Proving the combinatorial identity ${n \choose k} = {n-2\choose k-2} + 2{n-2\choose k-1} + {n-2\choose k}$

Hint: try writing the right side as $$\left[\binom{n-2}{k}+\binom{n-2}{k-1}\right]+\left[\binom{n-2}{k-1}+\binom{n-2}{k-2}\right]$$

The committee trick works pretty well. Say you want to choose a committee of $k$ people out of $n$. Among $n$ people there are two leaders (you and me). The terms on the RHS are number of possibilities: with both leaders, with one leader, and with no leaders.

Suppose you want to form a committee of $k$ people from a group of $n$. We know that we can do this in $\binom n k$ ways.

But suppose that two of the people are somewhat distinguished; let's call them A and B. So we want to count the cases based on these guys being in the committee or not separately:

We can have $\binom{n-2}{k-2}$ committees with both A and B.
We can have $\binom{n-2}{k-1}$ committees with A and without B.
We can have $\binom{n-2}{k-1}$ committees without A and with B.
We can have $\binom{n-2}{k}$ committees without A and without B.

The sum of these cases is equal to $\binom n k$.

You could also prove this equality without any computation by interpreting the meaning of $n \choose k$: it is the number of subsets of size $k$ (or $k$-subset) in a set $S$ of size $n$. Let us select two distinct elements $a$ and $b$ of $S$ and let $T = S-\{a,b\}$. The $k$-subsets of $S$ can be divided into four categories:

the $k$-subsets of $T$: ${n-2}\choose k$ possibilities,
the $k$-subsets of the form $R \cup \{a\}$, where $R$ is a $(k-1)$-subset of $T$: ${n-2}\choose {k-1}$ possibilities,
the $k$-subsets of the form $R \cup \{b\}$, where $R$ is a $(k-1)$-subset of $T$: ${n-2}\choose {k-1}$ possibilities
the $k$-subsets of the form $R \cup \{a, b\}$, where $R$ is a $(k-2)$-subset of $T$: ${n-2}\choose {k-2}$ possibilities.

Summing up, you get $$ {n \choose k} = {{n-2}\choose {k}} + {{n-2}\choose {k-1}} + {{n-2}\choose {k-1}} + {{n-2}\choose {k-2}} = {{n-2}\choose {k}} + 2{{n-2}\choose {k-1}} + {{n-2}\choose {k-2}} $$

Hint Use the standard and similar-looking Pascal's Identity: $${n \choose k} = {{n - 1} \choose k} + {{n - 1} \choose {k - 1}} .$$