JT
Posts:
1,150
Registered:
4/7/12


Re: 7!= 2*5! * 2*3! *7
Posted:
Jun 12, 2013 12:50 AM


On 11 Juni, 22:23, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote: > In article <22cede7aba1d4e0098298eef91562...@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 > lessthanorequal 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 1n.
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. n2 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

