
EXAMPLE

For n=2, we are constructing a square from 4 labeled linesegments with labeled endpoints. Solutions which differ by a rotation are considered equivalent, but solutions which are a reflection of each other are considered distinct (assume the square we are constructing is embedded in a plane, so we cannot flip it over to convert a lefthanded solution to righthanded solution). There are 6 ways to order the linesegments, and each linesegment can be oriented in 2 ways, so the total number of solutions is 6 * 2^4 = 96. For n=3, we are constructing a cube from 6 labeled squares with labeled vertices (assume we are confined to 3space, so we consider reflections of the cube to be distinct). Without loss of generality, we can pick one labeled square to serve as our face of reference. For this face, we must decide which side of the square will face the interior of the cube, but we do not care about which rotation we pick as these just translate into rotations of the cube. From this reference square, there are 5! ways to assign the remaining squares to the faces of the cube, and each square can be oriented in 8 ways (we can pick which side of the square will face the interior of the cube, and we can pick from 4 rotations). This gives 2 * 8^5 * 5! solutions. The factor of "2" comes from the choice of which side of the reference square will face the interior of the cube (a choice which would go away if we considered reflections to be equivalent).
