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:

:

:

kn

k

n xk

nx

0

)1(

nn

k k

n2

0

0)1(0

n

k

k

k

n

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) = 02C(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)

1

1

2

nn

k

nk

nk

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

ordered nonordered

repeated elements

nr 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: nr = 32 = 9

a a a b b b c c c

a b c a b c a b c

r-arrangements (no repetitions)

Example: A = {a,b,c}, r = 2.

P(3,2) = 9 – 3 = 6 = n!/(n-r)!=3 £ 2.

a a a b b b c c c

a b c a b c a b c

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!

Permutations - Continuation Function n! (n-factorial) has rapid

growth: Stirling approximation:

n n!

0 1

1 1

2 2

3 6

4 24

5 120

6 720

7 5040

8 40320

9 362880

nennn )/(2!

Permutations as functions Permutations can be regarded as

bijections of A onto itself. Example: A = {a,b,c}

a a a b b c c

b b c a c a b

c c b c a b a

r-selections (no repetitions)

Example: A = {a,b,c}, r = 2.

C(3,2) = (9 – 3)/2 = 3 = 3!/(2!1!)

a a a b b b c c c

a b c a b c a b c

r-selections with repetitions

Example: A = {a,b,c}, r = 2.

Answer:= (9 – 3)/2 + 3 = 6 = C(4,2)

a a a b b b c c c

a b c a b c a b c

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)

a a a b b b c c c

a b c a b c a b c

+ + + - - -

+ - - + + -

- + - + - +

- - + - + +

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 + ...