Probability Problem: Drawing balls from an urn. The problem is stated as such:  
"An urn contains $n$ red and $m$ blue balls. They are withdrawn one at a time until a total of $r$, $r \leq n$, red balls have been withdrawn. Find the probability that a total of $k$ balls are withdrawn. Hint: A total of $k$ balls will be withdrawn if there are $r − 1$ red balls in the ﬁrst $k − 1$ withdrawals and the $k$th withdrawal is a red ball."
The answer that I have come up with is
$$\frac{{n \choose r-1}{m \choose k-r}(n-r+1)}{{n+m \choose k-1}(n+m-k+1)}$$
which I am fairly confident in.
Several of the solutions I have found have claimed the solution to be
$$\frac{{n \choose r-1}{m \choose n-r}(n-r+1)}{{n+m \choose k-1}(n+m-k+1)}.$$
If these are indeed also correct, then I don't see where $m \choose n-r$ could be coming from. Would this not imply that $n=k$?

A numerical example shows that the second expression cannot be right. Take $n=m=2$ and $r=1$; clearly $k$ can only be $1,2$, or $3$. Then
$$\frac{\binom{n}{r-1}\binom{m}{n-r}(n-r+1)}{\binom{n+m}{k-1}(n+m-k+1)}=\frac4{\binom4{k-1}(5-k)}\;,$$
and
$$\sum_{k=1}^3\frac4{\binom4{k-1}(5-k)}=\frac44+\frac4{12}+\frac4{12}=\frac53\;,$$
which is nonsense: the sum has to be $1$.
Your expression is correct, and we may note as a small check that it passes this test:
$$\frac{\binom{n}{r-1}\binom{m}{k-r}(n-r+1)}{\binom{n+m}{k-1}(n+m-k+1)}=\frac{2\binom2{k-1}}{\binom4{k-1}(5-k)}\;,$$
so
$$\sum_{k=1}^3\frac{2\binom2{k-1}}{\binom4{k-1}(5-k)}=\frac24+\frac4{12}+\frac2{12}=1\;.$$
