Why binomial distribution doesn't count permutations? Why in Binomial distribution the formula starts with $n\choose k$ and not with something like $k!\over n!$? Isn't the order important? Or, it is important but due to the independence of the event?   
 A: From the relevant Wikipedia page: 

The binomial distribution is frequently used to model the number of successes in a sample of size $n$ drawn with replacement from a population of size $N$. 

Note two things: 


*

*What is key is the number of successes. So if with four tries  you have $(S,F,F,S)$ or $(F,S,F,S)$ is irrelevant, in both cases you have two successes. 

*Since one draws with replacement each try is independent from what happened before. Thus $(S,F,F,S)$ and $(F,S,F,S)$ certainly have the same probability. 
And, $\binom{n}{k}$ is the number of ways in which you can have $k$ successes among $n$ tries. 
If you do not draw with replacement, but without replacement, then the second point is no longer true and you get a different distribution, called hyper-geometric distribution. 
A: $$\frac{n!}{n_1!n_2!},\;\; n_1+ n_2=n,$$
counts the permutations of the sequence
$$\underbrace{p_1...p_1}_{n_1}\underbrace{p_2...p_2}_{n_2}.$$
Coincidentally, $$\frac{n!}{n_1!n_2!}=\frac{n!}{k!(n-k)!}={n \choose k}, \;\; n_1=k=\text{number of successes}.$$
Therefore, ${n \choose k}$, disguised as counting combinations, actually counts permutations.
