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February 23rd, 2013, 04:27 AM   #1
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Counting Problem

On a circle we have 2n points (n is a natural nomber). In how many ways we can construct n chords of the circle, which don't intersect each other, joining pairs of the 2n points?
Example: , and . (There are no 2 chords with a common point (including those 2n points on the circle), so there are exactly n chords)
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February 23rd, 2013, 01:19 PM   #2
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Re: Counting Problem

do you want a_n in terms of a_n-1, a_n-2 . .. a_1 ?
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February 23rd, 2013, 01:20 PM   #3
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Re: Counting Problem

Quote:
Originally Posted by limitkiller
do you want a_n in terms of a_n-1, a_n-2 . .. a_1 ?
in that case it is (a_0=1)
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February 24th, 2013, 08:52 AM   #4
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Re: Counting Problem

I want an explicit formula (Not dependig of any other term of the sequence)!
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February 25th, 2013, 06:13 AM   #5
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Re: Counting Problem

These are the Catalan numbers. There is a formula in terms of the binomial coefficient (or, equivalently, factorials).
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February 25th, 2013, 11:37 AM   #6
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Re: Counting Problem

Is this a mathematical Olympiad question?
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February 26th, 2013, 12:54 PM   #7
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Re: Counting Problem

Yes!
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