We have a crystall with $N$ atoms.

A Schottky defect is one of the atoms leaving their points in the lattice and going to the surface with $N' = cN^{2/3}$ possible atom places.

I want to compute the (Boltzmann) entropy of a crystal with $N_S$ Schottky defects, where $1 \ll N_S \ll N'$.

My first attempt:

We pick $N_S$ atoms out of the total $N$ atoms, which gives

$$\frac{N!}{N_S!(N-N_S)!}$$

possibilities.

Then we can place these $N_S$ atoms to the $N'$ points on the surface, which gives another

$$\frac{N'!}{N_S!(N'-N_S)!}$$

possibilities. In total we have

$$\frac{N'!}{N_S!(N'-N_S)!} \cdot \frac{N!}{N_S!(N-N_S)!} $$

possibilities to 'build' a crystall with $N_S$ defects.

However, in multiple sources i found the possibilities to 'build' a crystall with $N_S$ defects, to be given as:

$$\frac{(N+N_S)!}{N_S!N!}$$

I interpret this as picking $N$ atoms and placing them on a total of $N + N_S$ possible lattice points.

This doesn't make any sense to me. If we take $N$ atoms and place them along the total of $N + N_S$ lattice points, it includes the possibilities to just put the $N$ atoms back to their original places, which then would correspond to a crystall with $0$ defects. Also the latter formula includes the possibilities, where we place just one atom on the surface and so on...

So how can the latter formula give the correct number of possibilities to build a crystall with $N_S$ Schottky defects?

I thought maybe the latter formula is some limit of the first (using the conditions given at the beginning of the question), but i didn't manage to figure this out.

Where is the mistake?

I appreciate your help :)