I need some closure on this one: In how many ways can the vertices of an n-cube be labeled 0, … ,2ⁿ - 1 so that there is an edge between two vertices if and only if the binary representation of their labels differs in exactly one bit? Let G be an appropriately labeled n-cube. Pick an arbitrary vertex v in G. How many ways can one change labels of the vertices appart from v in G and still mantain the properties of the n-cube? Because of the nature of G, there are n incident edges on v with n adjacent vertices. The labels of these vertices differ from the label of v by one bit. If one swaps the labels of two of these vertices and makes the appropriate swaps elsewhere so as to maintain the n-cube (which can be done due to the symmetry of G), one obtains a new labeling. The number of combination of adjacent vertices that can be swapped is n(n - 1)/2, so the number of different labelings of the vertices in G without changing the label of v is n(n - 1)/2. The label of vertex v is arbitrary. Therefore, for every possible label of v there are n(n - 1)/2 labelings of the vertices in G. Because there are 2ⁿ - 1 possible labels for v, the number of possible labelings for the vertices in G is (2ⁿ - 1)(n - 1)n/2.