How do you prove ${n \choose k}$ is maximum when $k$ is $ \lceil \tfrac n2 \rceil$ or $ \lfloor \tfrac n2\rfloor $?

HINT:

As $\displaystyle \binom nk>0$ for $0\le k\le n$ where $n>0,k$ are integers

Check for $k$ such that $$\frac{\binom n{k+1}}{\binom nk}=\frac{n-k}{k+1}>=<1$$

I have done this proof in Metamath before; it may help to see the whole thing laid out.

The proof follows from the fact that the binomial coefficient is monotone in the second argument, i.e. ${n\choose k'}\le{n\choose k''}$ when $0\le k'\le k''\le\lceil\frac n2\rceil$, which can be proven by induction. Given this, you just set $k''=\lceil\frac n2\rceil$ and $k'=k$ or $k'=n-k$ depending on whether $k\le\frac n2$, and you get ${n\choose k}={n\choose n-k}\le{n\choose \lceil n/2\rceil}={n\choose \lfloor n/2\rfloor}$ (where the equalities are deduced by symmetry of the binomial coefficient under $k\mapsto n-k$).

To prove monotonicity, we prove ${n\choose k-1}\le{n\choose k}$ for $1\le k\le\lceil\frac n2\rceil$, and thus ${n\choose k}\le{n\choose k+1}\le\dots\le{n\choose l}$ for any $k\le l$ in the range. Now we have:

$${n\choose k-1}=\frac{n!}{(k-1)!(n-k+1)!}=\frac{n!}{k!(n-k)!}\frac{k}{n-k+1}={n\choose k}\frac{k}{n-k+1},$$

so ${n\choose k-1}\le{n\choose k}$ iff $\frac{k}{n-k+1}\le 1$. But that is equivalent to $$k\le n-k+1\iff 2k\le n+1\iff k\le \frac{n+1}2\iff k\le \left\lceil\frac{n}2\right\rceil,$$

and we are done.

Here's a combinatorial proof, by counting pairs of subsets contained in one another:

Any $k$-element subset is contained in $n-k$ different $(k+1)$-element subsets, whereas each $(k+1)$-element subset contains exactly $(k+1)$ different $k$-element subsets. So provided that $n-k > k+1$ we have that $\binom nk < \binom n{k+1}$, and an inequality the other direction in the other case. The fact that the maximum is in the middle follows immediately.