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Let $ ABCD \ $ be an finite chessboard ($n*n$ tiles) where $A$ is the left lower corner and $C$ its opposite. Each tile is denoted by a square with length $L=1$.

Our purpose is to determine the number of possible roads (denoted by $x_n$) from $A$ to $C$ knowing that we can ONLY travel from from a vertex to another through horizontal (strictly rightwards) and vertically upward lines.

Hint: for large enough $n$ the number of paths $x_n$ contains the number $ \sqrt {2\pi } $ and some power of $2$....

It's my first time seeing this kind of problem and I'd really appreciate if someone could shed some insight or maybe provide a solution..

user184588
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    Are you also restricted to only moving rightwards/upwards? Otherwise you don't have a finite number of ways to move between tiles. – Bruce Zheng Oct 15 '14 at 15:49
  • This is the exact text, but I think you're right, so we should consider the horizontal paths oriented rightwards – user184588 Oct 15 '14 at 15:52
  • We will be taking a sequence of $2n$ consecutive steps, $n$ rightward and $n$ upward. We can choose which $n$ steps in the sequence are rightward in $\binom{2n}{n}$ ways. The same idea works for an $m\times n$ board. – André Nicolas Oct 15 '14 at 16:24

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A legal path can be completely described by a word of length $2n$ made up of $n$ occurrences of the letter R (right) and $n$ occurrences of the letter U (up). Every path determines a unique word, and every word determines a unique path.

There are $\binom{2n}{n}$ such words. For there are $\binom{2n}{n}$ ways to choose the $n$ places in the word where R's occur.

Remark: The $\sqrt{2\pi}$ stuff in the hint refers to the Stirling approximation to the factorial.

The same idea can be used with a $m\times n$ board. In that case there are $\binom{m+n}{m}$ "up and/or right" paths from the bottom left corner to the top right.

André Nicolas
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  • First of all, thanks a lot for your beautiful explanation .I asked my teacher for the correct answer and I got $ 2^{2n + 1} \sqrt {2\pi } $ . I really don't get the sqrt(2pi) bit.. – user184588 Oct 15 '14 at 17:09
  • The number $\binom{2n}{n}$ is $\frac{(2n)!}{n!n!}$. By the Sirling approximation (please see Wikipedia) and some manipulation we find that $\binom{2n}{n}\approx \frac{2^{2n}}{\sqrt{\pi n}}$. The approximation to $\binom{2n}{n}$ that you quote is not correct. – André Nicolas Oct 15 '14 at 17:18
  • out of curiosity, what would be the error of this approximation ? – user184588 Oct 15 '14 at 18:17
  • Which approximation, Stirling or the wrong one? The article I linked to has information about the error in the Stirling approximation, and the error in the central binomial coefficient would be smaller. From the point of view of ratio the Stirling approximation is quickly very good. – André Nicolas Oct 15 '14 at 18:26
  • The Stirling one in this particular case – user184588 Oct 15 '14 at 18:27
  • There are many discussions of this, at various levels. For a fairly simple error estimate, please see the answers to this MSE question. – André Nicolas Oct 15 '14 at 18:33