Statistical Relationship Between Helmoltz Free Energy and Entropy

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Homework Statement


From Hill's "Introduction to Statistical Thermodynamics", question 3-4 reads:

Show that the substitution of of the most probable distribution in
S=k*ln(\Omega(C^{str})) leads to
A=-NkT*ln(\sum e^{\beta \epsilon_j})

(note "STR" denotes the case of most probable distribution and should be read as C*)

Homework Equations


The most probable distribution for a system of independent indistinguishable molecules:
\eta_i = \bar{C_j} /N =C^{str}/N = e^{-\beta \epsilon_j} / \sum_i e^{-\beta \epsilon_i}

Formula for accessible quantum states for an ensemble of independent indistinguishable molecules:
\Omega(C) = N! /( \prod_j C_j) !

Relationship between S and A
S = - (\partial A / \partial E) = kT ( \partial ln(Q) / \partial T )+ k ln(Q)


The Attempt at a Solution



First, I computed C in terms of the given equations. This is pretty straight forward:

C^{str}_j = Ne^{-\beta \epsilon_j} /{ \sum_i e^{-\beta \epsilon} }

Then, I computed the number of accessible quantum states with the above expression for C
\Omega (C^{str}) = { N! }/{\prod C^{str} }

Skipping a lot of algebra I arrived as this expression for ln(Ω(C)):
ln(C^{str}) = ln(N!)- \sum(ln(e^{-\beta \epsilon}) )- ln(\sum(e^{-\beta \epsilon})

Which can be reduced to:
ln(C^{str}) = ln(N!) + \sum(\beta \epsilon) - ln(Q) = ln(N!) + E \beta -ln(Q)

I am not sure exactly where I went wrong, as the formula I derived is very similar to the relationship between A and S. I think an approximation may have to be applied somewhere or perhaps the natural logs should not have been broken down into their operations. I am not sure, hence why I am posting here!

Thanks for the help.

PS, if needed I can post a small nomenclature of the variables.

EDIT:
Just realized I did not take into account the N factor in the equation for C, I believe this makes N! become (N-1)! in the following equations. Also, I did not show the additional formula if one were to expand \partial ln(Q) / \partial T
 
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