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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



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