MATH 310 FALL 2003 Combinatorial Problem Solving Lecture
MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 24, Wednesday, October 29
5. 5. Binomial Identities Continuation n Homework (MATH 310#8 W): • Read 6. 1. • Turn in 5. 5: 2, 4, 6, 8, 32 • Volunteers: • ____________ • Problem: 32.
Block Walking Model B(n, n) n C(p, n-p) n A(0, 0) 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).
Binomial coefficients n The following is true: C(n, r) = C(n, n-r) n C(n, r) = (n/r) C(n-1, r-1) n C(n, r) = ((n-r+1)/r)C(n, r-1) n
Newton’s Binomial Theorem n For each integer n: Proof (By induction). n Corollaries: n n :
Some Binomial Identities n C(n, 1) + 2 C(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) - 2 C(n, 2) - 3 C(n, 3) + 4 C(n, 4) +. . . +(-1)n n C(n, n) = 0 2 C(n, 0) + (22/2)C(n, 1) + (23/3)C(n, 2) + (24/4)C(n, 3)+. . . = (3 n+1 – 1)/(n+1) C(n, 0)2 + C(n, 1)2 +. . . + C(n, n)2 = C(2 n, n)
Proof Methods Equality rule (combinatorial proof) n Mathematical Induction n Newton’s Theorem (derivatives, integrals) n Algebraic exercises n Symbolic computation n Generating Functions (What is that? ) n
Arrangements and Selections n Choose r elements from the set of n elements repeated elements no repetitions ordered nonordered nr C(n+r-1, r) n!/(n-r)! C(n, r)
r-arrangements with repetirions n Example: A = {a, b, c}, r = 2. a a n a b a c b a b b b c Answer: nr = 32 = 9 c a c b c c
r-arrangements (no repetitions) n Example: A = {a, b, c}, r = 2. a a n a b a c b a b b b c c a c b c c P(3, 2) = 9 – 3 = 6 = n!/(n-r)!=3 £ 2.
Permutations P(n, r) = 0, for r > n. n Interesting special case n = r: n P(n) : = P(n, n) permutations. n P(0) = 1. n P(n) = n P(n-1). n In general: n P(n) = n(n-1). . . 2. 1 = n! n
Permutations - Continuation Function n! (n-factorial) has rapid n n! growth: 0 1 n Stirling approximation: 1 1 n 2 2 3 6 4 24 5 120 6 720 7 5040 8 40320 9 362880
Permutations as functions Permutations can be regarded as bijections of A onto itself. n Example: A = {a, b, c} n a b a a b b a c g b a d b c e c a z c b c a b a
r-selections (no repetitions) n Example: A = {a, b, c}, r = 2. a a n a b a c b a b b b c c a c b c c C(3, 2) = (9 – 3)/2 = 3!/(2!1!)
r-selections with repetitions n Example: A = {a, b, c}, r = 2. a a n a b a c b a b b b c c a c b c c Answer: = (9 – 3)/2 + 3 = 6 = C(4, 2)
r-selections with repetitions C(n+r-1, r) n Problem: Given p signs “+” and q signs “-”. How many strings (of length p+q) are there? n n Answer: C(p+q, p) = C(p+q, q).
r-selections with repetitions - Proof n n To each selection assign a vector: a a a b b b c c c a b c + + + - - - + + - - + - + - + + Answer: = (9 – 3)/2 + 3 = 6 = C(4, 2)
6. 1. Generating Function Models Algebra-Calculus approach. n We are given a finite or infinite sequence of numbers a 0, a 1, . . . , an, . . . n Then the generating function g(x) for a_n is given by: n g(x) = a 0 + a 1 x +. . . + a 2 xn +. . . n
- Slides: 18