Stuck in using Stirling's approximation to show and justify an approximation of the number of permutations with and without ordering

Question

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.

Answer 1

The approximation ${N+m-1\choose N}\approx \frac{m^N}{N!}$ is valid if $N$ is fixed and $m\to\infty$. Indeed, $$ {N+m-1\choose N}=\frac{(N+m-1)!}{N!(m-1)!}$$ and applying Stirling's formula to $(N+m-1)!$ and $(m-1)!$ yields $$ \frac{(N+m-1)!}{N!(m-1)!}\sim \frac{1}{N!}\Big(\frac{N+m-1}{m-1}\Big)^{\frac{1}{2}}e^{-N}\Big(\frac{N+m-1}{m-1}\Big)^{m-1}(N+m-1)^N$$

If $m\to\infty$ and $N$ is fixed, then $$\Big(\frac{N+m-1}{m-1}\Big)^{\frac{1}{2}}\to 1$$ and $$\Big(\frac{N+m-1}{m-1}\Big)^{m-1}=\Big(1+\frac{N}{m-1}\Big)^{m-1}\to e^N$$ while $$(N+m-1)^N=m^N\Big(1+\frac{N-1}{m}\Big)^N\approx m^N$$ for large $m$.

Combining all these estimates yields the desired result.