# Solution: 2022-15 A determinant of Stirling numbers of second kind

Let $$S(n,k)$$ be the Stirling number of the second kind that is the number of ways to partition a set of $$n$$ objects into $$k$$ non-empty subsets. Prove the following equality $\det\left( \begin{matrix} S(m+1,1) & S(m+1,2) & \cdots & S(m+1,n) \\ S(m+2,1) & S(m+2,2) & \cdots & S(m+2,n) \\ \cdots & \cdots & \cdots & \cdots \\ S(m+n,1) & S(m+n,2) & \cdots & S(m+n,n) \end{matrix} \right) = (n!)^m$

The best solution was submitted by 기영인 (KAIST 22학번, +4). Congratulations!

Other solutions were submitted by 김기수 (KAIST 수리과학과 18학번, +3), 김찬우 (연세대학교 수학과, +3). Late solutions were not graded.

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