This is a problem from my applied mathematics class where we are currently working on using Stirling's approximation which is:
$ n! \sim (\frac{n}{e})^n \sqrt{2 \pi n} $
and the context of this problem is combinatorics in counting microstates in statistical mechanics:
We are given a system of N particles each having m possible different states and we are asked the following:
a. We suppose the particles are distinguishable and we are asked the number of micro-states of the system? Here I thought since the N particles are different and we are allowed repetition the answer is $ m^N $
b. We suppose the particles are indistinguishable and we are asked the number of micro-states of the system? Here I thought since the N particles are not different and we are allowed repetition the answer is the number of possibilities to choose N from m with repetition and order is unimportant so the answer I thought of is $ {N+m-1} \choose{N} $.
c. Now we wish to approximate part b instead of using the more complicated exact result so we approximate part b's answer as $ \frac{m^N}{N!} $ and we wish to find the right conditions such that this approximation holds and we are instructed to approximate factorials via Stirling, that is $ \frac{m^N}{N!} \approx$ ${N+m-1}\choose{N}$
d. Finally, we are to find for which parameter (m or N) going to infinity the limit of the quotient of the exact part b and its approximation approaches 1, this was fair enough using Stirling and some limit arithmetic, so no problems.
Since I did parts a,b and d, my problem is only with c so I almost solved it and I am stuck to show and justify the approximation $ \frac{m^N}{N!} \approx$ ${N+m-1}\choose{N}$ via Stirling so that is where I need help in justifying the manipulations to present the approximation. And so I am writing here. I thank all helpers, and I would appreciate someone checking if my answers to a and b are correct.