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The following question is similar to this thread but I am not familiar with Catalan number. Any chance I could get an explanation for the given solution below? Thanks.

A $5\times5$ chessboard has a diagonal $PQ$. If an ant has to go from point $P$ to point $Q$, find the number of paths which do not cross the diagonal $PQ$ at any moment. (The ant can move only on the sides of the small squares ($1\times1$) either right or up).

Given Solution:

There are $\frac13{{10}\choose5}$ distinct paths.

My doubts:

I understand the ant has to travel on $10$ sides of $1$ unit. $5$ horizontal, $5$ vertical. Thus, $\frac{10!}{5!5!}$

But this includes even those paths which cross $PQ$.

My first instinct was to divide by $2$. One half below $PQ$, one half above it. But of course it's wrong.

My second attempt is as follows:

If the point $P$ is $(0,0)$, then the point $(1,1)$ is on $PQ$, let it be $R$.

The ant can either cross $R$ or, touch it and go right or, it can simply bypass it (by say moving just on $x-$axis and then on $x=5$ line).

Out of this, we are not okay with the crossing the diagonal case.

So, I thought the answer should be $\frac23{{10}\choose5}$ but again it's wrong.

How to approach this problem?

Edit:

As per this thread, seems like the answer is ${n \choose k} - {n \choose k-1}$ i.e. ${{10} \choose 5} - {{10}\choose 4}=\frac16{{10}\choose5}$. Is that so?

aarbee
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  • What do you not understand in the last link? – youthdoo Mar 01 '25 at 09:08
  • The second para says, "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 $(n-k,k)$. " I do not understand this. – aarbee Mar 01 '25 at 09:23
  • @youthdoo commented above. – aarbee Mar 01 '25 at 09:33
  • I think this is clear once you draw a diagram. For example, look at this one: https://wiki-images.artofproblemsolving.com//d/d0/IMG_4713.jpeg – youthdoo Mar 01 '25 at 10:58

1 Answers1

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A 'low-tech' way to solve this is to draw out the grid of number of unique paths.

               42
            14 42
         05 14 28
      02 05 09 14
   01 02 03 04 05
01 01 01 01 01 01

Since the ant can only travel right and up, the bottom row must all be 1. The first number of the next row must come from the number below it as that is the only allowed route. Remaining numbers come from the sum of the number to the left and the number below.

Therefore, the answer is 42 = $\frac{1}{6}C_{10,5}$.

Simon Goater
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  • The ant could also use the upper side so you should multiply by $2$. – ploosu2 Mar 03 '25 at 14:36
  • @ploosu2 That's probably the source of the ambiguity. Does that 'cross the diagonal'? A more rigorous definition of what it allowed is required here. – Simon Goater Mar 03 '25 at 19:13