TITLE: Combinatorial Proof for Binomial Identity: $\sum_{k = 0}^n \binom{k}{p} = \binom{n+1}{p+1}$
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: Here is another combinatorial approach using lattice paths. 

We can interpret $\binom{n+1}{p+1}$ as number of lattice paths of length $n+1$ containing $p+1$ horizontal $(1,0)$-steps and $n-p$ vertical $(0,1)$-steps.
This is valid because there are $\binom{n+1}{p+1}$ choices to select $p+1$ steps in horizontal direction leaving the remaining $n-p$ steps for the vertical direction.

$$ $$

Let's consider all paths from $(0,0)$ to $(p+1,n-p)$. We know there are $\binom{n+1}{p+1}$ different paths.
On the other hand each of these paths has to cross the vertical line going through $(p,0)$. The crossing points are $(p,0), (p,1), \ldots, (p,n-p)$. This way we partition the set of lattice paths into sets containing $\binom{k}{p}$ elements with $p\leq k \leq n$ establishing the identity
\begin{align*}
  \sum_{k=0}^n\binom{k}{p}=\binom{n+1}{p+1}\qquad\quad n\geq 0
  \end{align*}