Download presentation

Presentation is loading. Please wait.

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) A(0,0) C(p,n-p)

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 2 n-1 In other words: C(n,0) + (1/2)C(n,1)+... + (1/(n+1)) C(n, n) = (2 n+1 – 1)/(n+1) C(n,0) + 2 C(n,1) + C(n,2) + 2 C(n, 3) +... = 3 2 n-1 C(n,1) - 2C(n,2) - 3C(n,3) + 4C(n,4) +... +(-1) n n C(n, n) = 0 2C(n,0) + (2 2 /2)C(n,1) + (2 3 /3)C(n,2) + (2 4 /4)C(n,3)+... = (3 n+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 orderednonordered repeated elements nrnr C(n+r-1,r) no repetitions n!/(n-r)!C(n,r)

9
r-arrangements with repetirions Example: A = {a,b,c}, r = 2. Answer: n r = 3 2 = 9 aaabbbccc abcabcabc

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

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: nn! 01 11 22 36 424 5120 6720 75040 840320 9362880

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

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

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

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) aaabbbccc abcabcabc +++--- +--++- -+-+-+ --+-++

18
6.1. Generating Function Models Algebra-Calculus approach. We are given a finite or infinite sequence of numbers a 0, a 1,..., a n,... Then the generating function g(x) for a_n is given by: g(x) = a 0 + a 1 x +... + a 2 x n +...

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google