Difference between permutation and combination? 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.
 A: Permutations assume there are no repetition of the same objects. One way to see why the formula holds is as follows:
Say you have $n$ objects and want to choose $r$ from them.
You can choose any of the $n$ for the first one,
You can choose any of the $n-1$ for the second one,
.
.
.
You can choose any of the $n-r+1$ for the $r^{th}$ one.
Multiplying all these altogether gives you the number of possibilities:
$$
n(n-1)\cdots (n-r+1)=\frac{n!}{(n-r)!}
$$
A: There is some common confusion between these two terms.
The word "permutation" in general refers to one of three things depending on context. It can mean the order (arrangement) of a set as in combinatorics. Or it can refer to an arrangement of a subset of a given size as also in combinatorics. Or it can refer to an OPERATION of REarrangement in a space of 
such operations as in group theory, which is a very different thing.
"Combination" refers only to the makeup (constituency) of a subset with no concept of order. 
The term "combinations" refers to the number of subsets of a given size containing different constituents.
The combinatorial meanings of both of these terms are still valid when applied to sets with duplicate elements. A permutation (arrangement or rearrangement) can apply to a set or subset that contains duplicates. But "combination" usually assumes distinct elements in the subset, though the original set can contain duplicates. Of course, if so indicated, a combination could also contain duplicates. The presence of duplicates affects the combinatorial formulas for all of these.
A: 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 $n$.
A: I just want to demonstrate via a figure the difference between permutation and combination. Hope helpful for you. 

