Assume $s, j \in\mathbb{N}$. Define the set
$$\mathcal{A}\_{j,s}:=\{(n\_1,n\_2,\dots,n\_j)\in\mathbb{Z}\_{\geq0}^j\vert \,
n\_1+2n\_2+\cdots+jn\_j=j, \, n\_1+n\_2+\cdots+n\_j=s\}.$$ 



> 
> **Question.** Is there a combinatorial argument for this?
>  $$\sum\_{s=0}^j(-1)^{j-s}\sum\_{\mathcal{A}\_{j,s}}\binom{s}{n\_1,\dots,n\_j}=0.$$
>  I've an algebraic proof.
> 
> 
> 


**Update.** Here is the proof that I mentioned. Below, $[x^k]F(x)$ denotes the coefficient of $x^k$ in $F(x)$.


[This Multinomial Theorem](http://mathworld.wolfram.com/MultinomialSeries.html) with the specializations $a\_i=x^i,\, k=j$ and $n=s$ offers 
$$(x+\cdots+x^j)^s=\sum\_{n\_1+\cdots+n\_j=s}\binom{s}{n\_1,\dots,n\_j}x^{n\_1+2n\_2+\cdots+jn\_j}
=\sum\_{j} x^j\sum\_{\mathcal{A}\_{j,s}}\binom{s}{n\_1,\dots,n\_j}.$$
It follows that the coefficient of $x^j$ in $(x+\cdots+x^j)^s$ equals $\sum\_{\mathcal{A}\_{j,s}}\binom{s}{n\_1,\dots,n\_j}$. On the other hand, $(x+\cdots+x^j)^s=x^s(1-x^j)^s(1-x)^{-s}$. Consequently, the coefficient of $x^j$ is:
\begin{align} [x^j]\,\,\,x^s(1-x^j)^s(1-x)^{-s}&=[x^j]\,\,\,x^s\sum\_a(-1)^s\binom{s}ax^{aj}\sum\_b\binom{s+b-1}bx^b \\
&=[x^j]\,\,\,x^s\sum\_b\binom{s+b-1}bx^b=\binom{j-1}{j-s}=\binom{j-1}{s-1}. \end{align}
We proved $\sum\_{\mathcal{A}\_{j,s}}\binom{s}{n\_1,\dots,n\_k}=\binom{j-1}{s-1}$.
Binomial Theorem gives: $\sum\_{s=1}^jy^s\binom{j-1}{s-1}=(1+y)^{j-1}$. We conclude that
$$\sum\_{s=0}^j(-1)^{j-s}\sum\_{\mathcal{A}\_{j,s}}\binom{s}{n\_1,\dots,n\_k}=\sum\_{s=0}^j(-1)^{j-s}\binom{j-1}{s-1}=0.$$



In search of a combinatorial reasoning for a vanishing sum
For fixed $s,j>0$, $\sum\_{\mathcal{A}\_{j,s}}\binom{s}{n\_1,\dots,n\_j}$ enumerates the compositions of $j$ into $s$ positive parts by ordering the corresponding partitions. That is, we have
$$\sum\_{\mathcal{A}\_{j,s}}\binom{s}{n\_1,\dots,n\_j}=\binom{j-1}{s-1}$$
and the identity in question follows trivially.


