How to prove that for any $k$ between $2$ and $n-2$ we have $\binom{n}{k} \ \ge \binom{n}2$ I am trying to show that for any $k$ from $2$ to $n-2$ we have $\binom{n}{k} \ \ge \binom{n}2$.
What I tried
I started from the fact that $\binom{n}{k} \le \binom{n}{k+1}$ if $k \le \frac{n-1}{2}$ and by using the symmetry of binomial coefficients we also have this for $k \ge\frac{n+1}{2}$, but I don't know how to proceed from here on out and tie this to $k$ being between $2$ and $n-2$.
Any help would be greatly appreciated.
 A: Intuitively, choosing 2 objects out of $n$ and choosing $n-2$ objects out of $n$ are the same thing; you're really just sorting them into two groups, so you can interpret $\binom n 2$ or $\binom n {n-2}$ as being the number of distinct ways to form the smaller group. If $n < k < n-2$ then the smaller group (the lesser of $k$ and $n-k$) will be at least 3, thus there are more ways to form it.
Perhaps more explicitly, if $2<k<n-2$ then $(n-k)! < (n-2)!$
$$\binom n k = \frac{n!}{k!(n-k)!}$$
$$\binom n 2 = \frac{n!}{2!(n-2)!}$$
$$\frac{\binom n k}{\binom n 2} = \frac{2!(n-2)!}{k!(n-k)!} = \frac{2 \cdot (n-2) \cdot (n-3) \dots 2 \cdot 1}{k \cdot (k-1) \dots 2 \cdot 1 \cdot (n-k) \dots 2 \cdot1}$$
$$= \frac{(n-2)\cdot(n-3)\dots(n-k+1)}{k\cdot(k-1)\dots4\cdot3} \geq 1$$
A: Proof by induction:
(1) Base case: $n=6$, easy to verify, $2<k=3<6-2$,
$$~~~~~~~~~~~~~\binom{6}{3}>\binom{6}{2},~~\text{True.}$$
(2) Assume it is true for $n=m$, where $m>6$ and $m\in \mathbb{N^+}$, so we have:
$$\binom{m}{k}>\binom{m}{2}$$
(3) Induction step: when $n=m+1$,
$$\begin{align}
\binom{m+1}{k}&=\frac{(m+1)!}{k!(m+1-k)!}=\frac{m+1}{m+1-k}\binom{m}{k}>\frac{m+1}{m+1-k}\binom{m}{2}\\
\\
&=\frac{m+1}{m+1-k}\cdot\frac{m!}{2!(m-2)!}=\frac{m-1}{m+1-k}\cdot\frac{(m+1)!}{2!(m-1)!}=\frac{m-1}{m+1-k}\binom{m+1}{2}
\end{align}$$
since $k>2,~~\rightarrow~~m-1>m+1-k$, therefore:
$$\frac{m-1}{m+1-k}\binom{m+1}{2}>\binom{m+1}{2}$$
So we have:
$$\binom{m+1}{k}>\binom{m+1}{2}$$
Proof is completed.
