1 / 18

MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 24, Wednesday, October 29

MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 24, Wednesday, October 29. 5.5. Binomial Identities - Continuation. Homework (MATH 310#8W): Read 6.1. Turn in 5.5: 2,4,6,8,32 Volunteers: ____________ ____________ Problem: 32. Block Walking Model.

aubrey-hunter
Download Presentation

MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 24, Wednesday, October 29

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 24, Wednesday, October 29

  2. 5.5. Binomial Identities - Continuation • Homework (MATH 310#8W): • Read 6.1. • Turn in 5.5: 2,4,6,8,32 • Volunteers: • ____________ • ____________ • Problem: 32.

  3. Block Walking Model • How many paths are there from A to C if we may walk only upwards (U) or to the right (R)? • Find a binomial identity (by moving point C alnog the diagonal). B(n,n) C(p,n-p) A(0,0)

  4. Binomial coefficients • The following is true: • C(n,r) = C(n,n-r) • C(n,r) = (n/r) C(n-1,r-1) • C(n,r) = ((n-r+1)/r)C(n,r-1)

  5. Newton’s Binomial Theorem • For each integer n: • Proof (By induction). • Corollaries: • : • :

  6. Some Binomial Identities • C(n,1) + 2C(n,2) + ... + n C(n, n) = n 2n-1 In other words: C(n,0) + (1/2)C(n,1)+ ... + (1/(n+1)) C(n, n) = (2n+1 – 1)/(n+1) C(n,0) + 2 C(n,1) + C(n,2) + 2 C(n, 3) + ... = 3 2n-1 C(n,1) - 2C(n,2) - 3C(n,3) + 4C(n,4) +... +(-1)n n C(n, n) = 0 2C(n,0) + (22/2)C(n,1) + (23/3)C(n,2) + (24/4)C(n,3)+... = (3n+1 – 1)/(n+1) C(n,0)2 + C(n,1)2 + ... + C(n,n)2 = C(2n,n)

  7. Proof Methods • Equality rule (combinatorial proof) • Mathematical Induction • Newton’s Theorem (derivatives, integrals) • Algebraic exercises • Symbolic computation • Generating Functions (What is that?)

  8. Arrangements and Selections • Choose r elements from the set of n elements

  9. r-arrangements with repetirions • Example: A = {a,b,c}, r = 2. • Answer: nr = 32 = 9

  10. r-arrangements (no repetitions) • Example: A = {a,b,c}, r = 2. • P(3,2) = 9 – 3 = 6 = n!/(n-r)!=3 £ 2.

  11. Permutations • P(n, r) = 0, for r > n. • Interesting special case n = r: • P(n) := P(n, n) permutations. • P(0) = 1. • P(n) = n P(n-1). • In general: • P(n) = n(n-1)... 2.1 = n!

  12. Permutations - Continuation • Function n! (n-factorial) has rapid growth: • Stirling approximation:

  13. Permutations as functions • Permutations can be regarded as bijections of A onto itself. • Example: A = {a,b,c}

  14. r-selections (no repetitions) • Example: A = {a,b,c}, r = 2. • C(3,2) = (9 – 3)/2 = 3 = 3!/(2!1!)

  15. r-selections with repetitions • Example: A = {a,b,c}, r = 2. • Answer:= (9 – 3)/2 + 3 = 6 = C(4,2)

  16. r-selections with repetitions • C(n+r-1,r) • Problem: Given p signs “+” and q signs “-”. How many strings (of length p+q) are there? • Answer: C(p+q,p) = C(p+q,q) .

  17. r-selections with repetitions - Proof • To each selection assign a vector: • Answer:= (9 – 3)/2 + 3 = 6 = C(4,2)

  18. 6.1. Generating Function Models • Algebra-Calculus approach. • We are given a finite or infinite sequence of numbers a0, a1, ..., an, ... • Then the generating function g(x) for a_n is given by: • g(x) = a0 + a1x + ... + a2xn + ...

More Related