# 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 $\left(\!\!\binom{10}{5}\!\!\right) = \binom{10+5-1}{5} = 2002$.
• The answer $\binom{14}{5}$ is given by the standard "stars-and-bars" method. Arrange nine "bars" (separators) and five "stars" (placeholders for elements of multisets) in a line. Any items to the left of the first bar are counted as zeros; items between the ith and i+1st bar are counted as i's; items after the ninth bar are 9's. – hardmath Nov 23 '13 at 16:11