TITLE: How to reason about the combinatorial term "$n$ choose $k$" (i.e., $\binom{n}{k}=\frac{n!}{k!(n-k)!}$)
QUESTION [6 upvotes]: The combinatorial number is the number of picking $k$ unordered outcomes out of $n$ possible choices. In that setting, we have a set $A$ with $|A|=n$ and the combinatorial numbers is just really the number of subsets $S\subset A$ with $|S| = k$.
This number is
$$\binom{n}{k} = \dfrac{n!}{k!(n-k)!},$$
but how can we reason about this? How can we derive this formula? I've seem some people reasoning about this in the following way: the number of ways to choose permutations with size $k$ among $n$ objects is
$$n(n-1)\cdots (n-k-1) = \dfrac{n(n-1)\cdots (n-k+1)(n-k)\cdots 1}{(n-k)(n-k-1)\cdots 1} = \dfrac{n!}{(n-k)!},$$
then we have to divide by $k!$ to disconsider the order. Why is that? Why dividing by $k!$ we get the number of subsets of $A$ with size $k$?

REPLY [2 votes]: This might be verbose but bear with me, imagine you have a bag with $10$ marbles and you want to figure out all the ways you can pick the $10$ marbles out of the bag (so lists of 10 marbles in different combinations), for the sake of clarity imagine every one of those marbles have a unique color. In your first choice you have $10$ different marbles to choose from, let's say you pick pink, now you have $9$ different choices, but what if you originally picked orange, now you still have $9$ different choices. What this shows is that for every first marble you choose you have 9 more to go for, or in other words $10\times 9$ choices.  If we continue for the third marble we have 8 possible choices after we removed the second marble, so again we have $9\times 8$ choices but remember we had 10 groups of choices from the beginning, so in total we have $10\times 9\times 8$ choices. This goes on until you have 1 marble left in which case you have a total choice of $10 \times 9 \times 8 \times \cdots \times 1 = 10!$ possible choices for our bag of 10 marbles. Now let's say we actually only want to pick 4 marbles, not all 10. That means out of the possible $10!$ choices for all $10$ we only get $10\times 9\times 8 \times 7$ or equivalently:
$$\frac{10!}{(10-4)!} = \frac{10!}{6!} = 7\times 8\times 9\times 10$$
Possible choices. What if though, we don't care about the order of choice, so the marbles we do pick, we can label them as $R,G,B,Y$ order does not matter. So, $R,G,B,Y$ is equivalent to $R,B,G,Y$. So, if want to choose $4$ marbles from our bag of $10$ and we don't care about the arrangement of our elements we need to divide it by $4!$ which represents our number of ways to arrange $4$ elements, hence:
$$\frac{10!}{4!\cdot(10-4)!} = 210$$