Suppose we need to choose sets of size $2$ out of $\{A, B, C\}.$ The answer is given by $\frac {n!}{ (n - k)! k!}$. So, $n! = \{\text {ABC ACB BCA BAC CAB CBA}\}.$ What lists do $(n - k)!$ and $k!$ represent?
edit: let me reword the question. Consider a group of $5$ men and $7$ women. We can choose a subgroup of $3$ men and $2$ women from the above-mentioned in $\binom 53 \cdot \binom 72$ ways. Now it might sound idle like, but i was wondering if it was possible to do it with counting lists instead of sets.
I suppose you could treat it as follows.
$n!$ counts the number of elements in the set $\Pi$, of all permutations of the set {1,2,3,\dots,n}.
Define an equivalence relation $\sim_1$ on $\Pi$ where $p\sim_1 q$ if $p(i)=q(i)$ for $i>k$. Each equivalence class has $k!$ elements.
Then, define another equivalence class $\sim_2$ on $\Pi/\sim_1$ where two are equivalent if $\{q(k+1),q(k+2),\dots,q(n)\}=\{p(k+1),\dots,p(n)\}$. Then each equivalence class has $(n-k)!$ elements. And $\Pi/\sim_1/\sim_2$ is counting the set you want.
Another way to think of it is as:
$$\binom{n}{k}\cdot k! \cdot (n-k)! = n!$$
Then we are saying: You can find any permutation of $n$ elements by first picking $k$ elements, then permuting those and then permuting the $n-k$ remaining elements.
So, let's take $n=4,k=2$. Then there are $\binom{4}{2}$ ways to pick two elements. Say we've picked $\{A,B\}$. Then there are $2$ ways to permute these, and two ways to permute the remaining $\{C,D\}$, so you get $4=2!2!$ cases:
$$ABCD, ABDC, BACD,BADC$$
Multiply by $\binom{4}{2}$ and we should get all permutations of $n$ elements, so:
$$\binom{4}{2}\cdot 2!\cdot 2! = 4!$$