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
----
.===================================.
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| Don't Fear the Penguin. |
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