Counting the number of microstates that there are for a given configuration. How to prove this result?

Question

I'm doing some statistical physics and I came across a result which I'm not sure how to derive. Any help?

The answer turns out to be:

Can anyone help with this derivation? Thank you :D

Answer 1

Can anyone help with this derivation?

There are $N!$ ways to arrange $N$ different objects, but you don't have $N$ different objects, you have $n_0$ indistinguishable objects in the ground state, $n_1$ indistinguishable objects in the first state, and so on. So that means, from your $N!$ different permutations (assuming all the objects are distinguishable) all the permutations where you mix around the $n_0$ objects in the ground state all look exactly the same. And there are $n_0!$ such permutations, so you have to divide by $n_0!$. The argument is the same for the objects in the first excited state, so you also have to divide by $n_1!$, and so on.

Here is an example. Suppose you have one letter "A" and two indistinguishable letters "B". But, suppose that somehow we actually have marked one of the "B"s with a "1" and one of the "B"s with a "2" so we actually can distinguish them. In that case there are $3!=6$ different configurations (see below)

A B1 B2
A B2 B1
B2 B1 A
B1 B2 A
B1 A B2
B2 A B1

But really, we can't distinguish between the "B"s, so these 6 "different" configurations really look like:

A B B
A B B
B B A
B B A
B A B
B A B

We can't really differentiate between the permutations that only mix up the "B"s. I.e., there are really only $3=\frac{3!}{2!}$ different configurations:

A B B
B B A
B A B