Yes, that's one way to do it. Another way to do it is calculate the probability of selecting $k-1$ of the $n-1$ not special balls (and one of the one special balls) out of all the ways to choose $k$ of $n$ balls. Then expand and cancel terms.
$$\begin{align}\mathsf P(A) & = \dfrac{\dbinom{1}{1}\dbinom{n-1}{k-1}}{\dbinom{n}{k}}
\\[1ex] & = \dfrac{(n-1)!/(k-1)!(n-k)!}{n!/k!(n-k)!}
\\[1ex] & = \dfrac k n
\end{align}$$
And in hindsight you should see that this is the probability that the ball will be among the first $k$ of $n$ places if you line the balls up such that the special ball has an equal probability of being in each place.
Keep this in mind for the future. Sometimes it is easier not to count the ways to select balls into places, but to count ways to select the places to for the balls to be put.
If things look complicated, take a step back and see if you can look at the problem from a different angle.