Let A = \ 1, 2, 3, 4\ . The number of relations on A which are...

Let $A = \{1, 2, 3, 4\}$. We need to find the number of relations on $A$ that are reflexive, symmetric, and contain $(1, 2)$, but are not transitive.

First, let us count the total number of relations on $A$ that are reflexive, symmetric, and contain $(1, 2)$.
For a relation to be reflexive, it must contain all diagonal pairs: $(1, 1), (2, 2), (3, 3), (4, 4)$.
For it to be symmetric, if it contains $(a, b)$, it must contain $(b, a)$. Since $(1, 2)$ is in the relation, $(2, 1)$ must also be in it.
The remaining non-diagonal pairs can be grouped into $5$ symmetric subsets: $\{(1, 3), (3, 1)\}$, $\{(1, 4), (4, 1)\}$, $\{(2, 3), (3, 2)\}$, $\{(2, 4), (4, 2)\}$, and $\{(3, 4), (4, 3)\}$.
Each of these $5$ subsets can either be included or excluded independently.
Thus, the total number of reflexive and symmetric relations containing $(1, 2)$ is $2^5 = 32$.

Next, we subtract the number of these relations that are also transitive.
A relation that is reflexive, symmetric, and transitive is an equivalence relation.
Equivalence relations on $A$ correspond to set partitions of $A$. Since the relation contains $(1, 2)$, the elements $1$ and $2$ must belong to the same block in the partition.
Treating $\{1, 2\}$ as a single element, we are looking for the number of partitions of a $3$-element set ($\{\{1, 2\}, 3, 4\}$).
The number of partitions of a $3$-element set is given by the Bell number $B_3 = 5$.

Therefore, the number of relations that are reflexive and symmetric but not transitive is $32 - 5 = 27$.

Answer: $27$