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

Question

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.

See http://math.stackexchange.com/questions/722952/how-do-you-prove-n-choose-k-is-maximum-when-k-is-lceil-frac-n2-rceil — 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
also here http://math.stackexchange.com/questions/823693/prove-that-binom-nk-frac-n-n-kk-viewed-as-a-function-of-k — 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

score 2 · Answer 1 · answered Jul 18 '15 at 16:11

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

1 Answers1