Counting sets by their connectedness Let $U = \{u_1, u_2, \ldots , u_m \}$ where each $u_i$ is an $r$-subset of $[n]$ and 
$\,\bigcup u_i \!=\! [n]$. Construct the intersection graph of $U$. That is, let node $i$ correspond to $u_i$ and add edge $(i,j)$ if $u_i \cap u_j \neq \emptyset$. For example, suppose $n=8$ and $U = \{123, 124, 234, 567, 568\}$. The intersection graph is
$\qquad \qquad \qquad \qquad \qquad \qquad \quad $ 
We notice there are two connected components, ie. $\mathcal{C}(U)=2$. Given $n,r,m$ how many sets have exactly $c$ connected components?
$$
\begin{array}{ll}
T_{n,r,m}(c) &= \displaystyle \left[x^c\right] \sum_{\left|U\right| = m} x^{\mathcal{C}(U)} \\
&= \displaystyle \left[y^m x^c\right] \sum_U y^{\left|U\right|} x^{\mathcal{C}(U)}
\end{array}
$$
Can we come up with an expression for $T_{n,r,m}(c)$ or at least some way to efficiently compute it? I would be happy with an algorithm where $n \le 50, r \le 10$ is tractable. I've enumerated all the nonzero values for $n\le6$ to use for verification.

I've posted a partial answer that reduces the problem to counting $T_{n,r,m}(1)$, the sets with a fully connected graph.

I recently came up with a faster solution to this problem that reduces the recurrence dimension by 1.
Let $T_r(n, m, c)$ represent the number we are looking for, ie. the number of ways to take $m$ subsets of $[n]$ such that each subset has size $r$, the subsets cover $[n]$, and their intersection graph has $c$ connected components. I made $r$ a subscript because it is held fixed throughout. The logic is similar to $T_n^{c \times \ell}$ in my answer above. Consider the connected component that covers the element "1" from the ground set. Suppose it consists of $s$ subsets and that it covers $k$ total elements. There are $\binom{n - 1}{k - 1}$ ways to choose the covered elements which leads to the recurrence for $c > 1$.  
$$T_r(n, m, c) = \sum_{k} \sum_{s} \binom{n - 1}{k - 1} \; T_r(n\!-\!k, \,m\!-\!s,\,c\!-\!1) \; T_r(k, s, 1) $$
For connected graphs $c = 1$ we get the same restriction formula from before: all the possible graphs given $m$ minus those that don't cover $[n]$ minus those that have more than 1 connected component.
$$ T_r(n, m, 1) \; = \; {{n \choose r} \choose m} \; - \; \sum_{k \, < \, n} \sum_{c} {n \choose k} T_r(k, m, c) \; - \; \sum_{c \, > \, 1} T_r(n, m, c)$$ 
