Permutation:
$$P(n,r) = \frac{n!}{(n-r)!}$$
Combination:
$$C(n,r) = \frac{n!}{(n-r)!r!}$$
Apparently, you use combination when the order doesn't matter. Great. I see how a combination will give you all the possible well, combinations. However, I don't see what exactly does a permutation do then.
Best Answer
If you see how combinations work then you're most of the way there. Say I want to pick 3 letters out of ABCDE. There are $C(5,3)$ ways of doing this. But if order matters, then several things that I counted as the same are now different. Picking $ABC$ now generates $ABC,ACB,BAC,BCA,CBA,CAB$ as different choices, when they weren't before. How many different choices are there? Well that's the number of ways I can rearrange the $r$ chosen letters, which is $r!$. So if permutations matter:
$$P(n,r)=r!\cdot C(n,r)=\frac{r!n!}{(n-r)!r!}=\frac{n!}{(n-r)!}$$
Permutations are the number of different ordered selections of $r$ elements from a set of $n$.