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 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)

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)

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

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) C(n,n) 2 = C(2n,n)

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

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)

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

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

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) = n!

Permutations - Continuation Function n! (n-factorial) has rapid growth: Stirling approximation: nn!

Permutations as functions Permutations can be regarded as bijections of A onto itself. Example: A = {a,b,c}  aaabbcc bbcacab ccbcaba

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

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

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).

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

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 +...