# Is $2\binom{d}{k} \le \binom{2d}{k}$ true?

I am quite certain that $$2\binom{d}{k} \le \binom{2d}{k}$$ holds for every positive integers $$k,d$$, where $$1 \le k \le d$$.

Is there a simple proof? A combinatorial proof ?

An immediate combinatorial argument gives $$2\binom{d}{k} \le \binom{2d}{2k}$$, as the LHS corresponds to choosing $$k$$ elements out of $$d$$, and then choosing additional $$k$$ elements out of a distinct new set of $$d$$ elements. The RHS correspond to an "unconstrained" choice of $$2k$$ out of $$2d$$.

(I checked the cases $$k=1,d$$, and various other examples with "small numbers". For "large" $$d$$ the asymptotics of the binomial coefficients seems compatible with this inequality).

• When choosing $k$ out of $2d$ elements, you can choose $k$ among the first $d$ or $k$ among the last $d$ (or some mix). This gives $2\binom{d}{k}$ as a lower bound. – Wojowu Feb 4 at 11:11

When chosing $$k \ge 1$$ elements from a $$2d$$ set ($$k \le d$$), we can chose $$j$$ ($$0 \le j \le k$$) elements from the first $$d$$ elements and $$k-j$$ from the last $$d$$ elements. The number of possibilities is therefore
$$\binom{2d}{k} = \sum_{j=0}^k \binom{d}{j} \binom{d}{k-j} = 2\binom{d}{k} + \sum_{j=1}^{k-1} \binom{d}{j} \binom{d}{k-j}$$
The "remaining" sum might be empty (if $$k=1$$) but is clearly not negative.
When choosing $$k$$ out of $$2d$$ elements, we can choose $$k$$ among the first $$d$$ or $$k$$ among the last $$d$$ (or some mix). This gives $$2\binom{d}{k}$$ as a lower bound.