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 February 23rd, 2013, 04:27 AM #1 Member   Joined: Jan 2013 Posts: 96 Thanks: 0 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: $a_1=1$, $a_2=2$ and $a_3=5$. (There are no 2 chords with a common point (including those 2n points on the circle), so there are exactly n chords)
 February 23rd, 2013, 01:19 PM #2 Member   Joined: Feb 2011 Posts: 68 Thanks: 0 Re: Counting Problem do you want a_n in terms of a_n-1, a_n-2 . .. a_1 ?
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 $a_n= \sum_{i=0}^{n-1} a_i * a_{n-1-i}$(a_0=1)

 February 24th, 2013, 08:52 AM #4 Member   Joined: Jan 2013 Posts: 96 Thanks: 0 Re: Counting Problem I want an explicit formula (Not dependig of any other term of the sequence)!
 February 25th, 2013, 06:13 AM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Counting Problem These are the Catalan numbers. There is a formula in terms of the binomial coefficient (or, equivalently, factorials).
 February 25th, 2013, 11:37 AM #6 Member   Joined: Feb 2011 Posts: 68 Thanks: 0 Re: Counting Problem Is this a mathematical Olympiad question?
 February 26th, 2013, 12:54 PM #7 Member   Joined: Jan 2013 Posts: 96 Thanks: 0 Re: Counting Problem Yes!

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