What is the probability that we get exactly two heads when we flip a coin five times? The answer is apparently $5\choose2$, i.e. $\frac {5!}{2!3!} $
But I the way I learned it, the "choose" notation is to be used when counting possible subsets of a set. For example, $5\choose2$ is the number of ways to choose a $2$-length subset of a $5$-length set.
How can I reconcile that definition with the application here? The only thing relating this problem to a set-counting problem is writing out the outcomes as a $5$-length set. For example, {H,H,T,T,T} represents the event of getting two heads.
But there are repetitions in that "set", so why do we use the choose notation for it?
Furthermore, it would seem that {H,H,T,T,T} is a different event than {H,T,T,T,H}. Is that true? Because if so, the "set" is ordered, which is another reason why it doesn't make sense to use $n\choose k$ here: we are a this point dealing with lists (which are ordered and allow repetition).
Sorry if this is a stupid question, I've been struggling with this for a long time and I'd hugely appreciate any help.
Sets are not ordered. However, the outcomes in question are finite sequences of heads and tails. The sequence $(H, H, T, T, T)$ is different from $(H, T, T, T, H)$ since the heads and tails appear in a different order.
The number $\binom{5}{2}$ counts the number of sequences of length five in which exactly two heads appear. What we are choosing is the positions where the heads appear in the sequence. However, $\binom{5}{2} = 10$ is not a probability since a probability must lie between $0$ and $1$ inclusive.
If we make the assumption that heads and tails are equally likely outcomes, then there are $2^5 = 32$ equally likely possible sequences of heads and tails since there are two equally likely ways to fill each of the five positions in the sequence.
Under this assumption, the probability that exactly two heads appear in five tosses of the coin is $$\Pr(\text{exactly two heads}) = \frac{\binom{5}{2}}{2^5} = \frac{10}{32}$$