How do I prove :


$\sum\_{j=2}^{n} (-1)^j {\frac {M(n+j,j;2)}{j!}} = (-1)^n n! + 1$?


where $M(n+j,j;2)$ is the multinomial sum $M(n+j,j;2) = \sum\_{t\_1 + t\_2 + \dotsc + t\_j = n+j, t\_k \geq 2} {n+j \choose t\_1 \dotsc t\_k}$ which denotes the number of surjective functions from $n+j$ points to $j$ points with at least two elements in each pre-image. 


I have tried using the recursion formulas of counting doubly surjective functions which didn't help much. My thought was the alternate signs appear for some Moebius maps on some partial order on the disjoint set of surjective maps from $n+j$ points to $j$ points up to permuting the image points (which explains division by $j!$). My definition of reshuffling images which give a partial order which give $0$ to distant points. 



Some strange multinomial averaging
Denote by $f(n,m)=M(n+m,m;2)/m!$ the number of partitions of $\{1,2,\dots,n+m\}$ onto $m$ subsets of size at least 2 (observation of Darij Grinberg). Your identity rewrites as $A(n):=\sum\_{j\geqslant 1} (-1)^jf(n,j)=(-1)^nn!$. We have $f(n,m)=mf(n-1,m)+(n+m-1)f(n-1,m-1)$: the first summand corresponds to the partitions in which $n+m$ belongs to a subset of size at least 3; the second to the partitions in which $n+m$ belongs to a subset of size 2. Substitute this to $A(n)$, we get $A(n)=-nA(n-1)$ and your identity follows by induction (with base $n=1$, say).


