The overall number of ways to put $n$ balls in $m$ holes corresponds
to the number of weak compositions of $n$ into (exactly) $m$ parts.
A weak composition of $n$ is a $m$-tuple which is a integral solution to
$$
\left\{ \matrix{
0 \le x_{\,j} \hfill \cr
x_{\,1} + x_{\,2} + \cdots + x_{\,m} = n \hfill \cr} \right.
$$
and $x_k$ represents the number of balls in hole $k$, which can be even $0$.
The number of solutions is given by
$$
N_{ \le \,m} (n,m) = \left( \matrix{
n + m - 1 \cr
m - 1 \cr} \right)
$$
The standard combinations are instead the $m$-tuples with strictly positive parts
$$
\left\{ \matrix{
1 \le y_{\,j} \hfill \cr
y_{\,1} + y_{\,2} + \cdots + y_{\,m} = n \hfill \cr} \right.
$$
corresponding to the case in which no hole is empty and their number
is
$$
N_0 (n,m) = \left( \matrix{
n - 1 \cr
m - 1 \cr} \right)
$$
Clearly we have
$$
N_{ \le \,m} (n,m) = \left( \matrix{
n + m - 1 \cr
m - 1 \cr} \right) = \sum\limits_{0\, \le \,k \le \,m} {\left( \matrix{
m \cr
k \cr} \right)\left( \matrix{
n - 1 \cr
m - k - 1 \cr} \right)} = \sum\limits_{0\, \le \,k \le \,m} {N_{k} \left( {n,m} \right)}
$$
and $N_k$ represents the number of compositions with exactly $k$ zeros.
Now you have all the components to solve your problem, either under the meaning of
"exactly one" or "at least one" empty hole .
Note
What said above, when translated into probability dividing by $N_{ \le \,m} (n,m)$, will assume that each composition is equiprobable.
That it the case for instance for points in a $m$-D space, uniformly distributed on the diagonal plane $x_1+x_2+\cdots +x_m=n$.
But randomly throwing $n$ balls into $m$ holes prefigures a different probability space.
Consider in fact the balls to be labelled according to the throwing sequence.
Then you have a total of $m^n$ equiprobable layouts of "balls-in-holes", differing by occupancy number
or by the labeling of the occupants. However, in each hole, the balls will be arranged bottom-up (consider the hole diameter equal to that
of the balls) in an increasing order.
So their disposition will correspond to the number of ways to partition the set ${1,2,\cdots ,n}$ into $j$ non-empty subsets
(= occupied holes), multiplied by the falling factorial ${m^{\;\underline {\,j\,} } }$, the number of ways to place the $j$ subsets
into $m$ places.
Thus
$$
m^{\,\,n} = \sum\limits_{0\, \le \,j \le \,\min (n,m)} {\left\{ \matrix{
n \hfill \cr
j \hfill \cr} \right\}m^{\;\underline {\,j\,} } }
$$
So in this scheme the number of configurations with exactly $k$ empty holes is
$$
N_{\,k} (n,m) = \left\{ \matrix{
n \cr
m - k \cr} \right\}m^{\;\underline {\,m - k\,} }
$$
