Combination with repetition and no order. I'm looking for a way to calculate the number of combination of 10 choose 5 => $_5^{10}C$ while allowing any number repetitions and counting 12345 and 54321 only once (order isn't important, ie I count 11355 but not then 35115).
I think this number is majored by $10^5$, but how to remove ordering number ?

(Note: I'm using the notation $\binom{n}{k}$ here instead of $^n_kC$.)
A multiset is a collection of things that allows repetition but ignores order. The number of multisets of size $k$ with elements from a set of size $n$ is
$$\left(\!\!\binom{n}{k}\!\!\right) = \binom{n+k-1}{k} = \frac{(n+k-1)!}{k! (n-1)!}.$$
In your case, you'll have $\left(\!\!\binom{10}{5}\!\!\right) = \binom{10+5-1}{5} = 2002$.
