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So it's clear that the total number of shortest routes in a lattice path given a $mxn$ grid is $\binom{s}{r}$ where $s$ is the total number of steps, and $r$ is the total number of right steps. But this is simply to easy. So consider instead this question.

Given a shortest path $p$, how does this determine some $r$-element subset of an $s$-element set $A_p$?

Note: Edited for clarification.

Andrés E. Caicedo
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Azovax
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  • @mjqxxxx I edited it. Forgot to introduce what $r$ & $s$ were. Maybe that provides a little clarification? I'm completely lost though. – Azovax Oct 24 '13 at 03:15
  • Hm... The way I know that the number of lattice paths is ${s \choose r}$ makes explicit use of the fact that it's an $r$-element subset of a particular set. How did you come across that formula? – Eric Stucky Oct 24 '13 at 03:22
  • @EricStucky don't know if this helps, but the $s$ and $r$ are really only making things more complicated. $r$ in this case could be either $m$ or $n$, and $s=m+n$. – Azovax Oct 24 '13 at 03:26
  • That might help, but it doesn't answer the question: how do you know that ${s \choose r}$, or ${m+n \choose n}$ if you prefer, is the number of lattice paths? – Eric Stucky Oct 24 '13 at 03:42
  • @EricStucky It's because of binomial coefficients, which is proved here – Azovax Oct 24 '13 at 03:47

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The linked question from the comments explains what the set and the subset are:

The reason is quite simple: you must make a total of $m+n$ moves, consisting of $m$ moves down and $n$ to the right, in any order, and there are $\binom{m+n}m$ ways to choose which of the $m+n$ moves are down (or, equivalently, $\binom{m+n}n$ ways to choose which $n$ of them are to the right).

If you want to use an arbitrary subset $A$, rather than the subset of moves, this is also allowed, you just need a bijection $f$ between that set and the set of moves, then any path corresponds to a $n$-element subset, where $a\in A$ is in the subset iff $f(a)$ moves right.

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Eric Stucky
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