Timeline for Maximum number of subsets in which people co-exist with their friends

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Jan 28, 2021 at 23:46	comment	added	Max Alekseyev		@mgus: That's because by symmetry $\sum_{i\geq r'} \binom{a}{i}\binom{a}{r-i} = \sum_{i<r'} \binom{a}{i}\binom{a}{r-i}$, and therefore they both are equal to the half of $\sum_{i=0}^a \binom{a}{i}\binom{a}{r-i} = \binom{2a}{r}$.
Jan 28, 2021 at 20:27	comment	added	mgus		Hi, I understand your proof and thanks. But, can you explain why $\sum_{i=r'}^a \binom{a}{i}\binom{a}{r-i}=\frac12\binom{2a}{r}$ perhaps writing a couple of steps? I have tried to show this using the identity $\sum_{k=0}^{n}{{r}\choose{k}}{{s}\choose{n-k}} = {{r+s}\choose{n}}$ but I cannot figure it out.
Jan 28, 2021 at 19:55	vote	accept	mgus
Jan 27, 2021 at 22:14	history	answered	Max Alekseyev	CC BY-SA 4.0