How many ways are there to place $n$ identical balls into $m$ numbered bags from 1 to $m$, so that none of the bags are empty?
We draw the $n$ balls in a line. Placing them into $m$ numbered bags is equivalent to placing $m-1$ bars separating the $m$ bags, with $n-1$ available positions, without placing two bars in the same position. Thus, we have $\binom{n-1}{m-1}$ possibilities.

Note that if we do not require the bags to be non-empty, this is equivalent to the same problem if we add one ball to each bag. Therefore, the number of ways to place $n$ identical balls into $m$ numbered bags from 1 to $m$ that can be empty is $\binom{n+m-1}{m-1}$.