King, Peter R
peter.king at imperial.ac.uk
Tue Sep 1 17:08:54 UTC 2009
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.
From: ringing-theory-bounces at bellringers.net [ringing-theory-bounces at bellringers.net] On Behalf Of Fred Bone [Fred.Bone at dial.pipex.com]
Sent: Tuesday, September 01, 2009 5:49 PM
To: ringing-theory at bellringers.net
Subject: Re: [r-t] Cattermole
On 01 September 2009 at 11:14, Robin Woolley said:
> Reading the late Paul Cattermole's obituary in the paper on Saturday
> reminded me of a problem he set some years ago. The problem was: How many
> derangements are there on n bells, where a derangement is a row with no
> bell in its own position?
The number of derangements of n items is (n!/e). The Wikipedia article on
derangements has a good derivation of this (basically, consider swapping
the nth item with the item that belongs in the nth place: the result can
be reduced either to a derangement of n-1 or one of n-2, in n-1 ways
each; then massage the recurrence relation).
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