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!

Here is the best solution of problem 2022-15.

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

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