# 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.

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Counts the number of ways of choosing and ordering the choices. –  Arturo Magidin Jul 6 '12 at 2:44

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 $n$.

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I just want to demonstrate via a figure the difference between permutation and combination. Hope helpful for you.

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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,

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You can choose any of the $n-r+1$ for the $r^{th}$ one.

Multiplying all these altogether gives you the number of possibilities: $$n(n-1)\cdots (n-r+1)=\frac{n!}{(n-r)!}$$

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