We know ${n \choose k}$ means the number of of size $k$ of a set of $n$ objects. Why is that? The argument is: First we pick the first element. We have $n$ choices to do that. Now, we pick the second member and we have $n-1$ choices and so on until we get to the $k$ member and to do this we get $n-(k-1)$. So,
$$ n(n-1)(n-2)\cdots(n-k+1)$$
is the number of ways to pick $k$ members from a set of $n$ members. My question is: Why do we have to divide this by $k!$?? I am having a hard time understanding this concept.
$$ 6\times 5\times 4 = 120 $$ If the six objects are called $A,B,C,D,E,F,$ then the three chosen objects may be $A,D, F.$ There are three of those and we divide by $3!=3\times2\times 1 = 6.$ Here's why: $$ \begin{array}{rl} 1 & ADF \\ 2 & AFD \\ 3 & DAF \\ 4 & DFA \\ 5 & FAD \\ 6 & FDA \end{array} $$ That set of objects appears $6 = 3\times2\times1=3!$ times. But we want to count it only once, so we divide by $6.$