1 / 5

MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 22, Friday, October 24

Explore binomial coefficients and Pascal's Triangle applications in solving combinatorial problems. Study and implement algorithms for bonus points. Homework instructions included.

thomasrichardson
Télécharger la présentation

MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 22, Friday, October 24

An Image/Link below is provided (as is) to download presentation 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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. MATH 310, FALL 2003(Combinatorial Problem Solving)Lecture 22, Friday, October 24

  2. 5.5. Binomial Identities • Homework (MATH 310#7F): • Read Supplement (pp. 230-239) • Do 5.5: All odd numbered exercises. • Turn in 5.5: 10,12,20,26 • Volunteers: • ____________ • ____________ • Problem: 20. • Bonus! Study 5.6 and implement any of the four algorithms. (up to 5 points/program).

  3. Binomial Coefficients - Revisited • C(n, r) = P(n, r)/P(r) = n!/(r!(n-r)!) • C(n, 0) = 1 • C(n, n) = 1 • C(n, r) = C(n-1, r) + C(n-1, r-1). • Combinatorial Proof of line 4.

  4. Pascal Triangle r = 2 n = 1 C(5,2)=10 n = 5

  5. Power of Combinatorics – The Birthday Paradox. • Do we have two people with the same birthday? • Let n be the number of persons. Let P(n) denote probability that all birthdays are distinct. • For n=2: • P(2) = 364/365. • For n=3: • P(3) = 364/365 363/365. • For general n: • P(n) = 365 364 ... (365-n+1)/365n.

More Related