Why do we divide Permutations to get to Combinations?

Maybe, looking at an example clarifies this best :

You have $20$ objects and have to choose $5$ of them. How many possibilities are there ?

You have $20,19,18,17,16$ choices explaining how $\frac{20!}{15!}$ comes into the play.

Now each combination can appear in $5!$ possible orders which correspond to the same combination. Therefore we have to divide by $5!$ to find the number of combinations.

This can be generalized to arbitary numbers explaining how the binomial coefficient emerges.

Let's bunch together permutations that have the same content. Each resulting equivalence class has $k!$ members. Therefore, the number of such classes is $k!$ times smaller than the number of permutations. A combination is just an equivalence class or, if you prefer, one nominated element thereof.

Assume that you want to know the number of combinations of $k$ elements out of $n$ elements. One way to pick such a combination is the arrange all $n$ elements in any arbitrary order (there are $n!$ ways to do so) and pick the first $k$ of them as "your selection". As the remaining $n-k$ elements can be arranged in any order, each ordered selection of the first $k$ elements appears $(n-k)!$ times in the set of all permutations of the $n$ elements, so you have to divide $n!$ by $(n-k)!$ to count the different (ordered) "selections" of $k$ elements. Now, as you are not interested in the order of the $k$ elements, all $k!$ permutations of a "selection" represent the same combination, so you still have to divide the number by $k!$, which makes a total of $\frac{n!}{k!(n-k)!}$ combinations. So, in short, this dividing "removes the order" when counting (here: the order of the first $k$ elements and the order of the remaining $n-k$ elements).