Role of binomial coefficient in binomial distribution As I can verify, the binomial distribution $W_N(n) = \binom Nn p^n q^{N-n}$, is indeed a probability distribution:
$$\sum_{n=1}^N W_N(n) = (p+q)^N = 1\ ,$$ with $q = 1 -p$.
In the context of statistical physics, I don't understand why it is essential that we take into account that there are $\binom Nn$ possible ways to pick n molecules from $N$. Yes, we do not take the order into account when we pick the molecules, but what does the binomial factor actually mean in $W_N(n)$? What would happen if we leave it out, other than the fact that $W_N(n)$ then fails to be a probability distribution?
I have been ignoring this confusion for a while now, so a really down to earth answer would be appreciated. :)
Edit.
I understand the isolated concept of the binomial coefficient $\binom nk$; the number of ways to choose k objects out of n objects without ordering; the numerator accounts for the number of ways we can order the $k$ and $n-k$ objects respectively: $$\frac{1}{n!(n-k)!} \ .$$
What I don't understand is its role in the binomial distribution $W_N(n)$. 
In the context of my question; the probability that we find $n \ (< N)$ molecules in a subvolume $V_1 \subset V$; what is wrong with the following reasoning that I'm tempted to have?:
The probability of finding $n$ molecules inside a subvolume $V_1$ is equal to $p^n q^{N-n}$, where $p$ and $q$ are the probabilities of finding a molecule inside and outside the subvolume $V_1$ respectively.
Now, I do have a vague notion of that it is essential to take into account the fact that we are choosing $n$ molecules out of $N$ without ordering. But I don't understand the key concept of how this is resolved by just multiplying by $\binom Nn$, and what it would mean conceptually if we didn't.
If it is evident from the way I phrased the question that I'm missing a different unrelated concept, do tell. :) 
 A: The binomial coefficient is what gives it its bell-like shape so they are quite essential. Taking a quick look at Pascal's triangle one can clearly see the maximum being roughly at the centre. If you estimate the binomial coefficient using something like Sterling's Approximation for large values (in Statistical Mechanics you probably deal with somewhere around $10^{23}$ particles) you will probably still maintain the overall shape.
As a side-note, for large numbers, the original purpose of the Poisson Distribution was to approximate the Binomial Distribution by reducing the number of factorials needed to be calculated.
A: I understand a Binomial Experiment as $N$ independent Bernoulli-experiments. A Bernoulli-experiment is a experiment with two possible events. Success and No-Success. Like throwing a coin.
I don't know your context or setting, but if one use binomial distribution there is always place for this kind of interpretation. And in this interpretation the $ \binom{N}{n} $ is essential. If we fix $n$ molecules out of the $N$, then the probability, that these and only these fixed ones "succeed" is of course $p^n(1-p)^{N-n}$. But by $W_N(n)$ we are asking for the probability, that out of $N$ molecules exactly $n$ "succeed" without fixing any. So we have to take account of all possibilities for fixing $n$ out of $N$ and we came to
$$\binom Nn$$ 
A: Consider an example of coin flips. Label $H = 0$ and $T = 1$. How many ways are there of obtaining $2$ Tails? There are $\binom{5}{2}$ ways:
11000 
10100 
10010 
10001 
01100 
01010 
01001 
00110 
00101 
00011 
If you omit the $\binom{5}{2}$ term, you don't account for all of these possibilities. 
The same logic applies for the other terms.
A: Illustrate by simple example.  Coin flip 4 times.  Possible outcomes as follows:
tttt,ttth,ttht,thtt,httt,tthh,thth,thht,htth,hhtt,htht,etc.
Notice that there is only $\binom{4}{4}=1$ way to get 4 tails, $\binom{4}{3}=4$ ways to get 3 tails, $\binom{4}{2}=6$ ways to get 2 tails, etc.
A: Expanding on @Falrach's answer; if we, indeed, fix $r$ events who "succeed" amongst $n$ events, where the probability of succeeding resp. failing are $p$ and $q = 1-p$, we have the following argument for $n=5$ and $r=2$.
Denote $P(i)$ for the probability of $i$ succeeding events and $\hat{P}(k,l)$ for the probability of events $k$ and $l$ succeeding.
We have $P(2) = \hat{P}(1,2) + \hat{P}(1,3) + \ldots + \hat{P}(4,5)$. Since we can pick $r$ events from $n$ events in $\binom n r$ ways, we have
$P(2) = \binom 5 2 \hat{P}(1,2)$.
So we multiply by the binomial in if the succeeding events are indistinguishable.
