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[组合] n个顶点形成的多边形的个数 旋转等价

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hbghlyj Posted at 2023-1-25 18:21:09 |Read mode
Last edited by hbghlyj at 2025-3-21 01:20:28$n$个顶点形成的多边形的个数为$$(n-1)!\over2$$此处除以2因为1个cycle对应于2个多边形(两个绕行方向).
${(3-1)!\over2}=1$3个点只能形成1个三角形.
${(4-1)!\over2}=3$4个点可以形成 $\large\square\LARGE \bowtie ⧖$这3个四边形.
${(5-1)!\over2}=12$5个点可以形成12个五边形: StirlingNumberFirstKind_700.svg

问:有多少种不旋转等价的多边形?

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2025-4-21 01:14 GMT+8

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