Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: permutations?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg46474] Re: [mg46466] permutations?
  • From: "Sseziwa Mukasa,,(978) 536-2359" <mukasa at jeol.com>
  • Date: Fri, 20 Feb 2004 00:29:18 -0500 (EST)
  • References: <200402190802.DAA14953@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On Feb 19, 2004, at 3:02 AM, sean kim wrote:

> I think this is a permutation problem. but i'm not sure. do correct
> me.

It's not really a permutation problem.  A permutation is a reordering 
of a set without changing elements.  You are substituting new elements 
into the set.

[deleted]

> I have 32 from doing it the brute force way by hand.

If you think about _ and _' as being two states of the same element 
you'll realize you are simply counting in base 2.  ie, abcde=00000 
(base 2)=0 (base 10), abcde'=00001 etc.  You have 5 elements with two 
states each, hence 2^5=32 combinations.

>   how do i do that
> with Mathematica? I don't think it's fully connected graph.  because a 
> only
> connects to downstream b or b'

I'm not sure what you are getting at here.  You could use IntegerDigits 
to generate the binary representation of a number then substitute the 
first element with a or a' depending on whether it's 0 or 1, then the 
second with b or b' etc.

> also... what kinda problem is described above?

It's counting as far as I can tell.

Regards,

Ssezi


  • References:
  • Prev by Date: Re: permutations?
  • Next by Date: Re: Help Browser issue in 5.0.1 on Mac OS X
  • Previous by thread: permutations?
  • Next by thread: Re: permutations?