- 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
- From: firstname.lastname@example.org (sean kim)