Prove combinatorial Identity using a combinatorial argument. I have proved by induction on k that : $$ \binom{n}{n} + \binom{n+1}{n} + \binom{n+2}{n}+ ... + \binom{n+k}{n} = \binom{n+k+1}{n+1} $$  But now, I need to prove it using a combinatorial argument. Could someone give me a hint ? :/  Thank you !
Combinatorial Proof for Binomial Identity: $\sum_{k = 0}^n \binom{k}{p} = \binom{n+1}{p+1}$ I am studying combinatorics and I came across the identity $$\sum\limits_{k=0}^n \binom kp =\binom {n+1}{p+1}.$$ I have read the algebraic proof and it does not appeal to me. Is there an elegant counting trick we can use to arrive at this identity?
Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$ Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$     I've seen many complex proofs. I am looking for an elementary proof. I know the fact that $\binom{2000}{0} + \binom{2000}{1} + \binom{2000}{2} + \cdots \binom{2000}{2000} = 2^{2000}$. This may help here.