- 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
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.
> 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
> 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.
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