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This question already has an answer here:

You start at coordinate (0, 0) on a grid and want to reach position (3, 3). On this grid, you can only move right or up, not diagonally. How many paths are there?

This is a question I am trying to solve - would the answer to this be 20? I have done this graphically, writing out all the different ways this journey could be done and arrive at 20 - however, is there a simpler way to calculate this, with a formula?

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marked as duplicate by Mike Earnest, Xander Henderson, Lee David Chung Lin, Lord Shark the Unknown, Eric Wofsey Feb 12 at 4:25

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The number of ways to get to $(a, b)$ is equal to the number of ways to get to $(a-1, b)$ plus the number of ways to get to $(a, b-1)$.

Draw a grid and place the number $1$ at the origin. That's because there is exactly one way to reach the origin.

Now take the points $(0, 1)$ and $(1, 0)$ and write how many ways you can get to each of the points. Then take $(0, 2), (1, 1)$ and $(2, 0)$ and see how many ways you can get there. Then take $(0, 3), (1, 2), (2, 1)$ and $(3, 0)$. And so on. You should get your answer within a minute or so. No need for formulas, although there is one, and it's rather simple.

Bonus: If you stand at the point $(4, 4)$ and look at your grid as you draw it, can you recognize the process? That's where the formula comes from.

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  • $\begingroup$ Thank you for the help - I did something similar to get to 20 but I want to know if there's a formula to calculate it without having to draw it all out $\endgroup$ – faboys Feb 11 at 13:59
  • $\begingroup$ @faboys If you do the entire grid this way, filling in all points you can get to (not just the square that's relevant to your problem), doing it systematically by the diagonals as I outlined here, but turn the paper around so you're looking from the direction of the point $(4, 4)$, then you ought to recognize the process and find the formula quite easily from there. But if not, then $(3, 3)$ is so close to the origin that you don't really lose much time doing it the "hard" way. $\endgroup$ – Arthur Feb 11 at 14:01
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    $\begingroup$ Would the formula be 6!/(3! * 3!)? $\endgroup$ – faboys Feb 11 at 14:08
  • $\begingroup$ @faboys Yup. That's the answer. So yeah, 20 is what you get in the end. $\endgroup$ – Arthur Feb 11 at 14:09
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    $\begingroup$ Thanks - got the lines wrong when doing it graphically. $\endgroup$ – faboys Feb 12 at 10:27

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