Thread: Combinatorics
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Old 2006-05-09, 08:38   #4
Kees's Avatar
Dec 2005

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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 (k-1)! available to one of the (k-1)! permutations we get a k-subset twice, this would mean that another k-subset 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 k-1 elements.There are however quite some subsets of k-1 elements. Are we still getting distinct k-subsets ?
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