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I'm currently studying general mechanics where the following problem came up:

Assume we have the space $\Gamma = \mathbb{R}^6$ which we are dividing in small cells $C_i$ . Let $f(\vec{x}, \vec{v})$ be a probability distribution on our $\Gamma$. The amount of elements in a cell $C_i$ is (not exactly) given by: $N_i = \int_{C_i} f(\vec{x}, \vec{v}) d^3x d^3v$.

Since it doesn't matter in my problem how my elements are numbered, as long as there are the same amount of particles in each cell, I'm interested in finding out how many possibilities for one configuration of particles in cells there are.

Let $N$ be the total amount of particles i.e. $N = \sum N_i$. My initial guess is, that there are $N!$ different possibilities because there are $N$ slots. So the first particle has $N$ possible places, the second one has $N-1$ possible places $\implies$ $N \cdot (N-1) \cdot ... \cdot 1$ = $N!$

But this isn't correct yet, because if I have a cell that has, lets say, $2$ elements, there are $2!$ possible ways for the elements to get labeled which still give the same overall state, which means I have to divide by $2!$. This makes me say the correct result would be $\frac{N!}{N_1! \cdot N_2! \cdot ... \cdot N_k!}$ ($k$ = amount of cells), which I know is the right solution but I'd like to know it with some more mathematical rigor.

Anybody care to help me?

Cheers!

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This is pretty standard; look up "multinomial coefficient". –  ShreevatsaR Jul 10 '13 at 14:42

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I don't know this way has the rigour you want, but here goes:

You have $N$ total particles, $N_1$ go to the first cell. The number of ways to achieve this is

$${N\choose N_1} = \frac{N!}{(N-N_1)!N_1!}$$

You are left with $N-N_1$ remaining particles, and now $N_2$ of these go to the second cell. The number of ways this can be done is

$${N-N_1\choose N_2} = \frac{(N-N_1)!}{(N-N_1-N_2)!N_2!}$$

Continue this way till the end; the overall number of ways to do this exercise is

$${N\choose N_1}{N-N_1\choose N_2} \cdots {N-N_1-N_2\cdots -N_{k-1} \choose N_k}= \frac{N!}{N_1!N_2!\cdots N_k!} $$

All the factors other than $N!$ and $N_i!$ get cancelled due to successive multiplication.

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That's mathematical rigor in my physical book :) Thanks, that makes sense! –  user17574 Jul 10 '13 at 13:42

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