TITLE: Prove the following by two different methods, one combinatorial and one algebraic
QUESTION [6 upvotes]: Reading through my textbook I came across the following problem, and I am looking for some help solving it. I am asked to prove the following by two different methods, one combinatorial and one algebraic. If I could get help with either or both it would be great, thanks!
Prove that this identity is true,
$$\binom{n}{k} -\binom{n-3}{k} =\binom{n-1}{k-1} + \binom{n-2}{k-1} + \binom{n-3}{k-1}$$

REPLY [3 votes]: Combinatorial proof  without words:
              
  \begin{align*}
\hline
\end{align*}

Or with some words:
We consider the lattice paths of length $n$  from $(0,0)$ to $(k,n-k)$ consisting of $(1,0)$-steps and $(0,1)$-steps  only. The number of these paths is $$\binom{n}{k}$$ since we have to choose precisely $k$ $(1,0)$-steps out of $n$ steps.

The identity
  \begin{align*}
\underbrace{\binom{n}{k}}_{\color{blue}{\textbf{P}}} -\underbrace{\binom{n-3}{k}}_{\color{red}{\textbf{Q}}} =\underbrace{\binom{n-1}{k-1}}_{\textbf{R}} + \underbrace{\binom{n-2}{k-1}}_{\textbf{S}} + \underbrace{\binom{n-3}{k-1}}_{\textbf{T}}
\end{align*}
  is valid, since 

the number of paths from $(0,0)$ to $\color{blue}{P=(k,n-k)}$ which do not pass $\color{red}{Q=(k,n-3-k)}$

is the same as the number of paths from $(0,0)$ which go either

to $R=(k-1,n-k)$
or $S=(k-1,n-1-k)$
or $T=(k-1,n-2-k)$.

Note: Each of the paths to $R,S,T$ can be uniquely extended to $P$ via a $(1,0)$-step followed by zero, one or two $(0,1)$ steps.