Re: Combinatorics question




Proginoskes wrote:
> Chris Patil wrote:
> > Hi there.
> >
> > I'm a biologist who's stumbled into a (computational) problem for which
> > I need to understand combinatorics a bit better than I do.
> >
> > I was wondering whether anyone could point me in the right direction
> > (e.g., intro texts) to solve this sort of problem:
> >
> > Suppose that I have two bins of finite size N. Each bin contains a
> > specific set of letters (N each). There is no order. Within a bin, a
> > given letter can appear zero, 1, or many times (up to and including N)
> > in its bin.
>
> Your bin is modeled by a mathematical object called a multiset.
>
> > E.g.:
> > Bin 1 = "Q Q D D"
> > Bin 2 = "N N E E"
> >
> > Pairs of letters are drawn by taking one letter from each bin, until
> > all the letters are gone.
>
> To make each pair, are you taking one letter from Bin 1 and one letter
> from Bin 2? Or are you taking two letters out of Bin 1 and Bin 2
> (collectively)? I suspect the former, since this looks like a model of
> matching genes from two parents.
>
> > E.g. for the bins above, there are three distinguishable sets of letter
> > pairs:
> >
> > a. "QD QE ND NE"
> > b. "QD QD NE NE"
> > c, "QE QE ND ND"
>
> As has been pointed out, the only pairs you can get are QN, QE, DN, and
> DE. It looks like you're starting with the sets Q Q N N and D D E E
> (and not Q Q D D and N N E E, like you stated).
>
> Let's continue assuming that you're starting with QQNN and DDEE.
>
> > I'm interested in the number of ways that each distinct set of pairs
> > could be formed -- in general, for any letters in the bins. In the
> > example, I know that there are four times as many ways to get (a) as
> > (b) and (c), and that there are an equal number of ways to get (b) and
> > (c) -- 36 ways total to pull out letters,
>
> You shouldn't get a total of 36. This is because you can think of
> assigning each of the letters in one multiset to one of the letters in
> the second multiset. (You can change from multisets to sets by
> numbering the Q's, N's, D's, and E's as Q1 Q2 N1 N2 D1 D2 E1 E2.) Such
> an assignment is called a 1-to-1 function, which can be represented as
> a permutation.
>
> The number of permutations of n objects is denoted by n! ("n
> factorial"), and is the product of the integers between 1 and n. Thus,
> in your case, 4! = 4*3*2*1 = 24.

That's the number of ways the letters can be selected from just
bin 1. Bin 2 is independent, isn't it? So isn't the actual number
of possible results is 4! * 4! = 576. That's what I get from
my database.

>
> If you order the objects, you can deal with permutations of the set
> {1,2,...,N}. So the elements of the first bin could be ordered as Q1 Q2
> N1 N2, and the elements of bin 2 as D1 D2 E1 E2.
>
> One permutation is 1234; i.e., element #1 of bin 1 is matched to
> element #1 of bin 2, element #2 of bin 1 is matched to element #2 of
> bin 2, etc. This gives you the pairing Q1-D1 Q2-D2 N1-E1 N2-E2, and
> when we get rid of the labels, we see we have QD QD NE NE.
>
> Another permutation is 1243; this is the same as 1234, except element
> #3 of bin 1 is matched with element #4 of bin 2, and element #4 of bin
> 1 is matched with element #3 of bin 2. This gives us the matching Q1-D1
> Q2-D2 N1-E2 N2-E1, which (after removing the numeric lables) is a
> pairing of the type QD QD NE NE.
>
> We can generate a complete list -- there are programs already encoded
> which will give all possible permutations -- and then you can count how
> many QD QD NE NE 's you have, etc. (I know you're looking for a formula
> for this, but I don't know one right off, and the procedure I've listed
> can be programmed to work with any set of bins.)
>
> > 24 that give (a), 6 for (b), and 6 for (c).
>
> I suspect it's actually 16, 4, and 4, for each of these sets.

I get 384, 96, 96 (still 4:1 ratio).

>
> > But I got that by working it out by hand, which is not
> > useful for the real examples I'm going to be working with, in which N
> > will be >10, and any of 21 letters can appear in any proportion.
> >
> > So I'm not sure how to generalize this for larger N, more complex sets
> > of letters, etc.
> >
> > Is there a formula, or must it be solved algorithmically?
>
> I suspect there is a formula. Probably involving determinants or
> perminants of a certain matrix.
>
> --- Christopher Heckman
>
> > I can imagine an inefficient algorithmic way to do it, like so:
> >
> > 1. Determine all possible ways to draw letter pairs.
> > 2. Determine all possible distinguishable sets of pairs, and then go
> > through the list for of all possible ways and count.
> >
> > I could implement that, and it would solve my problem, but I suspect
> > that if I were just a little better at combinatorics, I would know how
> > to come up with a general formula. I've been reading up on this all day
> > and most of the examples of combinations and permutations involve a
> > single bin, and I'm stuck.
> >
> > Can anyone give advice?
> >
> > Thanks,
> >
> > Chris

.



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