The number of relations on the set $$A = \{1, 2, 3\}$$ containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetric, is ______.

The set has three elements, so there are $$3^2 = 9$$ ordered pairs in the universal relation on $$A$$.
For a relation $$R$$ to be

• reflexive   ⇒   $$(1,1),(2,2),(3,3) \in R$$ (always present)
• required   ⇒   $$(1,2) \in R$$ (given in the question)
• of size $$\le 6$$   ⇒   we may choose at most two more pairs
• not symmetric   ⇒   there exists at least one ordered pair whose reverse is absent
• transitive   ⇒   whenever $$(a,b),(b,c)\in R$$, then $$(a,c)\in R$$.

Let

$$R_0=\{(1,1),(2,2),(3,3),(1,2)\} \qquad\text{(4 mandatory pairs)}$$

The five remaining off-diagonal pairs are

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

We may add 0, 1 or 2 of them. Each choice is examined below.

Only $$R_0$$ itself. No new two-step paths are created, so $$R_0$$ is transitive. Because $$(2,1)\notin R_0$$, it is not symmetric. Hence one relation is obtained.

Test every single pair from $$P$$.

1. Add $$(2,1)$$ ⇒ $$(1,2),(2,1)$$ are both present, so the relation becomes symmetric ⇒ reject.

2. Add $$(1,3)$$. Two-step paths are
    $$(1,3)\,(3,\_)$$ gives only loops, and $$(1,2)\,(2,\_)$$ gives existing pairs. Hence transitive and not symmetric ⇒ accept.

3. Add $$(3,1)$$. Path $$(3,1),(1,2)$$ demands $$(3,2)$$, which is missing ⇒ violates transitivity ⇒ reject.

4. Add $$(2,3)$$. Path $$(1,2),(2,3)$$ demands $$(1,3)$$, missing ⇒ reject.

5. Add $$(3,2)$$. Path $$(1,2)$$ is unchanged; path $$(3,2)$$ followed by loops requires nothing new. Relation is transitive and not symmetric ⇒ accept.

Thus two relations are obtained in this case.

Choose two pairs from $$P$$ (there are $$\binom{5}{2}=10$$ possibilities). Only the following three choices satisfy transitivity and lack of symmetry.

(i) $$\{(1,3),(2,3)\}$$ : path $$(1,2),(2,3)$$ already gives $$(1,3)$$, present ⇒ transitive.

(ii) $$\{(1,3),(3,2)\}$$ : path $$(1,3),(3,2)$$ already gives $$(1,2)$$, present ⇒ transitive.

(iii) $$\{(3,1),(3,2)\}$$ : path $$(3,1),(1,2)$$ gives $$(3,2)$$, present ⇒ transitive.

In each of the three relations at least one ordered pair lacks its reverse, so none is symmetric.

No other pair-sets work: each of the remaining seven choices either forces a missing third pair (breaking transitivity) or makes the relation symmetric.

Hence three relations arise in this case.

Adding up all cases: $$1 + 2 + 3 = 6.$$

Therefore, the number of required relations is $$\mathbf{6}.$$