Put 13 identical balls in 8 different holes. What is the probability that there's one empty hole? Put $13$ identical balls in $8$ different holes. What is the probability that there is one empty hole?
For all the options it's easy. divide into $7$ partition, and - $\binom{8+13-1}{13}=\binom{20}{13}$
Now I want to calculate the options for one empty hole. What I did:
I left out one of the holes, and divided into $6$ partitions. This way there will be one empty hole for sure. In order to get ONLY one empty hole, I put in each hole 1 ball, and then I get: $\binom{12}{6}$
The probability:
$P(A)=\frac{\binom{12}{6}}{\binom{20}{13}}$
I'm not sure about this answer, because I checked only for $1$ specific hole. Should I multiple it by $8$?
 A: 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
$$
N_{\,k} (n,m) = \left\{ \matrix{
  n \cr 
  m - k \cr}  \right\}m^{\;\underline {\,m - k\,} } 
$$
