Prove the following by two different methods, one combinatorial and one algebraic

Combinatorial: Rearrange to $${n\choose k}={n-1\choose k-1}+{n-2\choose k-1}+{n-3\choose k-1}+{n-3\choose k}$$

Choose $k$ objects out of $\{1,2,\ldots, n\}$. This is counted by LHS. We can also partition such choices into four parts:

(a) $1$ is the smallest item in our set of $k$.

(b) $2$ is the smallest item in our set of $k$.

(c) $3$ is the smallest item in our set of $k$.

(d) Our set of $k$ contains none of $\{1,2,3\}$.

The four terms of the RHS count just these four cases.

Hint: Put $\binom{n-3}{k}$ to the other side and collect terms.

This way we obtain starting with the right-hand side \begin{align*} &\binom{n-1}{k-1}+\binom{n-2}{k-1}+\color{blue}{\binom{n-3}{k-1}+\binom{n-3}{k}}\\ &\qquad=\binom{n-1}{k-1}+\color{blue}{\binom{n-2}{k-1}+\binom{n-2}{k}}\\ &\qquad=\color{blue}{\binom{n-1}{k-1}+\binom{n-1}{k}}\\ &\qquad=\color{blue}{\binom{n}{k}} \end{align*}

Repeatedly, use the identity (Pascal's Identity), namely $$ \binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}. $$ Note that $$ \left(\binom{n}{k}-\binom{n-1}{k-1}\right)-\binom{n-2}{k-1}-\binom{n-3}{k-1}-\binom{n-3}{k} $$ equals $$ \binom{n-1}{k}-\binom{n-2}{k-1}-\binom{n-3}{k-1}-\binom{n-3}{k} $$ which equals $$ \binom{n-2}{k}-\binom{n-3}{k-1}-\binom{n-3}{k} $$ which equals $$ \binom{n-3}{k}-\binom{n-3}{k}=0 $$ as desired.