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. :) 

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.
