Introductory Combinatorics
Introductory Combinatorics
5th Edition
ISBN: 9780134689616
Author: Brualdi, Richard A.
Publisher: Pearson,
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Chapter 8, Problem 1E

Let 2n(equally spaced) points on a circle be chosen. Show that the number ofways to join these points in pairs, so that the resulting nline segments do notintersect, equals the nth Catalan number Cn.

Expert Solution & Answer
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To determine

To show: That the number of ways to join the 2n (equally spaced) points in pairs.

Explanation of Solution

The number of ways to join the 2n (equally spaced) points in pairs, so that the resulting n line segments do not intersect, equals the nth Catalan number Cn. Let 2n (equally spaced) points on a circle be chosen.

The C0=1. Now assume that n1. Label the points 1,2,...,2n clockwise around the circle.

Let M=Mn denote the set of matchings of the 2n points counted by Cn.

For a matching in M let t denote the point matched to point 1. Note that t is even. For 1sn let M(s) denote the set of matchings in M such that point 1 is matched with point 2s.

The sets {M(s)}s=1n partition M, so |M|=s=1n|M(s)|.

For 1sn to compute |M(s)|. To construct a matching in M(s), there are Cs1 ways to match points 2,3,...,2s1 and there are Cns ways to match points 2s+1,2s+2,...,2n.

Therefore |M(s)|=Cs1Cns.

By these comments Cn=s=1nCs1Cns.

Cn=1n+1(2nn)

Where, n=0,1,2,....

Consider the generating function

g(x)=n=0Cnxn

Using the Catalan and obtain

xg(x)2=g(x)1

Using the quadratic formula

g(x)=1±(14x)122x

In other words

xg(x)=1±(14x)122x

Using Newton’s binomial theorem this becomes

xg(x)=1±n=0(12n)(4)nxn2

For this equation at x=0 the left-hand side is zero, so the right-hand side is zero. Therefore,

xg(x)=1n=0(12n)(4)nxn2=n=1(12n)(4)nxn2=n=1(12n)(4)nxn12=n=0(12n+1)(4)n+1xn2

Consequently, for n1.

Cn=(12n+1)(4)n+12=1n+1(2nn)

Note that, Cn=(2nn)(2nn1), where, n=1,2,3,....

Hence, proved.

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