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What is the number of ways to choose $n$ digits from $\{0,1,2,3,4,5,6,7,8,9\} $, where repetition of digits is allowed?

For example, if $n=3$, $111$ and $121$ are valid but $321$ and $123$ are considered same and not different. Similarly, $121,211,112$ are considered same.

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If order of digits is not an issue, you can use stars and bars to find the answer, since it is then counting multisets with $n$ elements from the set $\{0,1,2,3,4,5,6,7,8,9\}$.

If an all-zeroes selection is allowed, the answer is thus $\binom {n+9}{9}$; otherwise subtract $1$ to remove that case.

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