All possible combinations algorithm
Jordan Gunderson
jordy at gundy.org
Wed Jun 15 09:42:10 MDT 2005
Here it is with tardmode taken off. The wording is a little better.
> Hey Dan,
>
> Thanks for the math question. If the number of elements is E, the number
> of possible arrangements is equal to the E! (read "E factoral"). It's E
> times all of the integers between E and zero. Be careful though, because it
> gets really big really fast.
>
> In this example, I started out with one element in the first sequence. In each following
> sequence I placed the next element in all of the possible positions from
> the possibilities (separated by dashes) of the preceding sequence, starting in the last
> position and moving forward.
>
> Thanks for the fun question. Jordy
> ______________________________
>
> A 1 Elements = 1! = 1 = 1
>
> AB 2 Elements = 2! = 2*1 = 2
> --
> BA
>
> ABC 3 Elements = 3! = 3*2*1 = 6
> ACB
> CAB
> ---
> BAC
> BCA
> CAB
>
> ABCD 4 Elements = 4! = 4*3*2*1 = 24
> ABDC
> ADBC
> DABC
> ----
> ACBD
> ACDB
> ADCB
> DACB
> ----
> CABD
> CADB
> CDAB
> DCAB
> ----
> BACD
> BADC
> BDAC
> DBAC
> ----
> BCAD
> BCDA
> BDCA
> DBCA
> ----
> CABD
> CADB
> CDAB
> DCAB
> ----
> .===================================.
> | This has been a P.L.U.G. mailing. |
> | Don't Fear the Penguin. |
> | IRC: #utah at irc.freenode.net |
> `==================================='
More information about the PLUG
mailing list