In many textbooks when they introduce the concept of combinations, it is usually right after introducing permutations, so the usual interpretation they give to combinations is something along the lines of:
"You have $n$ items and need to pick $k$ out of them. If we care about the order, we use $\frac{n!}{(n-r)!}$. However, since the order does not matter, and there are $k!$ ways to arrange your selection of $k$ objects, you divide the number of permutations by $k!$, giving $\frac{n!}{k!(n-k)!}$".
This makes perfect sense for me and it is the intuition I always had in my mind.
However, in the context of the Binomial distribution, one uses the combination formula to justify the different ways you can arrange $k$ "Heads" and $n-k$ "Tails" such as:
"You have $n$ flips of a coin where $k$ are Heads and $n-k$ are Tails. You need to count the different ways to arrange them. If $n=5$ and $k=2$, then one possible way is $(H,H,T,T,T)$ and another one is $(H,T,H,T,T)$. The different ways to arrange your $k$ Heads and $n-k$ Tails is just $\frac{n!}{k!(n-k)!}$."
In fact, if you look at Khan's Academy, he introduces combinatorics in the first way, but when he talks about the Binomial Distribution, he simply takes it as granted the second interpretation and never uses the first one in his explanation.
What confuses me is that these two different ways to interpret the formula for a combination $_nC_k$ are not consistent with each other (to my head). One is talking about having an urn of $n$ objects and selecting $k$, while the other is saying the ways to arrange $n$ objects where $k$ are of one kind and $n-k$ are of another. How do you apply the interpretation of select $k$ objects from an urn of $n$ to the case of arranging $0$ and $1$s needed for the binomial distribution?
