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My understanding is that a permutation is used to find the number of rearrangements of different elements, taking into account possible orders. A combination is used to find the number of rearrangements of different elements leaving out order.

If you have a combination with repetition, isn't that the same thing as a permutation? What exactly is being repeated?

I was hoping someone could clear this up for me. Thanks.

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In a permutation you have only one each of the available elements. So if you are taking permutations of three elements out of $\{1,2,3,4\}$ some of the legal ones are $(1,2,3), (3,1,4), \text {and} (2,3,1)$. There are a total of $24$ of them. As you say, combinations do not care about order. There are only four taking combinations of three out of the same set: $\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\}$. Combinations with repetition would allow multiples, so $\{1,2,2\}$ becomes legal, but is still the same as $\{2,1,2\}$. These are called multisets.

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Permutation P(n,r) is used when order matters. Ex. How many arrangements of ABCDE if A goes first.

Permutation with repetition still order matters, but use $n^r$. Ex. You have three slots to fill up three numbers in, and they can be repeated. $10 * 10 * 10$ or $10^3$.

Combination C(n,r) or $n\choose r$ with order does not mater. Ex. Pick 7 books out of 10 books on the shelf, the books are not the same.

Combination with repetitions C(n+r-1, r) = $n+r-1\choose r$ . Ex. You need 10 screws and you have 18 screws to choose from. (It is obvious you can have the same screw more than once).

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  • $\begingroup$ It is clear that combinations with repetition is no the same as permutation. $\endgroup$ – Kourosh Feb 19 '17 at 8:40

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