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Explanation: Combinatorial Interpretation of Stirling Numbers of the Second Kind
(related to Part: Stirling Numbers)
Let $n,r\ge 0$ be integers. The Stirling number of the second type $\left\{\begin{array}{c}n\\r\end{array}\right\}$ can be interpreted as the number of ways to arrange $n$ objects into $r$ nonempty subsets.
Examples
 $\left\{\begin{array}{c}4\\2\end{array}\right\}=7,$ since there are $7$ ways to split $4$ objects into $2$ nonempty subsets: $$\begin{array}{cccc}
\{a,b,c\}\{d\}&\{a,b,d\}\{c\}&\{a,c,d\}\{b\}&\{b,c,d\}\{a\}\\
\{a,b\}\{c,d\}&\{a,c\},\{b,d\}&\{a,d\}\{b,c\}
\end{array}$$
 $\left\{\begin{array}{c}n\\1\end{array}\right\}=1,$ since there is only one way to split $n$ objects into one nonempty subset.
 $\left\{\begin{array}{c}n\\r\end{array}\right\}=0$ for $r > n$ since there is no way to split $n$ objects into a nonempty subset of $r > n$ elements.
 $\left\{\begin{array}{c}n\\0\end{array}\right\}=0$ for all $n > 0$ since there is no way to split $n$ objects into a nonempty subset of $0$ elements.
 $\left\{\begin{array}{c}0\\0\end{array}\right\}=1$ since there one way to split $0$ objects into a nonempty subset of $0$ elements.
 $\left\{\begin{array}{c}n\\2\end{array}\right\}=2^{n1}1$ since if a set of $n>0$ objects is split into two nonempty subsets, one of them contains the last element and some subset of the first $n1$ elements. There are $2^{n1}$ ways to choose a subset from the $n1$ first elements, but we want to exclude the subset containing all $n1$ elements, since otherwise the other split part would be empty. Therefore we have to subtract $1.$
Convention
$\left\{\begin{array}{c}n\\r\end{array}\right\}$ can be verbalized as "$n$ subset $r$."
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References
Bibliography
 Graham L. Ronald, Knuth E. Donald, Patashnik Oren: "Concrete Mathematics", AddisonWesley, 1994, 2nd Edition