# Prove that $\binom{n}{r} \leq \binom{n}{\lfloor{\frac{n}{2}}\rfloor}$ is true [duplicate]

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.

Thank you.

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

• – lab bhattacharjee Jul 18 '15 at 15:38
• Try to show that $\binom{n}{k} < \binom{n}{k+1}$. When it holds? – Michael Galuza Jul 18 '15 at 15:39
• – user2838619 Jul 18 '15 at 15:43
• People pointing out other questions: if you know the question is a duplicate, flag it as a duplicate! – Steven Stadnicki Jul 18 '15 at 15:56

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, $(n-1)!(n+1)! = \dfrac{n+1}{n-1} n!n!>n!n!$.