JT
Posts:
1,448
Registered:
4/7/12


Combinatorics
Posted:
Jun 12, 2013 2:15 AM


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 <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
I will try to implement, in javascript.

