Let $n$ be an even positive integer and $K\_{2n}$ be the complete graph on $2n$ vertices. There are $\dfrac{1}{2}{{2n}\choose n}={{2n-1}\choose n}$ subgraphs of $K\_{2n}$ which is isomorphic to $K\_{n,n}$ and we use $E\_1,E\_2,\dots,E\_{{2n-1}\choose n}$ to denote the edge sets of the ${2n-1}\choose n$ subgraphs respectively.


My question is whether I can partition the edge set of $K\_{2n}$ to $S\_1,S\_2,\dots,S\_{2n-1}$ such that $|S\_i|=n$ for every $1\leq i\leq2n-1$ and $\{\cup^{n}\_{k=1}S\_{i\_k}:1\leq i\_1<i\_2<\dots<i\_n\leq 2n-1\}=\{E\_1,E\_2,\dots,E\_{{2n-1}\choose n}\}$?



The existence of a specific partition of the edge set of $K_{2n}$
**No,** this is not possible. Towards a contradiction, suppose such a partition $S\_1, \dots, S\_{2n-1}$ of $E(K\_{2n})$ exists. I first claim that each $S\_i$ must be a matching. If not, then some vertex $v$ has degree at least 2 in say $S\_1$. But now if we sort the $S\_i$ according to the degree of $v$ in $S\_i$ and pick the $n$ largest ones, then the union of these $S\_i$ will not be isomorphic to $K\_{n,n}$ since $v$ will have degree larger than $n$. 


Since each $S\_i$ is a matching, each $S\_i$ determines a bipartition of $V(K\_{n,n})$. We may suppose that $S\_1$ and $S\_2$ determine different bipartitions. But now, no collection of $S\_i$s containing both $S\_1$ and $S\_2$ can induce a copy of $K\_{n,n}$. 


