Bourbaki

Exercise 3, § 5

Tag 03MF, permanent. Cite it as [Bourbaki, Tag 03MF].

Theory of Sets, III, 235

Prove the relations

(n0)(np)+(n1)(n−1p−1)+(n2)(n−2p−2)+⋯+(np)(n−p0)=2p(np),\binom{n}{0}\binom{n}{p} + \binom{n}{1}\binom{n-1}{p-1} + \binom{n}{2}\binom{n-2}{p-2} + \cdots + \binom{n}{p}\binom{n-p}{0} = 2^p\binom{n}{p},
(n0)(np)−(n1)(n−1p−1)+(n2)(n−2p−2)−⋯+(−1)p(np)(n−p0)=0.\binom{n}{0}\binom{n}{p} - \binom{n}{1}\binom{n-1}{p-1} + \binom{n}{2}\binom{n-2}{p-2} - \cdots + (-1)^p\binom{n}{p}\binom{n-p}{0} = 0.

(Consider the subsets of pp elements of [1,n][1, n] which contain a given subset of kk elements (0⩽k⩽p)(0 \leqslant k \leqslant p), and use Exercise 2 for the second formula.)

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