I have a graph G of n vertices and with a k-list color assignment for each vertex out of $\sigma$ colors. If a choose at random all k colors for each list assignment I can model this with a k-uniform multihypergraph T chosen uniformly at random,
$V(T) = \{0,..\sigma\}$ being the available colors and $e \in E(T) \Leftrightarrow e\equiv \{c_1, \dots c_k\}$ represents a color list assignment $L_u=\{c_1, \dots c_k\}$for some vertex $u$. What I want to show is that the expected number of pairs of multiple edges in T, is: \begin{equation}\begin{pmatrix} \sigma \\ k \end{pmatrix} \:\begin{pmatrix}\begin{pmatrix} \sigma \\ k \end{pmatrix} +n -3 \\ n-2 \end{pmatrix} \begin{pmatrix}\begin{pmatrix} \sigma \\ k \end{pmatrix} +n -1 \\ n\end{pmatrix}^{-1} \end{equation} I understand that if I consider all k-subsets of sigma I study each k-subset repetition as a list by separate, but I dont see how to obtain the other two terms, I would appreciate any indication on how to proceed or see this result!
Choosing at random all $k$ colors for each of the $n$ vertices of a graph is not equivalent to choosing $T$ uniformly at random from all multigraphs. When we choose a random list assignment for $G$, the $n$ objects we're assigning a list to are distinguishable. When we choose a uniformly random multigraph, they're not - we just have a set of $n$ hyperedges.
If we take the graph-centric approach, the expected number of pairs should just be $$ \binom n2 \binom \sigma k^{-1} $$ There are $n$ vertices in $G$, so there are $\binom n2$ pairs of vertices, and each pair is assigned the same color with probability $\binom \sigma k^{-1}$.
If we take the multihypergraph-centric approach, then the total number of multihypergraphs is $$ \left(\!\!\binom{\binom \sigma k}{n}\!\!\right) = \binom{\binom \sigma k + n - 1}{n} $$ where $\left(\!\binom rs\!\right)$ is the multiset number, which we reduce to an ordinary binomial coefficient by the stars-and-bars method. The number of multihypergraphs containing a specific type of pair is $$ \left(\!\!\binom{\binom \sigma k}{n-2}\!\!\right) = \binom{\binom \sigma k + n - 3}{n-2} $$ since once we assign two edges to that type, there are $n-2$ edges left to distribute. (This overcounts triple and higher edges in just the right way: if there are $t$ edges of the same type, then there are $\binom t2$ pairs, and $\binom t2$ ways to represent that as "two edges of that type, and a hypergraph with $n-2$ more edges".) Finally, there are $\binom \sigma k$ different types of edges. So the expected number of pairs of multiple edges should be $$ \binom \sigma k \left(\!\!\binom{\binom \sigma k}{n-2}\!\!\right)\left(\!\!\binom{\binom \sigma k}{n}\!\!\right)^{-1} $$ as desired.