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I encounter a problem where we need to prove the identity $\sum\limits^n_{i = 0} \binom {i} {k} = \binom{n+1} {k+1}$ by counting the lattice paths. I just can't find a way to do it.

Any help would be appreciated!

Martin Sleziak
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metricspace
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    Here you go: http://math.stackexchange.com/q/1490794/321264. – StubbornAtom Sep 04 '16 at 07:24
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    You're taking $k+1$ steps east and $n-k$ steps north, right? So split into cases according to on what level the last of the east steps is taken. – Hans Lundmark Sep 04 '16 at 08:58
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    The question to which StubbornAtom linked is not directly useful, since the answers do not include one by counting lattice paths. Use the suggestion by Hans Lundmark. – Brian M. Scott Sep 04 '16 at 19:03

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We can interpret $\binom{n+1}{k+1}$ as number of lattice paths of length $n+1$ containing $k+1$ horizontal $(1,0)$-steps and $n-k$ vertical $(0,1)$-steps.

This is valid because there are $\binom{n+1}{k+1}$ choices to select $k+1$ steps in horizontal direction leaving the remaining $n-k$ steps for the vertical direction.

$$ $$                     enter image description here $$ $$ Let's consider all paths from $(0,0)$ to $(k+1,n-k)$. We know there are $\binom{n+1}{k+1}$ different paths.

On the other hand each of these paths has to cross the vertical line going through $(k,0)$. The crossing points are $(k,0), (k,1), \ldots, (k,n-k)$. For each path we take as crossing point the node $(k,y)$ which belongs to the path and which has maximum $y$.

This way we partition the set of lattice paths into sets containing $\binom{i}{k}$ elements with $k\leq i \leq n$ establishing the identity

\begin{align*} \sum_{i=0}^n\binom{i}{k}=\binom{n+1}{k+1}\qquad\quad n\geq 0 \end{align*}

Markus Scheuer
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  • While this is the right idea, the formulation is not precise. Notably you do not explain what you mean exactly by crossing that vertical line, in case the path contains more than one point on that line, as it may very well do (the example drawn conveniently does not, and therefore sheds no light on this matter). – Marc van Leeuwen Nov 18 '16 at 05:07
  • @MarcvanLeeuwen: Thanks for your comment. I've updated my answer accordingly and formulated the partitioning more precise. – Markus Scheuer Nov 18 '16 at 16:45