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Discrete Mathematics with Applications

FISHAN KONDOWE NICHOLAS SIAME ADAM

JEREMOT MASOAMBETA

November 6, 2017

Preface

i

Dedication

iii

TABLE OF CONTENTS TABLE OF CONTENTS

Table of Contents Preface i Acknowledgments ii

• 1 Unit 1 : Counting Techniques Dedication iii
  • 1.1 Lesson 1: Counting Principles
    • 1.1.1 Multiplication Rule or Product Rule
  • 1.2 Lesson 2: Permutation
  • 1.3 Lesson 3: Combination
  • 1.4 Lesson 4: Binomial Coefficients
  • 1.5 Lesson 5: Multinomial Theorem
  • 1.6 Lesson 6: Sum Rule
  • 1.7 Lesson 7: Inclusion-Exclusion Principle
• 2 Unit 2 : Difference Equations
  • 2.1 Lesson 1: Generating Functions
    • 2.1.1 Counting with Generating Functions
  • 2.2 Lesson 2: Formulating Relations
    • 2.2.1 Methods of Solving Recursive Equations
• 3 Unit 3 : Pigeon-hole Principle
  • 3.1 Lesson 1: Pigeon-hole Principle (PHP)
  • 3.2 Lesson 2: Simple Form of PHP
  • 3.3 Lesson 3: Regular Form of PHP
  • 3.4 Lesson 4: Extended Regular Form of PHP
  • 3.5 Lesson 5: Generalized PHP
• 4 Unit 4: Graph Theory Basics
  • 4.1 Lesson 1: Vertex, Edge and Valence
  • 4.2 Lesson 2: Walks, Paths’s, Trails
  • 4.3 Lesson 3: Disconnectedness of Graphs
  • 4.4 lesson 4: Vertex Colouring & Time Tabling
    • 4.4.1 The Greedy Algorithm For Colouring Vertices
    • 4.4.2 Map Colouring

1.1 Lesson 1: Counting Principles 1 UNIT 1 : COUNTING TECHNIQUES

1.1.1 Multiplication Rule or Product Rule The product rule involves a possibility of two events occurring at the same time. Consider the following Figure:

How many ways can we switch on the bulb? Obviously the answer is 6. Thus, we have multiplied the number of switches in A and number of switches in B. In other words, if we let set A = {a, b, c} and set B = { 1 , 2 }, then the number of ways of choosing a pair of members such that one is from A and another from set B is 6. Thus, (a, 1), (a, 2), (b, 1), (b, 2), (c, 3), (c, 4) = 3 × 2 = 6 We can now generalize the problem above with the following Theorem.

Theorem 1.1 Product Rule of Counting/ Multiplication Rule Let A and B be two finite sets. Then the number of ways of choosing one member from A and another from B is n(A) × n(B)

Example 1.

a) How many two digit positive integers are there such that 2 does not appear on the units (ones)?

Solution

There are nine positive integers:{ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } from where two sets of integers; the tens and the units (one) are to be generated. We are restricted that 2 cannot appear on the units(ones). Hence, there remains 8 options of generating a unit integer. There is no restriction on the tens integers set hence there are 9 integers. By product rule, there are 9 × 8 = 72 two positive integers such that 2 does not appear on the units.

b) In Malawi, each automobile license plate number consists of two letters followed by a four- digit number. How many distinct license plate numbers can be formed in Malawi? Solution

1.1 Lesson 1: Counting Principles 1 UNIT 1 : COUNTING TECHNIQUES

We are choosing 2 letters from 26 letters of English alphabet. In the ordered arrangement, a letter may be repeated. Thus, we have 26 × 26 = 26^2 = 676 possible letters.

Similarly, we are choosing 4 digits from 10 digits which are 0, 1, 2, 3, 4, 5, 6, 7, 8 ,9. In the ordered arrangement, a digit may be repeated. Thus, we have 10× 10 × 10 ×10 = 10^4 = 10, 000 possible digits.

Therefore, by product rule of counting, there are

262 · 104 = 676 · 10000 = 6, 760 , 000

distinct license plate numbers which can be formed in Malawi.

The product rule is not just restricted to two sets, we can have as many sets as possible. The idea is now generalized in the following theorem.

Theorem 1.2 Generalized Product Rule

Let A 1 , A 2 , A 3 , · · · , Ak be finite sets. Then the number of ways of choosing k members one from each set is n(A 1 ) × n(A 2 ) × n(A 3 ) × · · · × n(Ak)

Example 1.

a) How many rows of T and F can one have for the statement (proposition)

q 1 ∨ q 2 ∨ q 3 ∨ · · · ∨ qn?

Solution

Each of the proposition can be evaluated to be T or F. Putting the responses on the rows, we end up having two rows. Thus, (^) ( T F

for n times. Hence, we have 2n^ rows.

b) How many 4 digit positive integers are there using the digits 1, 2 , 3 , · · · , 9 such that no digit is repeated.

