2

Consider a grid in $\mathbb{N}_0^2$. We can draw a path in it by traveling from point to point via a horizontal line segment to the right or vertical line segment going up. Let $k,n \in \mathbb{N}$ and such that $k \leq n/2$. I try to compute the number of paths from $(0,0)$ to $(n-k,k)$ that lie (non-strictly) under the diagonal and do not cross it. (That is, they may touch a point $(i,i)$, but not $(i,i+1)$.)

I have read in an article that the solution is ${n \choose k} - {n \choose k-1}$, but I have no idea how to prove this.

Math Student 020
  • 1,599
  • If we forget about the restriction that we may not cross the diagonal, how many paths would then be possible? – Daniel Fischer May 06 '15 at 11:07
  • I'd say a path has length $n$ and we can choose to go up $k$ times, giving a total of ${n \choose k}$ possible paths. – Math Student 020 May 06 '15 at 11:12
  • Right. So we must now find an argument that shows that the number of paths that cross the diagonal is $\binom{n}{k-1}$. I'm not seeing a good argument yet, but I'm sure there is one. – Daniel Fischer May 06 '15 at 11:17
  • 1
    Have you looked into the Catalan numbers? They count something similar to this (walks from $(0,0)$ to $(n,n)$ that do not cross the diagonal). Can the proof be appropriately modified? – TravisJ May 06 '15 at 11:21
  • What does this question have to do with combinatorial game theory? – bof May 06 '15 at 11:42
  • 1
    @DanielFischer Hmm, possibly one can count those based on the length of the path lying strictly above the diagonal, though I am also not seeing a nice way to do it. – Tobias Kildetoft May 06 '15 at 12:15

1 Answers1

4

If we take a path from $(0,0)$ to $(n-k,k)$ which crosses the diagonal, it must at least touch the line $y=x+1$, so let P be the point where it first touches this line.

If we reflect the portion of the path from $(0,0)$ to P around the line $y=x+1$, we get a path from $(-1,1)$ to $(n-k,k)$.

Conversely, any path from $(-1,1)$ to $(n-k,k)$ must cross the line $y=x+1$, so if we let P be the point where the path first touches this line and reflect the portion of the path from $(-1,1)$ to P about this line, we get a path from $(0,0)$ to $(n-k,k)$ which crosses the diagonal.

Since this gives a bijection between the two types of paths, and since the number of paths from $(-1,1)$ to $(n-k,k)$ is given by $\binom{n}{k-1}$, we can conclude that the number of paths from $(0,0)$ to $(n-k,k)$ which do not cross the diagonal is given by $\binom{n}{k}-\binom{n}{k-1}$.

user84413
  • 27,211