Study Materials for Applied Combinatorics | MATH 3012, Study notes of Mathematics

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Math 3012: Applied Combinatorics notes

Alex Whitney

June 29, 2007

Contents

• 1 May 14,
  • 1.1 Introductory Notes
  • 1.2 Fundamental Rules of Counting
    • 1.2.1 The Rule of Sums
    • 1.2.2 The Rule of Products
  • 1.3 Permutations
• 2 May 16,
  • 2.1 Combinations
    • 2.1.1 Binomial Coefficient
    • 2.1.2 Multinomial Coefficient
  • 2.2 Binomial Theorem
  • 2.3 Multinomial Theorem
• 3 May 19,
  • 3.1 Combinations and Permutations with Repetition
• 4 May 21,
  • 4.1 Induction
  • 4.2 Division Algorithm
• 5 May 23,
  • 5.1 Division Algorithm (continued)
  • 5.2 Euclidean Algorithm
  • 5.3 Additional Proofs
• 6 May 25,
  • 6.1 Cartesian (cross) Product
• CONTENTS
  • 6.2 Relations
  • 6.3 Functions (Mapping)
    • 6.3.1 One-to-One (Injective)
    • 6.3.2 Onto functions
  • 6.4 Bijection
  • 6.5 Pigeonhole Principle
• 7 May 30,
  • 7.1 Principle of Inclusion and Exclusion
    • 7.1.1 Set Notation
  • 7.2 June 1,
• 8 June 4,
  • 8.1 Derangements Continued
  • 8.2 Rook Polynomials
  • 8.3 Mathematical Definition
  • 8.4 Computing Rook Polynomials
• 9 June 6,
  • 9.1 Arrangements with Forbidden Positions
• 10 June 13, 2007 (Reformat)
  • 10.1 Introduction to Generating Functions
  • 10.2 Formulas for Series
• 11 June 15, 2007 (Reformat)
  • 11.1 Partial Fraction Decomposition
  • 11.2 Application of Partial Fractions
• 12 May 18,
  • 12.1 Exponential Generating Functions
  • 12.2 More Infinite Series Formulas
• 13 June 20,
  • 13.1 The Summation Operation
• 14 June 22, 2007 (Reformat)
  • 14.1 Recurrence Relations
  • 14.2 Non-Constant Recurrence Relations
• CONTENTS
  • 14.3 ???
• 15 June 25, 2007 (Reformat) - tions 15.1 Second Order Linear Constant Coefficient Recurrence Rela-
• 16 June 27, 2007 (reformat)
  • 16.1 Review
  • 16.2 Non-Homogeneous Recurrence Relations - lation 16.2.1 Steps to solving a non-homogeneous recurrence re-
• 17 June 29, 2007 (reformat)
  • 17.1 Combining Generating Functions and Recurrence Relations
    • Solve 17.2 Using Generating Functions to Problems We Can’t Already
  • 17.3 Additional Formulas

Lecture 1

May 14, 2007

1.1 Introductory Notes

• Formula sheets are allowed on exams, but no calculators
• Test one covers chapters 1, 4, 5, and 8 (Basic Counting Methods)
• Test two covers chapters 9, 10, and 11 (Advanced Counting Methods, basic graph theory)
• The final covers 12 and 13, in addition to the previous materials (Optimization on Graph theory)

1.2 Fundamental Rules of Counting

1.2.1 The Rule of Sums

Rule of Sums If the possible outcomes of A can be divided into two dis- joint sets, B and C, then |A| = |B| + |C|.

Example 1.2.1. There are two barrels of apples. The first barrel has 10 apples, and the second barrel has 8 apples. A person walks by and selects an apple from a barrel. How many different apples could the person have selected?

Solution The apples in each barrel are disjoint - that is to say that any apple in the first barrel cannot be in the second barrel and vice versa. The

LECTURE 1. MAY 14, 2007 6

Example 1.3.1. A coin is flipped and the result is recorded. The coin is flipped again and the result recorded. List all of the permutations for this procedure.

Solution {HH}, {HT}, {TH}, {TT}

r-permutations of n The number of permutations of all subsets of r objects from a set of n distinct objects is given by the equation:

P(n,r) =

n! (n − r)!

Lecture 2

May 16, 2007

2.1 Combinations

r-combinations of n An r-combination of n objects is a subset of r distinct objects from n.

Example 2.1.1. List all of the possible 2-combinations of { 1 , 2 , 3 }.

