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Ok, now consider joining $2n$ points on a circle using $n$ nonintersection chords.

How do I prove that the number of ways to join the points equals then $n$-th Catalan number $C_n$?

Thanks!

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    There are two basic approaches. (1) Find a bijection with something else that you already know is counted by the Catalan numbers. (2) Show that these numbers satisfy the Catalan recurrence $C_n=\sum_{k=0}^{n-1}C_kC_{n-1-k}$, assuming that you know that recurrence. To give more of an answer, I’d have to know more about what you already know about the Catalan numbers. – Brian M. Scott Apr 01 '13 at 05:20
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    @Brian M Scott I am struck on same problem asked above but I don't know about how Catalan numbers are strings of n pairs of correctly matched parenthesis. I have studied that Catalan numbers= no. Of sequences of 2n terms whose partial sums are always positive and also Catalan numbers also equal number of ways to divide a convex polygon with n+1 sides into triangles by diagonals that don't intersect each other . In brief, I have followed Brualdi. Can you please help me in proving this question? –  Feb 24 '20 at 17:00

2 Answers2

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André has sketched one of the two natural approaches. Here’s one of the more straightforward versions of the other.

If you know that $C_n$ is the number of strings of $n$ pairs of correctly matched parentheses, you can try to find a bijection between ways of joining your $2n$ points and correctly matched strings of $n$ pairs of parentheses.

Label the points $1,2,\dots,2n$ clockwise around the circle. Each chord then corresponds to a pair of distinct numbers in $\{1,\dots,2n\}$. Use the chords to lay out a row of $2n$ parentheses, $p_1,p_2,\dots,p_{2n}$, according to the following rule. If the points $k$ and $\ell$ are connected by a chord, and $k<\ell$, make $p_k$ a left parenthesis and $p_\ell$ a right parenthesis.

  • Show that the resulting string of parentheses is correctly matched. (The fact that the chords don’t cross one another is important here.)

  • Show that the proceduce is reversible: given a correctly matched string of $n$ pairs of parentheses, there’s a way to reconstruct a set of non-intersecting chords that give rise to it.

Brian M. Scott
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  • Right, +1. Actually once you've understood what is going on it is so obvious that I would find it hard to motivate myself to still go and prove anything ;-) – Marc van Leeuwen Apr 01 '13 at 09:14
  • @Marc: A familiar feeling, that! (Though I have to admit that I’ve been burnt by it once or twice.) – Brian M. Scott Apr 01 '13 at 19:03
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The most direct approach is to use the recurrence relation $C_{n+1}=\sum_0^n C_iC_{n-i}$ for the Catalan numbers.

Let $S_n$ be the number of ways to do your division. If you can show that the sequence $(S_n)$ satisfies the same recurrence as $(C_n)$, and the same initial condition, then you will have proved that $(S_n)=(C_n)$ for all $n$.

Getting the recurrence for the sequence $(S_n)$ is geometrically quite natural. You may want to look at the Wikipedia article on the Catalan numbers. Working directly with the closed form formula $C_n=\frac{1}{n+1}\binom{2n}{n}$ would be more difficult than using the recurrence suggested above.

André Nicolas
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