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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: , and . (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_n1, a_n2 . .. a_1 ?

February 23rd, 2013, 01:20 PM  #3  
Member Joined: Feb 2011 Posts: 68 Thanks: 0  Re: Counting Problem Quote:
 
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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