Solution { 1 , 2 }, { 1 , 3 }, { 2 , 3 }

2.1.1 Binomial Coe ffi cient

Theorem 2.1.1. Binomial Coefficient. The number of r-combinations of n dis- tinct objects is equal to:

n! r!(n − r)!

n r

2.1.2 Multinomial Coe ffi cient

Theorem 2.1.2. Multinomial Coefficient. Given n objects from k classes, if the objects in each class are identical, then the number of unique permutations of n is equal to:

Lecture 3

May 19, 2007

3.1 Combinations and Permutations with Repe-

tition

Example 3.1.1. Give the number of combinations of 5 objects from {s 1 , s 2 , s 3 }, with replacement.

Solution There are 3 classes and we are selecting 5 objects. We can rep- resent any given combination by using a notation of “crosses and bars”^1. For example,

|*|

would indicate that there are two objects from the first class, three objects from the second class, and zero objects from the third class. Looking at a combination in this manner, there are seven possible positions for either a star or a bar. In order to create a combination, we only need to fix the positions of bars or stars. Since there are 5 + 3 − 1 = 7 positions and we are fixing 3 − 1 = 2 bars, there are a total of

2

combinations.

This can be applied to a general case.

Theorem 3.1.1. The number of combinations of n objects from r classes is equal to:

(^1) I have also heard this called “stars and bars” and “balls and buckets”

LECTURE 3. MAY 19, 2007 10

n + r − 1 r − 1

Example 3.1.2. Find the number of combinations of 10 objects selected from 3 classes, with at most two objects from the first class.

Solution

Number in first class: 0:

By the rule of sums, the number of combinations of 10 objets from three classes is 11 + 10 + 9 = 30.

Example 3.1.3. Find the number of non-negative integer solutions to x 1 +x 2 +x 3 = 10 , x 1 ≤ 2.

Solution The solution to Example 3.1.3 is the same as the solution to Ex- ample 3.1.2. If x 1 , x 2 , x 3 are the classes, then we are selecting 10 objects from the three classes with at most 2 selections from the first class. Therefore, the answer is 30.

LECTURE 4. MAY 21, 2007 12

k(k + 1) 2

2(k + 1) 2 =

k(k + 1) + 2(k + 1) 2 =

(k + 1)(k + 1 + 1) 2

By induction, 0 + 1 + 2 + 3 +... + n = n(n 2 + 1), ∀ n ≥ 0 

Strong Principle of Induction To prove S(n), first prove S(n) for a base case n 0. Assume S(k) holds for all values between n 0 and k, then show that S(k + 1) holds. If S(n 0 ) and S(k + 1) hold, then S(n) holds for all n.

Example 4.1.2. Prove that every integer n ≥ 2 is a product of primes

Solution Proof by induction.

Hypothesis: P(n) = pq = n, where p, q are prime and k ≤ n ≥ 2, n ∈ Z

Base: n = 2 P(2) = 2 ∗ 1 = 2

Step: P(k + 1) = pq If k + 1 is prime, then k + 1 is a product of primes. If k + 1 is not prime, k + 1 = pq, where 2 ≤ (p, q) ≤ k. p and q are product of primes, so pq is a product of primes.

By induction, n is a product of primes ∀n ≥ 2 

4.2 Division Algorithm

“b divides a” Given a, b ∈ Z and b , 0, then b | a if a = mb, where m ∈ Z. If this is true, then it is said that “b divides a.”

Lemma 4.2.1. If b | a and c | b, then c | a.

LECTURE 4. MAY 21, 2007 13

Proof

Since b | a, ∃ m ∈ Z such that a = mb Since c | b, ∃ n ∈ Z such that b = nc Therefore, a = m(nc) = (mn)a, so c | a. 

Theorem 4.2.2. Division Algorithm. ∀ a, b ∈ Z with b > 0 , ∃ q, r ∈ Z such that a = q ∗ b + r with 0 ≤ r ≤ b.

LECTURE 5. MAY 23, 2007 15

a = qc + r, 0 < r < c (by division alg.) r = a − qc r = a − q(sa + tb) r = (1 − qs)a + (−qt)b r = s′a + t′b Contradiction 

∴ r ∈ S

Theorem 5.2.2. Euclidean Algorithm. Given a, b ∈ Z, Let r 0 = a, r 1 = b ri = qri+ 1 + ri+ 2 , 0 ≤ ri+ 2 < ri+ 1 gcd(ri, ri+ 1 ) = gcd(ri+ 1 , ri+ 2 ) This process is repeated until rk+ 2 = 0. Work backwards until you get back to gcd(a, b).

Example 5.2.1. Find the G.C.D. of 1024 and 28. Find s, t so that gcd(1024, 28) = 1024 s + 28 t.

