Configuration of particles

free_to_fly · #1 $Old$ November 27th, 2008, 09:49 AM

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.

awkward · #2 $Old$ November 28th, 2008, 04:14 PM

Quote:

Originally Posted by free_to_fly $View Post$

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,

$\binom{n}{n_1 \; n_2 \; \dots \; n_N} = \frac{n!}{n_1! \, n_2! \, \cdots \, n_N!}$ .

free_to_fly · #3 $Old$ November 29th, 2008, 10:22 AM

Thanks for the answer, but could you please explain where you got it from? I can't really see that. Thanks.

awkward · #4 $Old$ November 30th, 2008, 08:19 AM

The multinomial coefficient

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

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

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

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