Your results have gone too far. The graph corresponding to this point is not an equivalence problem. You finally 100! /2 only excludes two reverse permutations respectively. You think these pictures are different, but there are many similarities, because the circle itself is not only axisymmetric, but also centrally symmetric. You divide by two only to consider the symmetry of the circle. If the number of points you choose is 1, your 100! /2 includes the same figure that will include other odd-numbered points.