Probability of finding n particles in a volume v I'm trying to calculate the probability of finding $n$ particles in a certain volume $v$ in a system with a total of $N$ particles and total volume of $V$. My problem is that I've tried two approaches which both seem valid to me, but give differing answers.
One approach is to use binomial probability, where the probability of success (particle in the volume of interest) is $\frac{v}{V}$. Furthermore, the particles are indistinguishable, so it doesn't matter the order of "successes" and "failures". This gives:
$P=(1-\frac{v}{V})^{N-n}\,(\frac{v}{V})^{n}\,\frac{N!}{(N-n)!n!}$
My other approach is to say to start saying that any configuration (remembering particles are indistinguishable) has equal probability and so the probability for our event is simply $P=\frac{\mathrm{\#\ of\ configurations\ with\ n\ particles\ in\ the\ cell}}{\mathrm{\#\ of\ configurations}}$. Now from combinatorics, the number of configurations is $\binom{N+\frac{V}{v}-1}{N}$, and the number of configurations with $n$ particles in $v$ is $\binom{N-n+\frac{V}{v}-2}{N-n}$. This gives a probability:
$P=\frac{\binom{N-n+\frac{V}{v}-2}{N-n}}{\binom{N+\frac{V}{v}-1}{N}}=\frac{(\frac{V}{v}-1)!\,N!\,(N-n+\frac{V}{v}-2)!}{(N+\frac{V}{v}-1)!\,(\frac{V}{v}-2)!\,(N-n)!}$
Which definitively isn't the binomial distribution, I checked numerically as well. So what is the problem?
 A: The discrepancy results from the fact that the first approach implicitly assume distinguish-ability while the second doesn't. 
Let's take a simple example: $n=1$,$N=2$, and there are two cells($v=\frac{1}{2}V$), what's the probability to have 1 particle in the first cell? Now label the particles by $1$ and $2$(and remove the label in the end) and exam your two approaches.


*

*There are four configurations, $[(1,2),()]$, $[(1),(2)]$, $[(2),(1)]$, $[(),(1,2)]$. You treat them with equal probability $\frac{1}{4}$. There are two configurations with only 1 particle in the first cell. So the probability is $\frac{1}{2}$.

*There are only three configurations, because $[(1),(2)]$ and $[(2),(1)]$ are the same physical states. The probability is therefore $\frac{1}{3}$.


The second approach is correct, because only different physical  states are equally likely. 
EDIT Feb. 7, 2014: My conclusion was wrong, although particles should be indistinguishable. 
OP and I implicitly assume there is only 1 microstate in each cell, which is not realistic. Suppose there are $M$ sub-cells in the unit cell with volume $v$; equivalently there are $M$ different states in each cell $v$. Let $\frac{V}{v} = C$, then there are $(C-1)M$ sub-cells outside $v$ as shown in the figure. 

There are $\binom{M+n-1}{M-1}$ ways to distribute $n$ indistinguishable particles into $M$ sub-cells in $v$. Similarly there are $\binom{(C-1)M+N-n+1}{C(M-1) -1}$ ways to distribute the rest $N-n$ particles into $(C-1)M$ sub-cells outside $v$. So the total number of configurations for exactly $n$ particles in $v$ is
\begin{equation}
\Gamma(n) =\binom{M+n-1}{M-1} \binom{(C-1)M+N-n-1}{(C-1)M -1}
\end{equation}
while the total number of configurations is the ways to distribute $N$ particles into $CM$ sub-cells, 
\begin{equation}
  \sum_{n=0}^{N}\Gamma(n) = \binom{CM+N-1}{CM-1}
\end{equation}
The probability in question is then
\begin{equation}
  p_n = \frac{\Gamma(n)}{\sum_{i=0}^{N}\Gamma(i)} = \binom{N}{n} \frac{(M+n-1)!}{(M-1)!} \frac{[(C-1)M+N-n-1]!}{[(C-1)M-1]!} \frac{(CM-1)!}{(CM+N-1)!}
\end{equation}
Then we take the dilute limit, which means the number of sub-cells $M$ are large compared even with $N$. These factorials can be simplified,
\begin{equation}
p_n \approx \binom{N}{n} ( \frac{1}{C})^n (\frac{1-C}{C})^{N-n} =  \binom{N}{n} ( \frac{v}{V})^n (1- \frac{v}{V})^{N-n} 
\end{equation}
Oh God, I got the same answer as the first approach.
So there is no inconstancy for these two approaches after we take $M\rightarrow \infty$; that means particles even in $v$ can only occupy a small fraction of sub-cells, which somehow makes them effectively distinguishable.
