How many ways can n identical balls be distributed into k distinct boxes, such that at least one box is empty? This is a problem in my combinatorics book that uses the principle of inclusion-exclusion. I can follow almost all of what is said, except the book says that if we consider $A_{i}$ to be the set of solutions where box i is empty, then $|A_{i}| = {n-(k-1)-1 \choose k-1}$. 
The book does not explain why this is true. And I want to know why, since I thought that $|A_{i}| = {n+(k-1)-1 \choose k-1}$. 
So that you can get to the root of my misunderstanding, my reasoning was that a placement of n identical balls into k distinct boxes is the same as the number of nonnegative integer solutions to $x_{1}+\cdots+x_{k} = n$. 
Any help would be much appreciated!
 A: To begin with, I'd solve the problem differently: By a stars-and-bars, there are $n-1\choose k-1 $ ways to place $n$ balls into $k$ bins such that no bin is empty. Subtract this from the $n+k-1\choose k-1$ ways to place $n$ balls into $k$ bins  without restriction.
A: First of all, let's clear that, in common speaking
"distributing $n$ identical (undistinguishable) balls into $k$ distinct (distinguishable) boxes" 
does not have the same acception as
"launching $n$ identical (undistinguishable) balls into $k$ distinct (distinguishable) boxes" 
In the first case (distributing, pouring, ..) it is understood that you consider the number of different
occupation hystograms, i.e. the number of k-tuples $(x_1,x_2, \cdots , x_k)$ such that
$$
\left\{ \matrix{
  {\rm 0} \le {\rm integer}\;x_{\,j}  \hfill \cr 
  x_{\,1}  + x_{\,2}  + \; \cdots \; + x_{\,k}  = n \hfill \cr}  \right.
$$
which are  the "weak"  Compositions of $n$ into $k$ parts,
and in your particular case, the complement of the case in which no box is empty 
$$
\left\{ \matrix{
  {\rm 1} \le {\rm integer}\;x_{\,j}  \hfill \cr 
  x_{\,1}  + x_{\,2}  + \; \cdots \; + x_{\,k}  = n \hfill \cr}  \right.\quad  \Rightarrow \quad \left\{ \matrix{
  {\rm 0} \le {\rm integer}\;y_{\,j}  \hfill \cr 
  y_{\,1}  + y_{\,2}  + \; \cdots \; + y_{\,k}  = n - k \hfill \cr}  \right.
$$
which are  the  "standard" Compositions of $n$ into $k$ parts. 
As explained in the referenced article they are respectively
$$
N_w  = \binom{n+k-1}{n} \quad 
N_s  =\binom{n-1}{n-k}  \quad \left| {\;0 \le n,k} \right.
$$
note that this way of writing the binomials  extends the definition also to null values of the parameters.
The repartition of $N_w$ over the configurations with exactly $j$ empty boxes is given by
$$
N_w  = \binom{n+k-1}{n}
   = \sum\limits_{\left( {0\, \le } \right)j\left( { \le k} \right)} {\binom{k}{j} \binom{n-1}{n - \left( {k - j} \right) } }
$$
The number of configurations in which precisely the box $i$ is empty will then clearly be
$N_w(n,k-1)$ if the others may be empty or not, and $N_s(n,k-1)$ if the others are non-empty.
So you are right, and 
the global answer to your problem will
$$
N = N_w  - N_s  =  \binom{n+k-1}{n} - \binom{n-1}{n-k} 
$$
