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Per Alexandersson
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How can you prove that Sum (-1)^r * C( 2k-2,r)*(5k-2-r)^(2k-2) from r=0 to 2k-2 = (2k-2)!$$ \sum_{r=0}^{2k-2} (-1)^r \binom{2k-2}{r} (5k-2-r)^{2k-2} =(2k-2)! $$ This result I have obtained by comparing results of two different approaches for the partitioning of the set of vertices of a convex n-gon into nonintersecting polygons.

How can you prove that Sum (-1)^r * C( 2k-2,r)*(5k-2-r)^(2k-2) from r=0 to 2k-2 = (2k-2)! This result I have obtained by comparing results of two different approaches for the partitioning of the set of vertices of a convex n-gon into nonintersecting polygons.

How can you prove that $$ \sum_{r=0}^{2k-2} (-1)^r \binom{2k-2}{r} (5k-2-r)^{2k-2} =(2k-2)! $$ This result I have obtained by comparing results of two different approaches for the partitioning of the set of vertices of a convex n-gon into nonintersecting polygons.

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Partitioning the set of vertices of a convex n-gon into nonintersecting polygons

How can you prove that Sum (-1)^r * C( 2k-2,r)*(5k-2-r)^(2k-2) from r=0 to 2k-2 = (2k-2)! This result I have obtained by comparing results of two different approaches for the partitioning of the set of vertices of a convex n-gon into nonintersecting polygons.