Re: A Question about Combinatorica
- To: mathgroup at smc.vnet.net
- Subject: [mg102657] Re: [mg102627] A Question about Combinatorica
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 17 Aug 2009 08:07:06 -0400 (EDT)
- References: <200908170804.EAA27022@smc.vnet.net>
On 17 Aug 2009, at 10:04, Marwa Abd El-Wahaab wrote:
> Dear Sir,
> I have a question about having five letters like {A, B, C, D, E}. In
> order
> to get all possibilities, we have 5! possible cases like ABCDE,
> EABCD,.......etc
>
> The number of these possibilities are 120. How and why this number
> becomes
> 60 by dividing by 2 ?
>
> What are 60 possibilities & how extract them from 120?
>
> I used this function to get 120:
>
> MinimumChangePermutations[{A,B,C,D,E}]
>
> What should I do after this to get 60?
>
> Thanks too much
>
> I really need your help
>
> *Marwa Ali Abd El Wahaab*
> *Teaching Assistant*
> Faculty of Engineering
> Mansoura University
>
>
I find it impossible to understand what you are talking about ...
except one possibility: the number of even permutations (which make up
the so called "alternating" subgroup of the symmetric group) is half
the number of all permutations, i.e. n!/2, as you can see from
<< Combinatorica`
Length[AlternatingGroup[5]]
60
Is that what you had in mind?
Andrzej Kozlowski
- References:
- A Question about Combinatorica
- From: Marwa Abd El-Wahaab <m.a.elwahaab@gmail.com>
- A Question about Combinatorica