Number of ways to select subsets to cover the original set

Suppose $$S$$ is a set with $$n$$ elements. Let $$U=\{T|T\subseteq S\text{ and }|T|=k\}$$, i.e., $$U$$ is the set of all possible subsets of $$S$$ with cardinality $$k$$. Clearly $$|U|=\binom{n}{k}$$. Then how many ways are there to select $$m$$ subsets, such that the union of these subsets is $$S$$?

My attempt basically takes advantage of the inclusion-exclusion principle. The number of distinct ways to select $$m$$ subsets from $$U$$ is $$\binom{\binom{n}{k}}{m}.$$ We subtract the combination of subsets which does not include one element of $$S$$. For $$x\in S$$, the number of subsets which does not contain $$x$$ is $$\binom{n-1}{k}$$, so we should subtract $$n\binom{\binom{n-1}{k}}{m}.$$ We then add the combinations that does not include two elements of $$S$$. Similarly, we add $$\binom{n}{2}\binom{\binom{n-2}{k}}{m}.$$ Repeat this procedure until $$\binom{n-j}{k}, where $$j$$ is the number of elements in $$S$$ that are not contained in the subset. So, at last, the answer to the problem is $$\Large\sum\limits_{\begin{matrix}j=0\\\binom{n-j}{k}\ge m\end{matrix}}^{n}{(-1)}^j\binom{n}{j}\binom{\binom{n-j}{k}}{m}.$$ But I'm not quite satisfied to this answer. Is there a closed form of this expression? Or has this problem be well-studied? Thanks.

• I am pretty certain you have found the most satisfying form of the answer. There is one mistake; you forgot the $(-1)^j$ factor. Less importantly, there is no need to specify $\binom{n-j}k\ge m$ in the subscript; any terms with $\binom{n-j}k<m$ are automatically zero, so there is no harm in including them. Jul 8, 2023 at 16:19
• This answer comes to the same conclusion. Jul 8, 2023 at 16:21
• @MikeEarnest Thanks, I've fixed the formula.
– Soha
Jul 9, 2023 at 2:59