Solution 1024 = 36 ∗ 28 + 16 28 = 1 ∗ 16 + 12 16 = 1 ∗ 12 + 4 12 = 3 ∗ 4 + 0 ∴ gcd(1024, 28) = 4 4 = 16 − 1 ∗ 12 4 = 16 − 1 ∗ (28 − 1 ∗ 16) 4 = 2 ∗ (128 − 36 ∗ 28) − 1 ∗ 28

5.3 Additional Proofs

Lemma 5.3.1. If a, b ∈ Z+^ and p is prime, then p | ab if and only if p | a or p | b.

Proof If p | a, then p | ab. If p - a, then gcd(p, a) = 1 and ∃ s, t ∈ Z such that 1 = sp + ta. b = spb + tab Since p | spb and p | tab, then p | b 

LECTURE 5. MAY 23, 2007 16

Lemma 5.3.2. If a 1 ,... an ∈ Z+^ and p is prime, then p | (a 1 ∗... ∗ an if p | a 1 or p | a 2 or... or p | an.

Proof By induction. Hypothesis: S(k) = above. True ∀ k ≥ 1. Base: S(1) : p | a. p | a if p | a. Step: p | a 1... ak+ 1 → p | (a 1... ak)(ak+ 1 ) Proved by lemma 5.3.

Theorem 5.3.3. Any integer n ≥ 1 can be written as a product of unique primes.

Proof Uniqueness. n = pα 1 1 + pα 2 2 +... + pα nn

n = qβ 11 + qβ 22 +... + qβ nn

If p 1 | pα 1 1 + pα 2 2 +... + pα n n, then p 1 | qβ 1 1 + qβ 22 +... + qβ nn. By Lemma 5.3.1, we can assume without loss of generality that p 1 | q 1. Since p 1 and q 1 are prime, then p 1 = q 1.

pα 2 2 +... + pα nn = q β 2 2 +^...^ +^ q

βn n Continuing with the application of Lemma 5.3.1, p 2 = q 2 ,... pk = qk. 

Lemma 5.3.4.

3 is irrational.

Proof Proof by contradiction.

Suppose

p q gcd(p, q) = 1, p, q ∈ Z. 3 = p^2 q^2 3 q^2 = p^2

LECTURE 6. MAY 25, 2007 18

6.3 Functions (Mapping)

Function For any non-empty sets A, B, a function (or mapping) from A to B is a relation from A to B such that every element of A occurs in exactly one ordered pair in the relation. A function is written as: f : A → B, where A is the the domain and f (B) is the range.

Example 6.3.1. Is the relation in Example 6.2.1 a function? No, because 3 does not occur exactly once in every ordered pair.

6.3.1 One-to-One (Injective)

One-to-One (Injective) A function f : A → B is one-to-one if each element of B occurs at most once in the range of f (A).

Corollary 6.3.1. If f : A → B is a one-to-one function, then |A| ≤ |B|.

6.3.2 Onto functions

Onto (Surjective) A function f : A → B is onto if every element of B must occur exactly once in f (A).

Corollary 6.3.2. If f : A → B is an onto function, then |A| ≥ |B|.

6.4 Bijection

Bijection A function f : A → B is a bijection if and only if it is both one-to-one and onto.

Example 6.4.1. The number of subsets of { 1 ,... , n} = 2 n.

Solution Let A = {subsets of { 1 ,... , n} and B = {binary sequences of length n}.

LECTURE 6. MAY 25, 2007 19

The numbers in the set on the left represent the positions in the binary sequence where a 1 exists. Therefore, this mapping is both one-to-one and onto, so it is a bijection and |A| = |B|. Since |B| = 2 n, then |A| = 2 n^ 

6.5 Pigeonhole Principle

Pigeonhole Principle If m pigeons occupy n pigeonholes, and m > n, then at least one pigeonhole is occupied by at least two pigeons.

Example 6.5.1. Among any 15 integers, prove that there are at least two with the same remainder when divided by 14.

Proof Let B be the possible remainders are the set given by n mod 14, which are { 0 , 1 ,... , 13 }. Because |B| = 14, by the pigeonhole principle there must be at least two integers from any given set of 15 integers that have the same remainder when divided by 14. 

Example 6.5.2. Let m ∈ Z+^ where m is odd. Prove there exists a positive integer n such that m | 2 n^ − 1.

Proof Let A = { 21 − 1 , 22 − 1 ,... , 2 m+^1 − 1 }. |A| = m + 1. Let B = n mod m. |B| = m. By the pigeonhole principle, at least two elements of A have the same remainder when divided by m. Given two integers ∈ A, 2i^ − 1 and 2j^ − 1, i < j. m | (2j^ − 1 − 2 i^ + 1), so m | 2 i(2j−i^ − 1). Since m is odd, then m - 2 i, ∴ m | 2 j−^1 − 1.