Distribute $N$ indistinguishable balls, where $N$ is even, into $n$ distinguishable boxes if each box can contain at most $N/2$ balls How many ways are there to distribute $N$ ($N$ is an even number)  indistinguishable balls into $n$ distinguishable boxes if each box can contain at most $\frac{N}{2}$ balls.
My solution:
There are 
$${N+n-1}\choose{n-1}$$
different ways to distribute $N$ indistinguishable balls into $n$ distinguishable boxes.
But we must subtract the number of those cases in which there are more than $\frac{N}{2}$ balls in one of the boxes. 
If we put $\frac{N}{2}+1$ balls in the first box we get 
$${\frac{N}{2}-1+n-2}\choose{n-2}$$
different ways to  distribute $\frac{N}{2}-1$ balls into the boxes $2,\cdots , n$
Similary, for the case where we have $\frac{N}{2}+j$ balls in (exactly) one on the boxes, $j=1,..., \frac{N}{2}$ we get the formula
$${\frac{N}{2}-j+n-2}\choose{n-2}$$.
Then, the answer is
$${{N+n-1}\choose{n-1}}-n\sum_{j=1}^{\frac{N}{2}}{{\frac{N}{2}-j+n-2}\choose{n-2}}$$
 A: Since $N$ is even, let $N = 2k$.  Then the question can be rephrased as follows:  In how many ways can $2k$ indistinguishable balls be placed in $n$ boxes if at most $k$ balls can be placed in one box?
Clearly, $n \geq 2$, for otherwise we would not be able to distribute all the balls to the boxes given the restriction that at most $k$ balls may be placed in one box.
The number of ways of placing $2k$ balls in $n$ boxes is the number of solutions of the equation
$$x_1 + x_2 + x_3 + \ldots + x_n = 2k \tag{1}$$
in the nonnegative integers.  Since a particular solution of equation 1 corresponds to the placement of $n - 1$ addition signs in a row of $2k$ ones, equation 1 has 
$$\binom{2k + n - 1}{n - 1}$$
solutions since we must choose which of the $2k + n - 1$ positions (for $2k$ ones and $n - 1$ addition signs) will be filled with addition signs.
From these, we must exclude those cases in which one box receives more than $k$ balls.  Notice that there can be at most one such box since $2(k + 1) = 2k + 2 > 2k = N$.
Suppose that box 1 contains more than $k$ balls.  Then $x_1' = x_1 - (k + 1)$ is a nonnegative integer.  Substituting $x_1' + k + 1$ for $x_1$ in equation 1 yields 
\begin{align*}
x_1' + k + 1 + x_2 + x_3 + \ldots + x_n & = 2k\\
x_1' + x_2 + x_3 + \ldots + x_n & = k - 1 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with 
$$\binom{k - 1 + n - 1}{n - 1} = \binom{n + k - 2}{n - 1}$$
solutions.  By symmetry, there are an equal number of solutions of equation 1 for each of the $n$ variables that could exceed $k$.  Hence, we must exclude 
$$n\binom{n + k - 2}{n - 1}$$
solutions.  
Consequently, there are 
$$\binom{2k + n - 1}{n - 1} - n\binom{n + k - 2}{n - 1}$$
ways to distribute $N = 2k$ indistinguishable balls to $n$ boxes if no box can contain more than $N/2 = k$ balls.
