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How can I find the number of the shortest paths between two points on a 2D lattice grid?

If we have a point $P(x,y)$ in a coordinate system $[$with $x \geq 0$, $y \geq 0$; that is, in the 1st quadrant$]$

How can we find the number of ways of reaching $P$ from the origin $(0,0)$.

Ex: If P(2,1);

way1: 0,0 -> 1,0 -> 2,0 -> 2,1
way2: 0,0 -> 1,0 -> 1,1 -> 2,1 
way3: 0,0 -> 0,1 -> 1,1 -> 2,1

Is it possible to have a mathematical equation for it? and How? if we don't have one, what's the best possible way to find those.

Rules:

  1. You can move only in horizontal, vertical directions (diagonal is not possible)
  2. you can only move to the point $(a,b)$ such that $0 \leq a \leq x$ and $0\leq b\leq y$
  3. $a, b$ can be only natural numbers
Heath
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Surya Kasturi
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  • What counts as a way? From your examples, it's not clear what kinds of moves are allowed. For example, Does $(0,0) \rightarrow (42542, 152345) \rightarrow (2,1)$ count as a way to get to $(2,1)$ from the origin? – JohnJamesSmith Jan 29 '12 at 04:10
  • Or $:: (0,0) \to \left(\frac13,\operatorname{ln}(2)\right) \to (2,1) ::$? $;;;;$ –  Jan 29 '12 at 04:13
  • sorry for not mentioning rules, please look at the question, I added – Surya Kasturi Jan 29 '12 at 04:15
  • If there is no bound, you can go reach it in infinitely many ways, like specified above, $(0,0)\to(\infty,\infty)\to(2,1)$, do you take it to be a valid one ? , if not, specify some constraints and bounds @Surya – IDOK Jan 29 '12 at 04:17
  • only natural numbers are possible – Surya Kasturi Jan 29 '12 at 04:17
  • @iyengar please look at the rules. – Surya Kasturi Jan 29 '12 at 04:18
  • The move '0,1 -> 1,2' is diagonal. –  Jan 29 '12 at 04:23
  • @ByronSchmuland sorry, there was a mistake in example. However, I got the right answer. Please look at it now – Surya Kasturi Jan 29 '12 at 04:35
  • For those interested: I answered this question, but then noticed that the same question (written a little more precisely) appeared previously. – JavaMan Jan 29 '12 at 06:18

2 Answers2

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It is well known that the number of ways to get to the lattice point $(x,y)$ (supposing $x, y \geq 0$) by taking steps of one unit each either in the eastward or northward direction is exactly $$ {x + y \choose x} = {x+y \choose y} = \frac{(x+y)!}{x! y!}. $$

Such paths are called lattice paths. See, for example, here.

JavaMan
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  • Why the down vote? – JavaMan Jan 29 '12 at 04:55
  • Downvote ? , I gave an upvote. It wasn't me @JavaMan – IDOK Jan 29 '12 at 05:48
  • @iyengar: Thanks for the upvote, but I'm not too worried about who downvoted. I'm more worried about what part of my answer wasn't deemed satisfactory. – JavaMan Jan 29 '12 at 05:50
  • Haven't you noticed that your answer has been selected, your answer was very good, and seems to fit in a good sense, its only some fools around here, who unnecessarily down-vote the answer, without leaving any comments, it happened to me quite many times, even got the worst votes. But if someone down-votes the answers without a comment, it surely acts as a cause of annoying the person who answers the question. But let us leave this topic, and why should we bother about points here, if your answer is perfect. Does points earn us anything in reality apart from an apparent fame here . – IDOK Jan 29 '12 at 06:03
  • @iyengar: Just for the record, I never thought you downvoted, and actually I don't really care who downvoted just as I do not care about the reputation. I was merely more curious as to why the downvoter did so, and how I could improve my answer. – JavaMan Jan 29 '12 at 06:25
  • @JavaMan I wonder if you could answer another question which is based on this one. Link – Surya Kasturi Jan 29 '12 at 06:28
  • @JavaMan : I too was curious in knowing the problems, even I posted a feature request in Meta, I too suggested the same thing, I said I am not bothering about down-votes, I suggested that if the person who down-votes the answer leaves the reason for doing so, it would be helpful for person who posted the answer, both in formatting and correcting himself. But majority people said, we dont need to keep such feature. So I left, but wont it hurt you ( not only you but any one who answers and gets downvoted without a reason ) ?, I don't know about you, but it hurts me a lot. – IDOK Jan 29 '12 at 06:31
  • @JavaMan and : Do not worry about down or upvotes, I have seen many times other poeple's good answer (not only mine) have been downvoted for no reason, now a days I think it is comical for a good answer to get downvoted by some bozo without leaving a comment, I just do my bit and upvote, BTW : It wasn't me who downvoted :) – jimjim Jan 29 '12 at 08:01
  • @Arjang : But is there any solution to eradicate that comical "Down-voting" foolishness ? . – IDOK Jan 29 '12 at 10:40
  • @iyengar : Aaah yes, it has already been implemented, downvoting costs points, so it is not free to downvote! – jimjim Jan 29 '12 at 10:50
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It is a combinatorial question, where you have $x+y$ things and you have to pick $x$ (or $y$, both are symmetric) times when you can make a choice. In other words, you have $x$ ways to move in $x$ direction, $y$ way to move in $y$ direction. However, once you pick any $x$ direction, the choices for $y$ is fixed. Therefore, the total number of way you can do the above is $(x+y)$ choose $x$ (or $y$, respectively). Mathematically, it will be

$$\left( \begin{matrix} x+y \\ x \end{matrix} \right) = \left( \begin{matrix} x+y \\ y \end{matrix} \right).$$

Jalaj
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