On modified $10 \times 10$ chess board Alan and Peter starts moving towards each other with constant speed.Alan can move only two right and upwards. Peter can move only to left and downwards alone lines of chess board.Find number of ways Alan and Peter meet at some point during their trip.

My Try: I assumed they start at diagonally opposite corners.

So if Alan takes $x$ steps horizontally to right and $y$ steps vertically Up , Then to meet Alan ,Peter has to take $10-x$ steps horizontally left and $10-y$ steps Vertically down.

Hence totally they cover $10$ horizontal steps and $10$ vertical steps.

But now how to proceed?

  • $\begingroup$ What is a 'modified' chess board? Do Alan and Peter start in opposite corners? ( Alan in bottom left corner and Peter in top right corner?) Did you mean to write that Alan can only move to the right or upwards? Most importantly, what did you try? Any thoughts? HINT: where would be the possible meeting points? For each of those points, in how many ways can Alan get there? Also, use symmetry! $\endgroup$ – Bram28 Sep 1 '17 at 2:59

There are 11 possible meeting points, all along the diagonal from top left to bottom right.

There is only one way for Alan and Pater to get to and meet in the top left corner.

To get to the point that is one down and 1 right from there, Alan would need to go up 9 times and right 1 times, which can be done in $10 \choose 1$ ways. By symmetry, Peter has just as many ways to get to that point, so there are ${10 \choose 1}^2$ ways for them to get to and meet at that point.

FOr the point 1 down and 1 right from there, there are $10 \choose 2$ ways for Alan to get to that point, and same for Peter, so again multiply those.

I think you see the pattern now ...

Once you have computed the number of ways for them to get to and meet in the very middle point of the board (which is ${10 \choose 5}^2$), you are basically done, because by symmetry, the number of ways for the last 5 points is the same as for the first 5 points.

Or, you could just say that the number of ways is:

$$\sum_{i=0}^{10}{10 \choose i}^2$$

  • $\begingroup$ are you assuming that Alan starts at bottom left corner and Peter at Top right Corner? $\endgroup$ – Umesh shankar Sep 1 '17 at 3:26
  • $\begingroup$ @Umeshshankar Yes, that's right. $\endgroup$ – Bram28 Sep 1 '17 at 13:16

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