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Nov 1, 2017 at 12:08 comment added Fedor Petrov It is at least $2^d$ always by obvious reasons: $\binom{n}d/\binom{n/2}d=\prod_{i=0}^{d-1}\frac{n-i}{n/2-i}\geqslant 2^d$.
Nov 1, 2017 at 3:15 history answered Dustin G. Mixon CC BY-SA 3.0