Date: Jun 12, 2013 12:50 AM
Author: JT
Subject: Re: 7!= 2*5! * 2*3! *7
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