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May 12, 2022 at 19:08 comment added Max Alekseyev @FedorPetrov: Essentially this argument equivalent to the use of Vandermonde's identity: $$\binom{n+r-2}{r-2} = \sum_{k=1}^{r-1} \binom{n-1}{k-1}\binom{r-1}{r-1-k}$$ and noticing that $\binom{r-1}{r-1-k}\geq 1$. With a bit of extra effort I got a bit stronger bound.
May 12, 2022 at 18:52 comment added Fedor Petrov $\binom{n+r-2}{r-2} - \sum_{k=1}^{r-1} \binom{n-1}{k-1}$ is non-negative also by a simple combinatorial reasons: take $n-1$ boys and $r-1$ girls and choose $r-2$ children from them: if you fix the chosen $k-1$ boys, you may always add $r-k-1$ girls to them.
May 12, 2022 at 18:32 vote accept T. Amdeberhan
May 12, 2022 at 17:44 history edited Max Alekseyev CC BY-SA 4.0
edited body
May 12, 2022 at 17:34 history edited Max Alekseyev CC BY-SA 4.0
added 4 characters in body
May 12, 2022 at 17:25 history edited Max Alekseyev CC BY-SA 4.0
added 148 characters in body
May 12, 2022 at 17:10 history edited Max Alekseyev CC BY-SA 4.0
a bit better; added 3 characters in body; edited body
May 11, 2022 at 11:34 history answered Max Alekseyev CC BY-SA 4.0