Bourbaki

Exercise 6, § 5

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

Theory of Sets, III, 236

Prove the formula

(n+hh)=1+(h1)(n+h−1h)−(h2)(n+h−2h)+⋯+(−1)h(hh)(nh). \binom{n+h}{h} = 1+ \binom{h}{1}\binom{n+h-1}{h} - \binom{h}{2}\binom{n+h-2}{h} +\cdots +(-1)^h\binom{h}{h}\binom{n}{h}.

(If F denotes the set of mappings uu of [1,h][1,h] into [0,n][0,n] such that

∑x=1hu(x)≤n, \sum_{x=1}^{h}u(x)\leq n,

consider for each subset H of [1,h][1,h] the set of all u∈Fu\in F such that u(x)≥1u(x)\geq 1 for each x∈Hx\in H, and use Exercise 5.)

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