Why do we use ${n\choose k}$ for a binomial distribution instead of ${n+k-1\choose k}$? I am trying to get my head around this. In my understanding a binomial distribution uses replacement and ${n\choose k}$ precisely states that there's no repetition and that's not the case with a coin toss, for instance.
Thanks in advance.
 A: I can see why it's confusing. The reason it's like this is because the source of the "choosing without replacement" is in the mathematics of deriving the probabilities, not the actual process.
First off it should be clarified that $n\choose k$ does not actually "state" anything, it's just a number, equal to the number of different sets (or choices) of size k you can make from a collection of n different objects.
Why does this arise for a coin toss?
We're talking about a distribution, so let's take an example.
Say you're tossing a fair coin $n = 5$ times, so the number of heads $H$ is distributed as $Bin(5, 1/2)$.
Say we want to know the probability that $H = 2$.
Each sequence of 5 tosses is equally likely (here's where the fact that the process is with replacement comes in). That is, HHHHH, HHHHT, HHHTH, ... are all equally likely. And means that the probability of each of these outcomes is $(1/2)^5$.
Now we have to count how many of these outcomes has 2 heads. This is where "without replacement" comes in. Let's try to start enumerating them:
TTTHH
TTHTH
TTHHT
...
Hopefully you can see that there are 5 different positions that a heads or tails occurs, and we need to choose 2 of them to be H. So there are $5 \choose 2$ outcomes with 2 heads. And that means the total probability of this happening is ${5 \choose 2}\times(1/2)^5$.
