Help simplifying this using Stirlings' Formula?

[eprparadox]

Hey guys!

So I'm self studying Daniel Schroeder's Intro to Thermal Physics and we're to use Stirlings Approximation to simplify the following (assume that BOTH q and N are large, but we don't have a relationship between q and N):

<br /> \Omega \left( N,q\right) \approx \dfrac {\left( \dfrac {q+N} {q}\right) ^{q}\left( \dfrac {q+N} {N}\right) ^{N}} {\sqrt {2\pi q\left( q+N\right) / N}}<br />

The formula to find the multiplicity of an Einstein solid is:

<br /> \Omega \left( N,q\right) =\dfrac {\left( N-1+q\right) !} {q!\left( N-1\right) !}<br />

The hint in the text says to show that first:

<br /> \Omega \left( N,q\right) =\dfrac {N} {q+N}\dfrac {\left( q+N\right) !} {q!N!}<br />

With q and N being large, I thought I could immediately say that

<br /> \Omega \left( N,q\right) \approx \dfrac {\left( q+N\right) !} {q!N!}<br />

But I don't know where you get the

<br /> \dfrac {N} {q+N}<br />

I tried immediately applying Stirlings Approximation but got nothing like the hint or the answer we're looking for.

I'm a little lost on where to begin here. Any thoughts?

Thanks so much.

 

[Mentallic]

<br /> \Omega \left( N,q\right) =\dfrac {N} {q+N}\dfrac {\left( q+N\right) !} {q!N!}<br />

If you cancel the common factor of N in the numerator and N in the N! expression in the denominator, what do you get?

 

[eprparadox]

oh nice! got it.

Man, I'm dumb! Thanks so much.