# Configuration of particles

• Nov 27th 2008, 07:49 AM
free_to_fly
Configuration of particles
Suppose that n distinguishable particles are placed randomly in N boxes (states). A particular configuration of this system is such that there are ns particles in state s, where 1<=s<=N. If the ordering of particles in any particular state doesn't matter, find the number of ways of realising a particular configuration.

It's not at all obvious to me what I should do to tackle this question, any help would be hugely appreciated.
• Nov 28th 2008, 02:14 PM
awkward
Quote:

Originally Posted by free_to_fly
Suppose that n distinguishable particles are placed randomly in N boxes (states). A particular configuration of this system is such that there are ns particles in state s, where 1<=s<=N. If the ordering of particles in any particular state doesn't matter, find the number of ways of realising a particular configuration.

It's not at all obvious to me what I should do to tackle this question, any help would be hugely appreciated.

Hi Free To Fly,

If I understand your question correctly, the number of ways is the multinomial coefficient,

$\displaystyle \binom{n}{n_1 \; n_2 \; \dots \; n_N} = \frac{n!}{n_1! \, n_2! \, \cdots \, n_N!}$.
• Nov 29th 2008, 08:22 AM
free_to_fly
Thanks for the answer, but could you please explain where you got it from? I can't really see that. Thanks.
• Nov 30th 2008, 06:19 AM
awkward
The multinomial coefficient

$\displaystyle \binom{n}{n_1 \; n_2 \; \dots \; n_N}$

counts the number of ways to put n distinct objects in N boxes, with $\displaystyle n_1$ in the first box, $\displaystyle n_2$ in the second box, etc. See Multinomial theorem - Wikipedia, the free encyclopedia.

For a derivation, consider all $\displaystyle n!$ permutations of the n objects and consider the first $\displaystyle n_1$ to be in the first box, the next $\displaystyle n_2$ to be in the second box, etc. The order of the objects in the boxes is considered irrelevant, however, so divide by $\displaystyle n_1!$ to compensate for over-counting in box 1, divide by $\displaystyle n_2!$ to compensate for over-counting in box 2, ...

It's just like the binomial coefficient, only generalized to N choices instead of 2.