Fred.Bone at dial.pipex.com
Wed Sep 2 12:20:01 UTC 2009
On 01 September 2009 at 18:08, King, Peter R said:
> Can this be right? Is e here the basis of natural logs (2.718...)? If so
> this means the number of derangements is an irrational number rather than
> an integer. Surely this isn't correct. I think it is the nearest integer
> to n!/e.
The limit as n tends to infinity of (n! / derangements(n)) is e.
As the number of derangements is obviously an integer, I thought it
unnecessary to specify the rounding.
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