Show that $\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$ I cam across a question in The Book of Proof that states:
Show that $$\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$$
Which I found the answer to be:
Assume n ≥ 3. Then:
$$\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2} = \binom{n - 2}{3} + \binom{n-2}{2}+ \binom{n - 1}{2} = \binom{2}{2} + \binom{3}{2}+...+\binom{n-1}{2}$$
I have just learned about the binomial theorem, and I am not sure how we got to that answer.
I understand that $\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2}$ because that is the sum of the two previous rows in the Pascal triangle. But I can't find a simple explanation for the steps that follow or for why we assumed $n \geq 3$

$3$ is arbitrary here. You can apply the same idea of this proof to obtain that
$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-2}{k-1} + \dots + \binom{k-1}{k-1}  $$ with an anologous proof.
Also it's helpful to see the identity $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} $$ in a more combinatorial sense.
The number of ways of choosing $k$ objects out of $n$ is equal to the number of ways of choosing $k$ objects out of the first $n-1$ or choosing $k-1$ out of the first $n-1$ (therefore forcing your k^th choice to be the n^th object).
