Thread: Combinatorics
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Old 2006-05-09, 11:53   #5
Kees's Avatar
Dec 2005

3048 Posts

you are right, the second part of my answer is gibberish at best...

what jebeagles is showing is the following:

we have the set {1,...,k!+k-1} and we take some subset of k-1 elements.
say {1,...,k-1}.
To this subset we can 'add' k! different elements. Here we have our first problem. The number of sets {1,...,k-1} including permutations is (k-1)!

How to add the remaining elements? (we had k!=(k-1)! * k elements remaining after selecting k-1 elements). So which do we select ? And what happens to the selectionprocedures for other subsets (for instance {1,...,k-2,k}) after this selection. Can we still fill. I am not really satisfied with the answer, actually. All that seems to be shown is that the two sets have the same size (which I offered at the beginning of the problem). If that was the end to it, we would obviously have a bijection. But the extra condition makes this unclear for me.
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