No more than two of the pairs $\{v, x\}, \{v, y\}, \{w, y\}, \{w, z\}, \{x, z\}$ may be included in a 2-configuration of order 1, since otherwise at least one of $v, w, x, y, z$ would occur more than once. If exactly one is included, say $\{v, x\}$, then $w, y, z$ must be paired with $W, Y, Z$, respectively, and then $V$ and $X$ cannot be paired. So either none or exactly two of the five pairs above must be used. If none, then $v, w, x, y, z$ must be paired with $V, W, X, Y, Z$, respectively, and we have 1 2-configuration arising in this manner. If exactly two are used, we can check that there are 5 ways to do this without duplicating an element: $\{v, x\}, \{w, y\}$, $\{v, x\}, \{w, z\}$, $\{v, y\}, \{w, z\}$, $\{v, y\}, \{x, z\}$, $\{w, y\}, \{x, z\}$. In each case, it is straightforward to check that there is a unique way of pairing up the remaining elements of $A$. So we get 5 2-configurations in this way, and the total is 6.