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Topic: 7!= 2*5! * 2*3! *7
Replies: 14   Last Post: Jun 14, 2013 10:35 PM

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JT

Posts: 1,183
Registered: 4/7/12
Combinatorics
Posted: Jun 12, 2013 2:15 AM
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On 12 Juni, 06:50, JT <jonas.thornv...@gmail.com> wrote:
> On 11 Juni, 22:23, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
>

> > In article <22cede7a-ba1d-4e00-9829-8eef91562...@g9g2000vbl.googlegroups.com>,
>
> > JT  <jonas.thornv...@gmail.com> wrote:
> > >Well 5!*3!*7 = 5040 and 7!=5040, tomorrow i give it another try.
>
> > Obviously, since 7! = 5! * 6 * 7, and 3! = 6.  There is nothing
> > interesting in this - one of the numbers greater than 7 and
> > less-than-or-equal to 9 just happens to be a factorial.  It's no more
> > significant than that 5040! = 5039! * 7!.

>
> > -- Richard
>
> The reason i looking for a pattern is i will try implement a
> permutation algorithm for elements 1-n.
>
> I have seen there is  Steinhaus, Johnsson, Trotter algorithm that
> someone directed me to, my will probably be something similar working
> topdown from biggest to smallest using pairs.
> When looking at the pairs for the 5! permutation, it seem that each
> unique base pair 54,45,32 etc can produce 3! of subpairs.
>
> So for each of the 20 base pair
> 54,53,52,51,45,43,42,35,34,32,31,25,24,23,21,15,14,13,12,11 there is
> just 3! digits to play with, i was thinking this must be a working
> approach even for 7! or any number of permutations.
>
> First some combinatorial finding out how many ways to select 2 among
> 7, and from there is must follow that for each of those basepairs
> there can be 5! of combinatorial subpairs.
> I think that is the correct approach.
>
> 1. Find out how many ways 2 first digits can be combined within the
> the group of n digits and write out.
> 2. n-2 digit to calculate possible number of subpairs for each
> basepair, create write out subpair.
> 3. Repeat step 1 until n=0.
>
> 54 32 1
> 54 31 2
> 54 23 1
> 54 21 3
> 54 13 2
> 54 12 3


I will try to implement, in javascript.



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