I want to find the number of ways of selecting $k$-element subsets from an $n$-element set.
Suppose we select the elements one by one. There are $n$ ways of selecting the first one, leaving $n-1$ ways for the second, and so on, until we reach $n-k+1$ ways to select the $k$-th one, so there are:
$$n \cdot (n-1) \cdots (n-k+1)$$
ways. However, apparently we should divide by $k!$ Why? For what it's worth, I understand that we could select them two by two etc as well. I just don't understand how dividing by $k!$ fixes it all.
Best Answer
Dividing by $k!$ is needed because in your
$$n \cdot (n-1) \cdots (n-k+1),$$
you're actually permuting those $k$ element in a row. E.g. If you choose out
$$1,2,3$$
you will also count
$$1,3,2\\ 2,1,3\\ 2,3,1\\ 3,1,2\\ 3,2,1$$
since by your formula it just says $3\times2\times1=6,$ which includes all above shown, but they all represent the same set $\{1,2,3\}.$