Fock-space-related combinatorial problem v2 This problem comes from counting degrees of freedom in a system of bosons.
The integers $k_n$, $n=1..m$ are allowed to take values on the interval $-\Lambda \leq k_n\leq\Lambda,\,k_n\neq0$. I need to calculate the total number of $m$-tuples of $k_n$'s whose elements sum to zero:
$$
\underbrace{
\{k_1,k_2,\ldots,k_m\}:
\quad
k_1+k_2+\ldots+k_m = 0}_{\text{How many of those?}}
$$
Actually, even an asymptotic estimate would work for me. Can this be reduced to some well-known problem?
My idea is that we could probably shift everything by a constant and use the number of integer partitions?..
UPDATE
I forgot to mention that the order DOES matter in this problem ($m$-tuples differing only by order count as different ones), physically those correspond to different momentum modes.
 A: Let $X_1,X_2,\dots,X_m$ be iid random variables uniformly distributed over the nonzero integers in $[-\Lambda,\Lambda]$, and let $S_m=X_1+\dots+X_m$. You want the probability that $S_m=0$, times $(2\Lambda)^m$. 
As long as $m$ is large enough, then by the central limit theorem, $S_m/\sqrt{m} \approx N(0,\sigma^2)$, where $\sigma^2$ is the variance of $X_1$. Therefore, letting $\Phi(x)$ be the standard normal cdf, where $\Phi'(x)=(2\pi)^{-1/2}\exp(-x^2/2)$,  we have
\begin{align}
P(S_m=0) 
  &=P\left(-\frac{1/2}{\sigma\sqrt m}<\frac{S_m}{\sigma\sqrt{m}}<\frac{1/2}{\sigma\sqrt m}\right) 
\\&\approx\Phi\left(\frac{1/2}{\sigma\sqrt m}\right)-\Phi\left(\frac{1/2}{\sigma\sqrt m}\right)
\\&\approx\frac1{\sigma\sqrt{m}}\cdot \Phi'(0)
\\&=\frac1{\sigma\sqrt{2\pi m}}
\end{align}
Now, what is $\sigma$? Since $X_1$ is approximately a discrete uniform on the interval $[-\Lambda,\Lambda]$, which has variance $((\Lambda-(-\Lambda)+1)^2-1)/12$ (see Wikipedia: discrete uniform), it will be true that $\sigma\approx \Lambda/\sqrt{3}$. Therefore,
$$
\text{# of solutions}\approx \frac1{\Lambda \sqrt{\frac23 \pi m}}\cdot (2\Lambda )^m
$$
Furthermore, you can quantify the error in this approximation. The error in the approximating $S_m$ by a normal distribution is described by the Edgeworth series. Fortunately, since $X_1$ is symmetric and therefore has zero skewness, the error in this approximation is $O(1/m)$, so 
$$
\text{# of solutions}=(2\Lambda )^m\left(\frac1{\Lambda \sqrt{\frac23 \pi m}}+O(1/m)\right)
$$

You can also given an exact answer to this problem in terms of a double sum of binomial coefficients. This may be useful if you want to verify the quality of the previous asymptotic answer.
Let us first count tuples without the constraint $k_n\neq 0$. If you add the quantity $\Lambda+1$ to each entry, you get a tuple of positive integers between $1$ and $2\Lambda+1$ summing to $m(\Lambda+1)$. Referring to here, the number of solutions is
$$
\sum_{i=0}^m (-1)^i\binom{m}i\binom{m(\Lambda+1)-(2\Lambda+1)i -1 }{m-1}
$$
Here, we use the convention that $\binom{r}k=0$ when $r$ is negative. 
Now, to add back in the constraint $k_i\neq 0$, we use the principle of inclusion exclusion. Take all the solutions counted by the last equation, subtract the ones where some variable is zero, add back in the doubly subtract solutions, etc. The result is
$$
\sum_{k=0}^m (-1)^{m-k}\binom{m}k
\sum_{i=0}^k (-1)^i\binom{k}i
\binom{k(\Lambda+1)-(2\Lambda+1)i -1 }{k-1}
$$
A: Hint: If you can take any number in $[-\Lambda ,\Lambda]$(If you can not, i.e., your $k_n$ are fixed then this is a modification of subset sum problem in dynamic programming). You divide the $\{k_i\}_{i\in [m]}$ in two groups: the non-negatives and the negatives so that they add up to $0.$ So let $i$ the number of non-negatives, then the total number of tuples would be
$$\sum _{i=1}^m\binom{m}{i}\sum _{c=i}^{mi}\underbrace{P_{\geq 0,\leq \Lambda,i,c}}_{\text{non-negative}}\underbrace{P_{>0,\leq \Lambda,m-i,c}}_{\text{negatives}},$$
where $P_{\geq 0,\leq \Lambda,n,x}$ is the number of non-negative $n-$ tuples which add up to $x,$ and $P_{> 0,\leq \Lambda,n,x}$ is the number of positive $n-$tuples which add up to $x.$
You can compute this numbers using inclusion-exclusion and stars and bars.
