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#1
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Re: Maths problem
Here is a neat combinatorial argument .
Start from the origin and consider taking 2n steps on the lattice points of the line y=x or y=-x . Clearly the total number of paths is 2^(2n) . The number of ways of reaching the point (2n,0) is 2nCn . It is also true that the number of paths that do not intersect the x-axis is 2nCn . Since we can reflect all subsequent paths after the first step in the line y=1 or -1 to produce a bijection of paths that do not intersect the x-axis . Consider the last path that intersects the x-axis at the points (2k,0) . The number of paths is 2kCk . Therefore we require the last (2n-2k) to be above or below the x-axis . The total number of ways is equal to the number of ways of reaching the point (2n,0) which is just (2n-2k)C(n-k) . Now sum over everything and we get our identity . |
#2
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Re: Maths problem
One important detail I left out .
We move in the direction up-right or down-right . |
#3
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Re: Maths problem
To the original poster: Was this a homework problem in a combinatorics class? Just curious.
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#4
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Re: Maths problem
[ QUOTE ]
To the original poster: Was this a homework problem in a combinatorics class? Just curious. [/ QUOTE ] I made it up, but I'm sure this exact problem has been `made up' many times before. As above posters noted, it's closely related to generating function stuff for Catalan numbers. |
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