Prove that $\binom {n}{k} = \frac {n!} {(n-k)!k!}$, viewed as a function of $k$, has maximum at $k=\lfloor n/2 \rfloor, \lceil n/2 \rceil$. Prove that the binomial coefficient $\binom {n}{k} = \frac {n!} {(n-k)!k!}$, viewed as a function of $k$, has maximum at $k=\lfloor n/2 \rfloor, \lceil n/2 \rceil$ if $n$ is odd and maximum at $k=n/2$ if $n$ is even.
Also how do I see that $\binom {n}{k} = \frac {n!} {(n-k)!k!}$ is increasing on $[0;n/2]$ and decreasing on $[n/2;n]$ ?
I see that $\binom {n}{k} = \frac {n!} {(n-k)!k!} = \frac {n!} {(n-(n-k))!(n-k)!} = \binom {n}{n-k}$, so clearly the function is symmetric around $n/2$.
If it is possible, I would like an answer not depending on the derivative, but more on algebra.
 A: HINT:
Keeping integer $n(>0)$ constant,
$$\frac{\binom nk}{\binom n{k-1}}=\frac{n-k+1}k$$
Now this ratio will be $\displaystyle<=>1$ according as $\displaystyle\frac{n-k+1}k<=>1$
So, $\displaystyle\binom nk>\binom n{k-1}\iff \frac{n-k+1}k>1\iff k\le\frac{n+1}2$
A: Since you realize that $\binom{n}{k} = \binom{n}{n-k}$, then we just need to prove that the function is increasing from $k=0$ to $\frac{n}{2}$ or decreasing in the other half. Also, $\binom{n}{k}$ is number of ways of choosing(rejecting) $k$ objects from $n$ objects, or of rejecting(choosing) $n-k$ objects. This is increasing. You can choose 1 object in $n$ ways, then for two objects, the first in $n$ ways and second in $(n-1)$ ways, but since order doesn't matter we get $\frac{n(n-1)}{2}$ ways. Not third object can be chosen in $n-3$ ways and again ignoring order we get $f(n,2)*\frac{(n-2)}{3}$ ways etc. As is clear we are multiplying the previous number by a factor greater than 1, so from 0 upwards it is increasing. But once we reach $\frac{n}{2}$, we see that choosing $\frac{n}{2}+k, (0\le k\le \frac{n}{2})$ objects is same as rejecting $\frac{n}{2}-k$ objects, so since till this point the function was increasing, and from here the values are being reflected, we can conclude that the function is decreasing beyond this point.
