# The number of ways of placing M atoms on the interstices of a lattice

1. Dec 2, 2013

### ppy

Hi,

N atoms are arranged to lie on a simple cubic crystal lattice. Then
M of these atoms are moved from their lattice sites to lie at the
interstices of the lattice, that is points which lie centrally between the
lattice sites. Assume that the atoms are placed in the interstices in a
way which is completely independent of the positions of the vacancies.
Show that the number of ways of taking M atoms from lattice sites
and placing them on interstices is W = (N!/M!(N − M)!)2 if there
are N interstitial sites where displaced atoms can sit.

I literally do not know where to start with this question I know W is the number of ways of choosing M atoms from N atoms but I don't really know where to go with this.

Help would be great.

Thanks :)

2. Dec 2, 2013

### Staff: Mentor

What does the 2 mean? W = (N!/M!(N − M)!)2?

This can be expressed as (N choose M)2. And it does exactly what the name suggests: (N choose M) calculates the number of ways to choose M atoms out of N (to decide which atoms you move), and you get the same factor again for the choice of the interstitials.

3. Dec 3, 2013

### ppy

Hi yes the 2 should be a squared. This question is worth 6 marks. What would you have to do to get so many marks other than writing what you suggested?

4. Dec 3, 2013

### Staff: Mentor

I have no idea about the scale of the marks, but writing that down in a more ordered way should be sufficient.