1 / 16

# Some Parenthetical Remarks About Counting - PowerPoint PPT Presentation

Some Parenthetical Remarks About Counting. Dr. Henry Ricardo Hunter College High School October 12, 2012. Two Similar Problems In how many ways can we multiply n + 1 numbers two at a time?

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Some Parenthetical Remarks About Counting

Dr. Henry Ricardo Hunter College High School

October 12, 2012

Two Similar Problems

In how many ways can we multiply n + 1 numbers two at a time?

In how many ways can we arrange n left parentheses ( and n right parentheses ) as legitimate grouping devices?

“Legitimate” Parenthesization

• At any point in the process of counting from left to right, the number of (’s must be greater than or equal to the number of )’s.
• The total number of (’s must equal the total number of )’s.

Each arrangement of n left parentheses ( and n right parentheses ) is equivalent to a “mountain path”−−a sequence of n diagonal upward strokes / and n diagonal downward strokes \.

A valid arrangement of 2n parentheses corresponds to a mountain path that lies on or above the x-axis.

An invalid arrangement of parentheses corresponds to a mountain path that crosses the

x-axis.

A Mountain Path Correspondence

U D U U D D U D

( ) ( ( ) ) ( )

A Mountain Path Correspondence

U D D U U D D U

( ) ) ( ( ) ) (

A(n) = the number of all possible

mountain paths from (0, 0) to

(2n, 0)

G(n) = the number of mountain

paths from (0, 0) to

(2n, 0) which lie on or above

the x-axis

B(n) = the number of “bad”

mountain paths from

(0, 0) to (2n, 0)—those which

cross the x-axis

Then A(n) = G(n) + B(n),

or

Cn = G(n) = A(n) ­− ­ B(n)

An Equivalent Problem

Cn is the number of different ways a convex polygonwith

n + 2 sides can be cut into triangles by connecting vertices with straight lines.

The following hexagons illustrate the case n = 4

Some Other Equivalent Problems

The number of ways 2n people, seated around a round table, can shake hands without their hands crossing

The number of mountain ranges with n – 1 peaks such that they do not contain three consecutive upsteps or three consecutive downsteps

If a student wants to take n math courses m1, m2, . . ., mn and n computer courses c1, c2, . . ., cn , where mi is a prerequisite for mi +1 , ci is a prerequisite for ci + 1, and mi is a prerequisite for ci , then there are Cn ordered ways the student can take these 2n courses.

References

Fibonacci and Catalan Numbers: An Introduction by Ralph Grimaldi (Wiley, 2012)

Catalan Numbers with Applications by Thomas Koshy

(Oxford University Press, 2009)

Enumerative Combinatorics, Volume 2 by Richard P. Stanley (Cambridge University Press, 2001)

[Stanley has a set of exercises describing 66 problems equivalent to the parentheses problem.]