 2), (3, 3) \in R$.Given $(1, 2) \in R$ and $(2, 3) \in R$,by transi

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$3$

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(A) Let $R$ be a relation on $A = \{1, 2, 3\}$.Since $R$ is reflexive,$(1, 1), (2, 2), (3, 3) \in R$.Given $(1, 2) \in R$ and $(2, 3) \in R$,by transitivity,$(1, 3) \in R$.So far,$R$ must contain the set $S = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$.This set $S$ is already reflexive and transitive. It is not symmetric because $(1, 2) \in S$ but $(2, 1) 
otin S$.We can add other elements from $A 	imes A \setminus S = \{(2, 1), (3, 2), (3, 1)\}$ to $R$ while maintaining transitivity and ensuring it remains non-symmetric.If we add $(2, 1)$,then by transitivity $(1, 2) \in R$ and $(2, 1) \in R \implies (1, 1) \in R$ (already there) and $(2, 1) \in R$ and $(1, 3) \in R \implies (2, 3) \in R$ (already there).If we add $(3, 2)$,then by transitivity $(1, 2) \in R$ and $(3, 2) \in R$ is not possible,but $(2, 3) \in R$ and $(3, 2) \in R \implies (2, 2) \in R$ (already there).If we add $(3, 1)$,then $(2, 3) \in R$ and $(3, 1) \in R \implies (2, 1) \in R$.Testing combinations of $\{(2, 1), (3, 2), (3, 1)\}$:$1$. $R_1 = S \cup \{(2, 1)\}$ (Transitive,reflexive,not symmetric)$2$. $R_2 = S \cup \{(3, 2)\}$ (Transitive,reflexive,not symmetric)$3$. $R_3 = S \cup \{(2, 1), (3, 2), (3, 1)\}$ (This is the universal relation,which is symmetric,so exclude)$4$. $R_4 = S \cup \{(2, 1), (3, 1)\}$ (Transitive,reflexive,not symmetric)$5$. $R_5 = S \cup \{(3, 2), (3, 1)\}$ (Transitive,reflexive,not symmetric)Thus,there are $3$ such relations.

Let $A = \{1, 2, 3\}$. The number of relations on $A$ containing $(1, 2)$ and $(2, 3)$ which are reflexive and transitive but not symmetric is . . . . . . .

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