Show that $\binom{n}{k}< \binom{n}{k+1}$ if and only if $k < (n-1)/2$

Question

Show that $\binom{n}{k} < \binom{n}{k+1}$ if and only if $k < \frac{n-1}{2}$ and then use this to deduce that the maximum of $\binom{n}{k}$ for $k=0,1,\dots,n$ is $\binom{n}{\lfloor n/2\rfloor}$.

Answer 1

Hint: If you're able to use the formula $\binom{n}{k}=\frac{n!}{k! (n-k)!}$, then it's just arithmetic: $$\binom{n}{k}=\frac{n!}{k! (n-k)!}= \frac{k+1}{n-k}\frac{n!}{(k+1)! (n-k-1)!}=\underbrace{\frac{k+1}{n-k}}_{\substack{\text{when is this} \\ \text{$<1$?}}} \binom{n}{k+1}.$$

For the second part, we can use that $\binom{n}{k}=\binom{n}{n-k}$.

Answer 2

I like using

$${n\choose k}={n-1\choose k}+{n-1\choose k-1}$$

and induction. Then you check the base case.

The pictoral representation is Pascal's triangle, you are adding bigger numbers when you go farther along (up to the line of symmetry) so the values get larger. You even get the "only if" direction along the way thanks to the symmetry in the induction.