Prove the following by two different methods, one combinatorial and one algebraic 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,
$$\binom{n}{k} -\binom{n-3}{k} =\binom{n-1}{k-1} + \binom{n-2}{k-1} + \binom{n-3}{k-1}$$

The following  algebraic approach is some sort of overkill  and for  curiosity only.  We use the coefficient of operator  $[x^n]$     to  denote  the coefficient  of  $x^n$  in  a  series. This  way  we   can  write for instance
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n
\end{align*}

We start with the left-hand side and obtain
  \begin{align*}
\color{blue}{\binom{n}{k}-\binom{n-3}{k}}&=[x^k]\left((1+x)^n-(1+x)^{n-3}\right)\\
&=[x^k]\left((1+x)^3-1\right)(1+x)^{n-3}\\
&\color{blue}{=[x^k](x^3+3x^2+3x)(1+x)^{n-3}}\tag{1}
\end{align*}
The right-hand side admits the representation
  \begin{align*}
\color{blue}{\binom{n-1}{k-1} }&\color{blue}{+ \binom{n-2}{k-1} + \binom{n-3}{k-1}}\\
&=[x^{k-1}]\left((1+x)^{n-1}+(1+x)^{n-2}+(1+x)^{n-3}\right)\\
&=[x^{k-1}]\left((1+x)^2+(1+x)+1\right)(1+x)^{n-3}\\
&=[x^{k-1}]\left(x^2+3x+3\right)(1+x)^{n-3}\\
&\,\,\color{blue}{=[x^k]\left(x^3+3x^2+3x^2\right)(1+x)^{n-3}}\tag{2}\\
\end{align*}
Since (1) and (2) coincide the claim follows.

