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:

Step 1: Define Reflexive Property

A relation $R$ on a set $A$ is reflexive if every element in $A$ is related to itself. For the given set $A=\{1{,}2{,}3\}$, this means the pairs $(1{,}1)$, $(2{,}2)$, and $(3{,}3)$ must be present in the relation $R$.

Step 2: Apply Given Conditions and Reflexivity

We start by including the pairs required for reflexivity: $(1{,}1)$, $(2{,}2)$, $(3{,}3)$. We are also given that the relation must contain $(1{,}2)$ and $(2{,}3)$. So, our initial relation $R_0$ includes these five pairs.

Step 3: Apply Transitive Property

A relation is transitive if, whenever $(a,b)$ and $(b,c)$ are in the relation, then $(a,c)$ must also be in the relation. Since $(1{,}2)$ and $(2{,}3)$ are in $R_0$, the pair $(1{,}3)$ must also be in $R_0$ for the relation to be transitive. This adds $(1{,}3)$ to our relation.

Step 4: Form the Minimal Reflexive and Transitive Relation

After applying reflexivity and transitivity, we arrive at the minimal relation $R_1$. We check for any other transitive pairs: $(1{,}1)$ and $(1{,}2) \implies (1{,}2)$ (already present); $(1{,}2)$ and $(2{,}2) \implies (1{,}2)$ (already present); $(2{,}2)$ and $(2{,}3) \implies (2{,}3)$ (already present); $(1{,}1)$ and $(1{,}3) \implies (1{,}3)$ (already present); $(1{,}2)$ and $(2{,}3) \implies (1{,}3)$ (already present). All conditions are met.

Step 5: Check for Symmetry

A relation is symmetric if, whenever $(a,b)$ is in the relation, then $(b,a)$ must also be in the relation. In $R_1$, we have $(1{,}2)$ but not $(2{,}1)$. We also have $(1{,}3)$ but not $(3{,}1)$, and $(2{,}3)$ but not $(3{,}2)$. Therefore, $R_1$ is not symmetric, which satisfies the problem's condition.

Step 6: Consider adding more pairs while maintaining properties

We need to find relations that are reflexive and transitive but NOT symmetric. $R_1$ is one such relation. Now, we consider adding other pairs to $R_1$ while maintaining reflexivity and transitivity, but ensuring it remains non-symmetric. The pairs not in $R_1$ are $(2{,}1), (3{,}1), (3{,}2)$.

Step 7: Analyze adding pairs and their implications

We systematically consider adding the remaining pairs: $(2{,}1)$, $(3{,}1)$, and $(3{,}2)$. 1. If we add $(2{,}1)$ to $R_1$, the relation $R_2 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3), (2{,}1)\}$ is reflexive and transitive. It is not symmetric because $(1{,}3)$ is present but $(3{,}1)$ is not. 2. If we add $(3{,}1)$ to $R_1$, for transitivity, we must also add $(3{,}2)$ (from $(3{,}1)$ and $(1{,}2)$). The relation $R_3 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3), (3{,}1), (3{,}2)\}$ is reflexive and transitive. It is not symmetric because $(1{,}2)$ is present but $(2{,}1)$ is not. 3. If we add $(3{,}2)$ to $R_1$, the relation $R_4 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3), (3{,}2)\}$ is reflexive and transitive. It is not symmetric because $(1{,}2)$ is present but $(2{,}1)$ is not, and $(1{,}3)$ is present but $(3{,}1)$ is not. We can also add $(2{,}1)$ and $(3{,}2)$ to $R_1$. This requires adding $(3{,}1)$ for transitivity. The resulting relation is $R_5 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3), (2{,}1), (3{,}2), (3{,}1)\}$. This relation is reflexive and transitive. It is not symmetric because $(1{,}2)$ and $(2{,}1)$ are present, $(2{,}3)$ and $(3{,}2)$ are present, $(1{,}3)$ and $(3{,}1)$ are present. Wait, this relation IS symmetric. So this case is not valid. Let's re-evaluate. The pairs that can be added are $(2{,}1)$, $(3{,}1)$, $(3{,}2)$. Consider the minimal relation $R_1 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3)\}$. This is one such relation. Now, we can add elements to $R_1$ as long as the properties are maintained and it remains non-symmetric. Possible relations: 1. $R_1 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3)\}$ (Not symmetric, reflexive, transitive) 2. $R_1 \cup \{(2{,}1)\} = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3), (2{,}1)\}$. This is reflexive and transitive. It is not symmetric because $(1{,}3)$ is in it but $(3{,}1)$ is not. (2nd relation) 3. $R_1 \cup \{(3{,}2)\} = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3), (3{,}2)\}$. This is reflexive and transitive. It is not symmetric because $(1{,}2)$ is in it but $(2{,}1)$ is not. (3rd relation) 4. $R_1 \cup \{(2{,}1), (3{,}2)\} = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3), (2{,}1), (3{,}2)\}$. This is reflexive and transitive. It is not symmetric because $(1{,}3)$ is in it but $(3{,}1)$ is not. (4th relation) What about adding $(3{,}1)$? If $(3{,}1)$ is added, then for transitivity, $(3{,}1)$ and $(1{,}2) \implies (3{,}2)$ must be added. Also $(3{,}1)$ and $(1{,}3) \implies (3{,}3)$ (already present). So if $(3{,}1)$ is added, $(3{,}2)$ must also be added. Let's consider the pairs that would make the relation symmetric: $(2{,}1), (3{,}2), (3{,}1)$. We need to ensure that at least one of $(2{,}1), (3{,}2), (3{,}1)$ does not have its symmetric counterpart in the relation, or that if it does, another pair exists without its symmetric counterpart. Let $R_{base} = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3)\}$. This is reflexive and transitive, and not symmetric. We can add any subset of the remaining pairs $(2{,}1), (3{,}1), (3{,}2)$ as long as transitivity is maintained and the relation is not symmetric. Let's list all possible reflexive and transitive relations containing $(1{,}2)$ and $(2{,}3)$: 1. $R_1 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}3), (1{,}3)\}$ (Not symmetric) 2. $R_2 = R_1 \cup \{(2{,}1)\}$ (Not symmetric, since $(1{,}3)$ is in it but $(3{,}1)$ is not) 3. $R_3 = R_1 \cup \{(3{,}2)\}$ (Not symmetric, since $(1{,}2)$ is in it but $(2{,}1)$ is not) 4. $R_4 = R_1 \cup \{(2{,}1), (3{,}2)\}$ (Not symmetric, since $(1{,}3)$ is in it but $(3{,}1)$ is not) 5. $R_5 = R_1 \cup \{(3{,}1), (3{,}2)\}$ (Adding $(3{,}1)$ implies adding $(3{,}2)$ for transitivity with $(1{,}2)$). This relation is not symmetric because $(1{,}2)$ is in it but $(2{,}1)$ is not. 6. $R_6 = R_1 \cup \{(2{,}1), (3{,}1), (3{,}2)\}$ (Adding $(3{,}1)$ implies adding $(3{,}2)$ for transitivity with $(1{,}2)$). This relation is symmetric because all pairs $(a,b)$ have $(b,a)$ present. For example, $(1{,}2)$ and $(2{,}1)$, $(2{,}3)$ and $(3{,}2)$, $(1{,}3)$ and $(3{,}1)$. So this relation is symmetric and must be excluded. So, there are 5 such relations.

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