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A combination is a way of arranging items from a set when the order of the arrangement does not matter. An example would be choosing three random numbers from a list of nine. It would not matter if you chose $$1$$ then $$7$$ then $$4$$ or if you chose $$7$$ then $$1$$ then $$4$$.
A permutation is a way of arranging elements from a set when the order of the arrangement does matter. An example of this would be the code to a lock. If the code is $$1,7,4$$ then it cannot be entered as $$1,4,7$$ or $$4,7,1$$ or any other order.
As long as there is more than one item in a set, there will always be more permutations than combinations.

Both combinations and permutations can occur with or without repetition, meaning they either contain one or more items multiple times or they do not. Though this may not seem like it would make much of a difference, repeating items in a set quite drastically changes the way we should approach it.

Notations
$$n$$ usually represents the total number of items in a set.
$$k$$ usually represents the number of items in a selected subset.
$$C$$ usually represents combinations.
$$P$$ usually represents permutations.

$$P\left(n,k\right)$$ represents the number of different permutations of a subset ($$k$$) out of a larger set ($$n$$) and can also be written as:
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$$C\left(n,k\right)$$ represents the number of different combinations of a subset ($$k$$) out of a larger set ($$n$$) and can also be written as:
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This notation is also sometimes referred to as "n choose k".

Formulas
We use the factorial function when solving permutations and combinations.

Permutations with repetition
$$P\left(n,k\right)=n^k$$
E.G: How many different permutations of a subset of $$3$$ out of a total $$9$$ items are there when repetitions can occur?
$$P\left(9,3\right)=9^3=729$$

Permutations without repetition
$$P\left(n,k\right)=\frac{n!}{\left(n-k\right)!}$$
E.G: How many different permutations of a subset of $$3$$ out of a total $$9$$ items are there when repetitions cannot occur?
$$P\left(9,3\right)=\frac{9!}{\left(9-3\right)!}=\frac{9!}{6!}=\frac{9·8·7·6!}{6!}=9·8·7=504$$

Combinations with repetition
$$C\left(n,k\right)=\frac{\left(k+n-1\right)!}{k!\left(n-1\right)!}$$
E.G: How many different combinations of a subset of $$3$$ out of a total $$9$$ items are there when repetitions can occur?
$$C\left(9,3\right)=\frac{\left(3+9-1\right)!}{3!\left(9-1\right)!}=\frac{11!}{3!·8!}=\frac{11·10·9·8!}{3!·8!}=\frac{11·10·9}{3!}=$$
$$\frac{11·10·9}{3·2·1}=11·5·3=165$$

Combinations without repetition link to this drill
$$C\left(n,k\right)=\frac{n!}{k!\left(n-k\right)!}$$
E.G: How many different combinations of a subset of $$3$$ out of a total $$9$$ items are there when repetitions cannot occur?
$$C\left(9,3\right)=\frac{9!}{3!\left(9-3\right)!}=\frac{9!}{3!·6!}=\frac{9·8·7·6!}{3!·6!}=\frac{9·8·7}{3!}=\frac{9·8·7}{3·2·1}=3·4·7=84$$