At least you are trying for a solution. I wonder however whether I should accept this solution. It seems all right, but it silently assumes the following:
if during the process of 'adding' an element out of the (k1)! available to one of the (k1)! permutations we get a ksubset twice, this would mean that another ksubset does not get attributed and we should be able to settle that.
There is yet another snag: for the moment we have only looked at one subset of k1 elements.There are however quite some subsets of k1 elements. Are we still getting distinct ksubsets ?
