Binomial as a sum of tetranomials

[goofball]

Hello there,

I'm working on a kinetic theory of mixing between two species - b and w.

Now, if I want to calculate the number of different species B bs and W ws can form, I can use a simple combination:

(W+B)!/(W!B!)

Now, in reality in my system, ws and bs form dimers - ww, bb, wb and bw (since orientation matters).

The number of species we can build with these is

(ww+wb+bw+bb)!/(ww!wb!bw!bb!) summed over all possible combinations of ww, wb, bw and bb such that the number of ws and bs stays constant.

I have proved this numerically for up to b=100 and w=100, which in reality is all I care about, but I am interested if there is a general proof for this, and how you would go around it.

Many thanks!

 

[mathman]

Your description is a little unclear. Could you explicitly show the equation you are trying to prove.

 

[goofball]

Hi hi! Thanks for your reply, mathman.

It's all a bit confusing, I know. In my original problem, I had W white subunits and B black subunits, and when I wanted to look at the number of different combinations I could add them together, that would be
$\frac{(W+B)!}{W!B!}$

Now, as my next problem, I for multiple reasons have to treat these as 'dimer' - that is - units consisting of two 'subunits'. This means I can get white-white, white-black, black-white, and black-black dimers (white-black and black-white are different because I care about their orientation..)

Now, when I wan to calculate then number of different combinations of these, I am effectively counting

$\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}$ summed over all the possible combinations of ww, wb, bw and bb such that w and b stays constant.

A practical example - let's say B = 2 and W = 2,

in this case $\frac{(W+B)!}{W!B!} = \frac{(2+2)!}{2!2!} = 6$

Now, if we treat them as dimers, we have 4 different scenarios:

1) ww = 1, bb = 1

number of combinations is $\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}=\frac{(1+0+0+1)!}{1!0!0!1!}= 2$

2) wb = 2, bw = 0

number of combinations is $\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}=\frac{(0+2+0+0)!}{0!1!0!0!}= 1$

3) wb = 1, bw = 1

number of combinations is $\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}=\frac{(0+1+1+0)!}{0!1!1!0!}= 2$

4) wb = 0 bw = 2

number of combinations is $\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}=\frac{(0+0+2+0)!}{0!0!2!0!}= 1$

All of these sum to 6 which is the same result as from $\frac{(W+B)!}{W!B!}$.

This all actually intuitively makes sense, and I have proved this computationally from any arbitrary number of W and B, but was wondering if there was a mathematical proof for this for any general B and W (such that B+W is even so that we can form perfect dimers...)

Thanks again!

 

[mathman]

Your example is confusing. What is the relationship of W and B to ww, bb, wb, and bw? I don't understand why W+B isn't equal to ww+bb+wb+bw.