Provide a combinatorial argument that proves $\binom{n}{k}=\binom{n-2}{k}+2\binom{n-2}{k-1}+\binom{n-2}{k-2}$ 
$$\binom{n}{k}=\binom{n-2}{k}+2\binom{n-2}{k-1}+\binom{n-2}{k-2}$$ 

Someone please help me start this question, I am very lost and a bit confused ... 
Thank you!  
 A: On the lefthand side, you count the number of ways to choose $k$ items from a collection of $n$ distinct items (say, the set $\{1, 2, \dots n\}$).
For the righthand side, focus in on the elements $1$ and $2$. When you make a choice of $k$ elements, you must include 


*

*both $1$ and $2$, 

*exactly one of $1$ or $2$, or 

*neither $1$ nor $2$. 


Do you see how the righthand side enumerates these situations?
A: Use this combinatorial identity: ${{n}\choose{k}}={{n-1}\choose{k-1}}+{{n-1}\choose{k}}$ from Pascal's triangle. We then see that:
$${{n}\choose{k}}={{n-1}\choose{k-1}}+{{n-1}\choose{k}}$$
Now use the same identity and apply it to ${{n-1}\choose{k-1}}$ and ${{n-1}\choose{k}}$.
$${{n}\choose{k}}=({{n-2}\choose{k-2}}+{{n-2}\choose{k-1}})+({{n-2}\choose{k-1}}+{{n-2}\choose{k}})$$
$${{n}\choose{k}}={{n-2}\choose{k-2}}+2{{n-2}\choose{k-1}}+{{n-2}\choose{k}}$$
A: 
\begin{align*}
  (1+x)^{n} &= (1+x)^{2}(1+x)^{n-2} \\
  &=(1+2x+x^2) (1+x)^{n-2} \\
\end{align*}

Comparing coefficients of $x^{k}$ where $2\le k \le n-2$, the result follows.
