Octahedronumerical

Consider the dual cube to the octahedron. The vertices A, B, C, D, E, F, G,H of this cube are the centers of the faces of the octahedron (here ABCD is a face of the cube and (A, G), (B, H ), (C, E), (D, F ) are pairs of diagonally opposite vertices). Each assignment of the numbers 1, 2, 3, 4, 5, 6, 7, and 8 to the faces of the octahedron corresponds to a permutation of ABCDEFGH, and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. The problem requires us to count octagonal circuits that can be formed by eight of these diagonals. Six of these diagonals are edges of the tetrahedron ACFH, six are edges of the tetra-hedron DBEG, and four are long diagonals, joining opposite vertices of the cube. Notice that each vertex belongs to exactly one long diagonal. It follows that an octagonal circuit must contain either 2 long diagonals separated by 3 tetrahedron edges or 4 long diagonals alternating with tetrahedron edges. When forming a (skew) octagon with 4 long diagonals, the four tetrahedron edges need to be disjoint; hence two are opposite edges of ACFH and two are opposite edges of DBEG. For each of the three ways to choose a pair of opposite edges from the tetrahedron ACFH, there are two possible ways to choose a pair of opposite edges from tetrahedron DBEG. There are 3 · 2 = 6 octagons of this type, and for each of them, a circuit can start at 8 possible vertices and can be traced in two different ways, making a total of 6 · 8 · 2 = 96 permutations. An octagon that contains exactly two long diagonals must also contain a three-edge path along the tetrahedron ACFH and a three-edge path along tetrahedron the DBEG. A three-edge path along the tetrahedron the ACFH can be chosen in 4! = 24 ways. The corresponding three-edge path along the tetrahedron DBEG has predetermined initial and terminal vertices; it thus can be chosen in only 2 ways. Since this counting method treats each path as different from its reverse, there are 8 · 24 · 2 = 384 permutations of this type. In all, there are 96 + 384 = 480 permutations that correspond to octagonal circuits formed exclusively from cube diagonals. The probability of randomly choosing such a permutation is (480)/(8!)=1/84