   Chapter 9.3, Problem 44ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Note that a product x 1 x 2 x 3 may be parenthesized in two different ways: ( x 1 x 2 ) x 3 and x 1 ( x 2 x 3 ) . Similarly, there are several different ways to parenthesize x 1 x 2 x 3 x 4 . Two such ways are ( x 1 x 2 ) ( x 3 x 4 ) and x 1 ( ( x 2 x 3 ) x 4 ) . Two such ways are ( x 1 x 2 ) ( x 3 x 4 ) and x 1 ( ( x 2 x 3 ) x 4 ) . Let P n be the number of different ways to parenthesize the product x 1 x 2 ... x 4 . Show that if P 1 = 1 , then P n = ∑ k = 1 n = 1 P k P n − k for every integer n ≥ 2. (It turns out that the sequence P 1 ,   P 2 ,   P 3 ,   ... is the same as the sequence of Catalan numbers: P n = C n − 1 for every integer n ≥ 1. See example 5.6.4.)

To determine

To show that the different number of ways to parenthesize x1x2x3......xn is Pn=k=1n1PkPnk for all integers k2, if P1=1.

Explanation

Given information:

There are different ways to parenthesize the product for e.g. x1x2x3 can be parenthesized in two different ways x1(x2x3) and (x1x2)x3. Pn be the different number of ways to parenthesize x1x2x3......xn then Pn=k=1n1PkPnk for every integer n2 and P1=1.

Calculation:

Since it is given that Pn=k=1n1PkPnk and also observe that P1=1 and P2=1 .Since the product of one or two variables can’t be parenthesized in more than one way. So it is logically correct.

Now,

Pn=k=1n1PkP nkP3=k=131PkP 3kP3=k=12PkP 3kP3=P1P2+P2P1P3=1+1P3=2

Also the product x1x2x3 can be parenthesized in two different ways x1(x2x3) and (x1x2)x3

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