The above picture shows a sliding puzzle. This puzzle forms some Graph $G(V,E)$, where the number of vertices are given by all injective maps from the vertex set $V := \{1,2,3,4,5\}$ into the state set, such as $|V| = |V \hookrightarrow [2]\times[3]| = 6! = 720$. How many edges are there? I thought, the number of edges can be given by ${720\choose2}$, because each edge represents a transition of two states, but I'm missing something.
Once again as an answer, the set has $4$ vertices with a degree of $2$ and $2$ with a degree of three. This represents exactly $\frac{1}{3}$ and $\frac{2}{3}$ of the state space. As every edge can assign still a connection between two vertices only, the result is $\frac{(240 \cdot 3 + 480 \cdot 2)}{2}$.