On Stirling numbers of second kind

Question

I was studying Stirling numbers of second came across the general equation distribution of $n$ distinct things in $k$ identical subsets is $$S(n,k) = \frac{k^n - \binom{k}{1}(k-1)^n + \binom{k}{2}(k-2)^n - \cdots + (-1)^{k-1}\binom{k}{k-1}(1)^n}{k!}$$ but here every subset contains at least one thing. Is there is any equation which invalidates this condition? I mean $k$ subsets can include zero too.

Answer 1

Stirling's number $S_k^n$ represents the number of ways that $\{1, \ldots , n \}$ can be partitioned into exactly $k$ non-empty subsets.

If you allow for exactly one empty subset, then the number of allowable partitions grows to include partitions consisting of $\emptyset$ and all possible partitions of $k-1$ non-empty subsets. Thus the number becomes $S_k^n + S_{k-1}^n$.

If you allow for at most two empty subsets, the number becomes $S_k^n + S_{k-1}^n + S_{k-2}^n$.

If you allow for at most three empty subsets, the number becomes $S_k^n + S_{k-1}^n + S_{k-2}^n + S_{k-3}^n$, etc, etc.

If you want to consider $T^n$, the number of ways to partition $\{1,\ldots, n\}$ into $k$ (possibly empty) sets, you simply take $$T_k^n = \sum_{j=1}^k S_j^n.$$ Note that when $k = n$, this is equal to $B_n - 1$ where $B_n$ is the $n^{th}$ Bell number. Thus we might call these numbers incomplete Bell numbers, or something along those lines.