I had a problem that asked me to find which of the following is larger:
${2013 \choose 500}$ or ${2013 \choose 1500}$
Beneath is my proof. I think it is correct (though your verification and suggestions would nonetheless be appreciated). I haven't worked with factorials all that much, and am curious to see if it could be done any more swiftly. Are there easier ways to prove this proposition?
${\bf Proof:}$ $${n \choose k} = \frac{n!}{k!(n-k)!}$$ $${2013 \choose 1500} = \frac{2013!}{1500!(2013-1500)!} = \frac{2013!}{1500!\cdot 513!}$$ $${2013 \choose 500} = \frac{2013!}{500!(2013-500)!} = \frac{2013!}{500!\cdot 1513!}$$ Obviously, if $500!1513! > 1500!513!$, then ${2013 \choose 1500} > {2013 \choose 500}$ and vice-versa. Note that $\frac{500!1513!}{500!} = 1513!$, and $\frac{513!1500!}{500!} = 1500!\cdot(501\cdot 502 \cdot ... \cdot 513)$. We can rewrite $1513!$ as $1500 \cdot (1501 \cdot 1502 \cdot ... \cdot 1513)$. As both expansions $A = (501 \cdot ... \cdot 513), B = (1501 \cdot ... \cdot 1513)$ have 13 terms each, and $\forall a \in A, a < b$ for any $b \in B$, it must be that $\prod\limits_{a\in A} a < \prod\limits_{b \in B} b$. So $1513! > 1500!(501\cdot ... \cdot 513)$, so $500!1513! > 1500!513!$, so $\frac{2013!}{500!1513!}<\frac{2013!}{1500!513!}$, so ${2013 \choose 1500} > {2013 \choose 500}$. $\square$
The sequence $$\left\langle\binom{2013}k:k=0,1,\ldots,2013\right\rangle$$ is centrally symmetric and single-peaked, increasing from $\binom{2013}0=1$ to a maximum at $\binom{2013}{1006}=\binom{2013}{1007}$ and then falling to $\binom{2013}{2013}=1$. Since $500<513\le 1006$, we must have $$\binom{2013}{500}<\binom{2013}{513}=\binom{2013}{1500}\;.$$