Re: Please, urgent!!!

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From: Bev Miller
Date: Sun, 23 May 1999 00:00

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Sorry to sound basic, is n the same in each instance?
Samuel Sen� <rene.sene@wanadoo.fr> wrote in message
news:373D368A.6E302B6B@wanadoo.fr...
> Sorry for French people, je n'ai pas le courage de re-traduire le texte
> d'anglais que j'ai d�j� eu du mal � �crire.
>
> > I need help with this problem:
>
> I give a hint for the beginning of the problem. (I think I'm true but
there
> can be an error... but I'm sure it's something like that).
> Let's take s a permutation.
> s(1)<>1, ok. So s(1)=j (we have (n-1) choices). if s(j)=1, that's great
cause
> we have P(n-2) possibilities after. But if s(j)=k, k<>j,1 (n-2 choices),
we
> must go again.
> Let's think a little with that, we find:
> P_n=(n-1)*[P_(n-2)+(n-2)*[P_(n-3)+...]]...] and that's exactly:
> P_n=(n-1)(P_(n-1)+P_(n-2)).
> Great! And now, I'm sure (I didn't try) that a recurrence (I don't know
the
> english word) works to finish the first question. And the second musn't be
too
> hard (a sum, I think...).
> Samuel
>

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From: "Bev Miller" <bev@gwion.demon.co.uk>
Subject: Re: Please, urgent!!!
Date: 1999/05/23
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