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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: 1,388 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:
>
> 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

Date Subject Author
6/11/13 JT
6/11/13 JT
6/11/13 JT
6/11/13 Dirk Van de moortel
6/11/13 JT
6/11/13 Brian Chandler
6/11/13 JT
6/11/13 JT
6/11/13 JT
6/11/13 Richard Tobin
6/12/13 JT
6/12/13 JT
6/12/13 JT
6/11/13 Richard Tobin