I was trying to prove

$\displaystyle \binom{n}{r} \leq \displaystyle \binom{n}{\lfloor{\frac{n}{2}}\rfloor} $

where $r=0,1...,n$

I supposed that n is even and tried to divide: $\frac{\displaystyle \binom{n}{\frac{n}{2}}}{\displaystyle \binom{n}{r}}$ and ended up with this

$\frac{n!(n-r)!}{(\frac{n}{2})!\cdot (\frac{n}{2})!}$, but couldn't make any further progress.

Can you please help?

Thank you.


marked as duplicate by Steven Stadnicki, Micah, user147263, muaddib, Strants Jul 18 '15 at 20:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


We need to show that $(n-k)!(n+k)!$ or $(n-k)!(n+k+1)!$ are minimal when $k=0$.

Then $\dbinom{2n}{n}=\dfrac{(2n)!}{n!n!}$ and $\dbinom{2n+1}{n}=\dfrac{(2n+1)!}{n!(n+1)!}$ are maximal.

But this is obvious, as, for example, $(n-1)!(n+1)! = \dfrac{n+1}{n-1} n!n!>n!n!$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.