```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!.>> -- RichardThe reason i looking for a pattern is i will try implement apermutation algorithm for elements 1-n.I have seen there is  Steinhaus, Johnsson, Trotter algorithm thatsomeone directed me to, my will probably be something similar workingtopdown from biggest to smallest using pairs.When looking at the pairs for the 5! permutation, it seem that eachunique base pair 54,45,32 etc can produce 3! of subpairs.So for each of the 20 base pair54,53,52,51,45,43,42,35,34,32,31,25,24,23,21,15,14,13,12,11 there isjust 3! digits to play with, i was thinking this must be a workingapproach even for 7! or any number of permutations.First some combinatorial finding out how many ways to select 2 among7, and from there is must follow that for each of those basepairsthere 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 thethe group of n digits and write out.2. n-2 digit to calculate possible number of subpairs for eachbasepair, create write out subpair.3. Repeat step 1 until n=0.54 32 154 31 254 23 154 21 354 13 254 12 3
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