I would like to count the cardinality of the following set : Let $k,n$ be two integers such $k < n$ and $n_i$ be an integer for all $i$
$S = \{ (n_1, \ldots , n_k) \in \mathbb{Z}^k \ | \sum_{1 \leq \ i \leq k} n_i = n, \ n_1 > n_i$ for $i = 2, \ldots k \}$
What is $|S|$? Note that this is equivalent to counting the number of ways to place $n$ balls in $k$ boxes such that box number 1 contains more balls than all other boxes. I begun by calculating step by step and I find some formula however it is extremely complicated. Is there a simple way to count $|S|$? Thank you in advance.
Here's a solution as a double sum with binomial coefficients. Let $a_{n,k}$ be the number of non-negative solutions (as suggested in the comments) to your equation. You can introduce a new variable for the first value, say $m=n_1$, and then bound the rest of the values by $m$. So, \begin{eqnarray*} % \nonumber to remove numbering (before each equation) a_{m,n}&=&\sum_m[x^n]x^m (1+x+\dots+x^{m-1})^{k-1} &&\text{generating function solution} \\ &=&\sum_m [x^{n-m}]\left(\frac{1-x^m}{1-x}\right)^{k-1} &&\text{sum of finite geometric series} \\ &=& \sum_m [x^{n-m}]\frac{1}{(1-x)^{k-1}}\sum_\ell(-1)^\ell x^{m\ell}\binom{k-1}{\ell}&&\text{binomial theorem} \\ &=& \sum_{m,\ell} (-1)^\ell \binom{k-1}{\ell}[x^{n-m-m\ell}]\frac{1}{(1-x)^{1+k-2}}&&\text{combining powers}\\ &=&\sum_{m,\ell} (-1)^\ell \binom{k-1}{\ell}\binom{n-m-m\ell+k-2}{n-m-m\ell}&&\text{extraction of coefficient} \end{eqnarray*}