Disclaimer: I am a chemist with a strong interest in physics, and lately I have been toying with this problem in my spare time. I feel it is appropriate to ask here cause it involves some statistical mechanics and simple combinatorics, and not a whole lot of chemistry. Plus, I thought it would make for some interesting cross-disciplinary discussion. Correct me if I am wrong. This did not get any solutions on Physics.SE so I thought, I would ask it here. After all it is a combinatorics problem.
Background
Some molecules have a handedness; these two stereiosmers are mirror images of each other. In many experiments, it has been observed that molecules prefere to form homochiral clusters/assemblies.
Model
We chemists often label these mirror-image molecules with R- or S- labels. So we can just represent the molecules as coins with two faces-a black face, or a white face (each corresponding to one of the stereochemistry label). A cluster will be a collection of these coins.
Let us ignore all enthalpic terms, and focus entirely on entropy.
Say, we have a cluster of 9 molecules (a 9mer)-if it is derived from an enatiopure solution (i.e only R- or only S- present), we can only arrive at one possible structure-either all black, or all white coins.
If we have a racemic starting solution (50-50 black and white coins), we have $2^9$ possible structures, which ought to follow the binomial distribution.
Example: $1(9S), 9(8S/1R), 36(7S/2R), 84(6S/3R), 126(5S/4R), 126(4S/5R), 84(S/6R), 36(2S/7R), 9(1S/8R), and 1(9R)$.
The probability that a cluster of size N molecules, comprises of n molecules of a given enantiomer, $P = \frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}$
for an ideal racemic cluster p = 0.5
Now, we wish to account for the entropy related to the number of permutations that lead to equivalent structures by exchanging non-distinguishable molecules.
For a homochiral structure (all S or all R), there will be $N!$ states, where $N$ is the number of molecules.
For a racemic system, just doing some simple examples on paper, I feel there will also be $N!$ equivalent arrangements for a cluster comprising of N molecules. Basically, I am searching for the number of permutations that lead to equivalent structures.
(1) How could one show/prove this in general?
I think an illustration would be helpful at this point. This is the kind of coin based model I am referring to. For an 9mer, the first cluster is what I call homochiral. The second, third, fourth and fifth would be racemic clusters (an ideal racemic cluster will have 1:1 black and white coins, but that is not possible with 9 coins). Anyway, each of these structures have a multiplicity for example, structure 4 and 5 are essentially the same structure and can be obtained by a simple rotation. Two more structures exist in this family, thus structure 5 has a multiplicity of 4.
So a more realistic estimate would be $\frac{2^9}{4}$ (?) Not sure about this, I would love to hear someone's opinion on this.
(2) In general, for an N-mer, can we place an estimate on the multiplicity possesed by the largest set of structures.
As you state, the probability of selecting $n$ heads from $N$ trials with Bernoulli probability $p$ is $$P(X = n) = \binom{N}{n}p^n (1-p)^{N-n}$$
If we pre-determine the number of heads, that's equivalent to fixing the coin tosses, so we can eliminate the $p^n (1-p)^{N-n}$ and find that the number of arrangements is $\binom{N}{n}$. (That's actually backwards from the derivation of the binomial distribution, but never mind).
However, I'm not sure how relevant that is to the entropy, which is $$-\sum_{n=0}^N P(X = n) \lg P(X = n) = -\sum_{n=0}^N \binom{N}{n}p^n (1-p)^{N-n} \lg\left( \binom{N}{n}p^n (1-p)^{N-n}\right) \\ = -\sum_{n=0}^N \binom{N}{n}p^n (1-p)^{N-n} \left( \lg\binom{N}{n} + n\lg p + (N-n)\lg(1-p)\right) \\$$