Working on a problem in the symmetric group I have stumbled upon the following equation:


$$\sum\_{\substack{\pi=(1^{c\_1},2^{c\_2},\ldots,n^{c\_n})\\\textrm{partition of }n}}(-1)^{n-\sum\_{i=1}^nc\_i}\frac{n!}{\prod\_{i=1}^ni^{c\_i}c\_i!}\left(\sum\_{\substack{\eta=(1^{b\_1},2^{b\_2},\ldots,k^{b\_k})\\\textrm{partition of }k}}\prod\_{j=1}^k{c\_j\choose b\_j}\right)^\ell=0,$$
where $n,k$ and $\ell$ are positive integers and $1\le k,\ell\le n-1.$ (Here, $\pi=(1^{c\_1},2^{c\_2},\ldots,n^{c\_n})$ means that $n=1\cdot c\_1+2\cdot c\_2+\cdots +c\_n\cdot n$.)


I have two questions. First, whether this equality has appeared somewhere. (I have checked the book(s) of Stanley on Enumerative combinatorics, but with no luck.)
Second, given $n$ and $k$, I am intersted on the smallest value of $\ell$, where the equality above is not satisfied. For instance, when $n:=43$ and $k:=13$, with a computer computation one can check that the equality above is satisfied FOR EACH element in $\{1,2,3,4,5,6\}$ and $\ell:=7$ is the first time where this equality is not satisfied. However, I have no clue in how to get my hands on this value!



Identity involving binomial coefficients and partitions
If $A\subset B$, $b$ is a permutation (self-bijection) of $B$, and $a$ a permutation of $A$, we say that $a$ is a subpermutation of $b$ if any cycle of $a$ is a cycle of $b$. Your sum is the sum of ${\rm sign}(b)$ taken over all permutations $b$ of $B=\{1,2,\ldots,n\}$ and all subpermutations $a\_1,\ldots,a\_\ell$ of $b$ with $|a\_1|=|a\_2|=\ldots=|a\_\ell|=k$ (here $|a\_i|:=|A\_i|$ where $a\_i$ is a permutation of $A\_i\subset B$). Let us fix $A\_1,\ldots,A\_\ell$, and consider the partition $B=\sqcup\_{i=1}^m C\_i$ of $B$ generated by $A\_1,\ldots,A\_\ell$. Note that every set $C\_j$ must be invariant under $b$ and those $a\_i$'s for which $C\_j\subset A\_i$ (since for every $x\in C\_j$ the $b$-orbit of $x$ is contained in each set $A\_i$ containing $x$, thus in their intersection which is just $C\_j$), and the restrictions of $b$ and corresponding $a\_i$'s to $C\_j$ coincide.


If there exists $j$ such that $|C\_j|>1$, fix two elements $u\ne v$ and partition all tuples $(b,a\_1,\ldots,a\_k)$ onto couples multiplying $b$ and all $a\_i$'s with $A\_i\supset C\_j$ by the transposition of $u$ and $v$. This changes the parity of $b$, preserves the required property, and so proves that the corresponding part of your sum equals 0.


If $|C\_j|=1$ for all $j$, then $b$ and all $a\_i$'s must be identical permutations and this paet of your sum equals 1.


Therefore your question is equivalent to the following:


do there exist sets $A\_1,\ldots,A\_\ell\subset B$ with $|B|=n$, $|A\_i|=k$ for all $i$ such that for all $u\ne v$ in $B$ there exists $j\in \{1,\ldots,\ell\}$ such that $A\_j$ contains exactly one of $u$ and $v$ (in other words, the partition of $B$ generated by $A\_i$'s is the partition onto $n$ singletons)? Or, another reformulation: when the edges of the complete graph $K\_n$ may be covered by $\ell$ complete bipartite graphs $K\_{k,n-k}$?


Let me explain, for example, why for $n=43$, $k=13$ and $\ell=6$ such 6 sets $A\_1,\ldots,A\_6$ do not exist. For each element $u\in B$ consider the set $T(u)\subset \{1,2,3,4,5,6\}=\{i: u\in A\_i\}$. Such sets are mutually distinct, thus
$$6\cdot 13=\sum\_u|T(u)|\geqslant 1\cdot 0+6\cdot 1+15\cdot 2+21\cdot 3$$
(at most one set $T(u)$ is empty, at most 6 sets have size 1 etc),
that is not true.


