"be consistent"  
PerlMonks 
Re: A bad shuffleby QM (Parson) 
on Mar 24, 2005 at 00:28 UTC ( #441927=note: print w/ replies, xml )  Need Help?? 
The size of this sample space is NN. To each element of this space corresponds a permutation, but the size of the space of all possible permutations is N! , which is not only smaller than N^{N} for any N > 1, but more importantly, it is not a divisor of N^{N}, which means that it is impossible for the algorithm to give equal weight to all the permutations.If I'm not mistaken, N!  N^{N}, as N! = 1*2*...*N, and N^{N} = N*N*...*N. Therefore, N^{N}/N! = N^{N1}/(N1)! How does this affect your comment? Update: Oops, I was mistaken. QM
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