Consider the second-kind enthralling numbers,
\({\rm{S(n,j) = }}\frac{{\rm{1}}}{{{\rm{j!}}}}\sum\limits_{{\rm{i = 0}}}^{{\rm{j - 1}}} {{{{\rm{( - 1)}}}^{\rm{i}}}} \left( {\begin{array}{*{20}{l}}{\rm{j}}\\{\rm{i}}\end{array}} \right){{\rm{(j - i)}}^{\rm{n}}}\)
The number of possible distributions of \({\rm{n}}\) distinct objects into \({\rm{k}}\)indistinguishable boxes is then:
\(\sum\limits_{{\rm{j = 1}}}^{\rm{k}} {\rm{S}} {\rm{(n,j) = }}\sum\limits_{{\rm{j = 1}}}^{\rm{k}} {\frac{{\rm{1}}}{{{\rm{j!}}}}} \sum\limits_{{\rm{i = 0}}}^{{\rm{j - 1}}} {{{{\rm{( - 1)}}}^{\rm{i}}}} \left( {\begin{array}{*{20}{l}}{\rm{j}}\\{\rm{i}}\end{array}} \right){{\rm{(j - i)}}^{\rm{n}}}\)
We want to know how many different methods there are to distribute six distinct objects (temporary employees) into four indistinguishable boxes (identical offices).
\({\rm{n = 6}}\)
\({\rm{k = 4}}\)
Because none of the boxes can be empty, the Stirling number of the second kind \({\rm{S(6,4)}}\) is used to calculate the number of ways to distribute six identifiable objects into four indistinguishable boxes (by definition of the Stirling number of the second kind in the textbook).
Let's look at the second type of Stirling numbers:\(\begin{array}{c}{\rm{S(6,4) = }}\frac{{\rm{1}}}{{{\rm{4!}}}}\sum\limits_{{\rm{i = 0}}}^{\rm{3}} {{{{\rm{( - 1)}}}^{\rm{i}}}} \left( {\begin{array}{*{20}{l}}{\rm{4}}\\{\rm{i}}\end{array}} \right){{\rm{(4 - i)}}^{\rm{6}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{24}}}}{\rm{ \times }}\left( {{{{\rm{( - 1)}}}^{\rm{0}}}\left( {\begin{array}{*{20}{l}}{\rm{4}}\\{\rm{0}}\end{array}} \right){{{\rm{(4 - 0)}}}^{\rm{6}}}{\rm{ + ( - 1}}{{\rm{)}}^{\rm{1}}}\left( {\begin{array}{*{20}{l}}{\rm{4}}\\{\rm{1}}\end{array}} \right){{{\rm{(4 - 1)}}}^{\rm{6}}}{\rm{ + ( - 1}}{{\rm{)}}^{\rm{2}}}\left( {\begin{array}{*{20}{l}}{\rm{4}}\\{\rm{2}}\end{array}} \right){{{\rm{(4 - 2)}}}^{\rm{6}}}{\rm{ + ( - 1}}{{\rm{)}}^{\rm{3}}}\left( {\begin{array}{*{20}{l}}{\rm{4}}\\{\rm{3}}\end{array}} \right){{{\rm{(4 - 3)}}}^{\rm{6}}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{24}}}}{\rm{ \times }}\left( {{{\rm{4}}^{\rm{6}}}{\rm{ + ( - 4) \times }}{{\rm{3}}^{\rm{6}}}{\rm{ + 6 \times }}{{\rm{2}}^{\rm{6}}}{\rm{ + ( - 4)}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{24}}}}{\rm{ \times (4096 - 2916 + 384 - 4)}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{24}}}}{\rm{ \times (1560)}}\\{\rm{ = 65}}\end{array}\)
Hence, there are 65 waysto put six temporary employees into four identical offices so that there is at least one temporary employee in each of these four offices.