Solution

We can visualize the digits arrangement as:

1.2 Lesson 2: Permutation 1 UNIT 1 : COUNTING TECHNIQUES

Permutation is denoted by nPr or (^) nPr or P (n, r). They are all equivalent to saying “n permute r ”. Investigation: 5 P 2 Let A, B, C, D, E be different items. For instance, the first selection of the two objects would be to place item A in the first position and item B on the second position making AB. The second selection would be to put item B on the first position and item A on the second, BA. Thus, there are two possibilities of positioning a pair of items A and B. In the same way, there are two possibilities for each of the pairs : AC, AD, AE, BC, BD, BE, CD, CE, DE. In total, there are 20 possibilities of choosing two items from A, B, C, D, E and placing them in two fixed positions. Thus, 5 P 2 = 5 × 4 = 20.

Example 1.

a) Nicholas, Jeremot and Ellen were all born in the same ordinary year. In how many ways can these three persons have different birthdays?

Solution

An ordinary year has 365 days (different dates). Three birth dates are selected from a possible 365 days and assigned to three distinct persons. No, birth date may repeat itself within each year. The number of ways that this can occur is

P (365, 3) = 365 · 364 · 363 = 48, 228 , 180

b) How many 9-digit positive integers are there that use the digit 1, 2 , 3 , · · · , 9 where no digit is repeated? Solution

We want 9 P 9 = 9 × 8 × 7 × 6 × 5 × 4 × 3 × ×1 = 362, 880

Factorial Notation We were introduced to factorial notation in College Algebra as the product of integers from 1 to n inclusive occurring in n factorial ways, denoted n!. Thus, n! = n×(n−1)×(n−2)×· · ·× 3 × 2 × 1 For instance, 5! = 5 × 4 × 3 × 2 × 1 ≡ 5 × 4! and 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × × 1 Notice that 9 P 9 = 9!. Therefore, nPn = n! read “n factorial”.

Theorem 1.3 Permutation nPr = (^) (nn−!r)! Investigation of nPr

1.2 Lesson 2: Permutation 1 UNIT 1 : COUNTING TECHNIQUES

Consider the following Figure

We therefore conclude that nPr = n × (n − 1) × (n − 2) × · · · × n − (r − 1) ≡ n × (n − 1) × (n −

2. × · · · × n − r + 1.

If x 6 = 0, then xx = 1. Therefore,

nPr = n × (n − 1) × (n − 2) × · · · × n − r + 1 × 1

= n × (n − 1) × (n − 2) × · · · × (n − r + 1) × ((nn^ −−^ rr)) × ((nn^ −−^ rr^ −−^ 1)1) × · · · × 33 × 22 × (^11)

= (^) (n −n! r)!

Thus, 5 P 2 = (^) (5−5!2)! = 5 ×^4 3!× 3!= 20.

Permutations that have been performed so far are under the assumption that items are not re- peated at any of the positions. However, in some scenarios, permutation may involve repetition.

Permutation with Repetition

Suppose we are asked to find the distinct permutations from the word CHANCELLOR. How would you go about it?

We are interested with number of permutations of a multiset, that is a set of objects some which are alike. We observe that the word CHANCELLOR has 10 letters where the letters C and L appear twice, letters H, A, N, E, OR appear once.

Let P (n; n 1 , n 2 , · · · , nr) denote the number of permutations of n objects of which n 1 are alike, n 2 are alike, · · ·, nr are alike. Then the total permutations are calculated using the following theorem:

Theorem 1.4 Permutation with Repetition

P (n; n 1 , n 2 , · · · , nr) = (^) n n! 1!^ ×^ n 2!^ ×^ n 3!^ × · · · ×^ nr! where n = n 1 + n 2 + · · · + nr.

Using the given formula the total permutations on the word CHANCELLOR is:

P (10; 2, 1 , 1 , 1 , 1 , 2 , 1 , 1) = (^) 2! × 1! × 1! × 1!10! × 1! × 2! × 1! × 1! = 907, 200

Example 1.

1.3 Lesson 3: Combination 1 UNIT 1 : COUNTING TECHNIQUES

i. With replacement ii. Without replacement

Q.2. Three hungry Eagles are flying in the sky in search for the food. They suddenly see 200 Chambo fish being sun-dried in preparation for the market. In how many possible ways would these Eagles take away the fish?

Q.3. A row of six seats in a classroom is to be filled by selecting individuals from a group of ten students

i. In how many different ways can the seats be occupied? ii. If there are six boys and four girls in the group and if boys and girls are to be alternated, find the number of different seating arrangements.

Q.4. Find distinguishable permutations of letters in the words:

i. ARRANGEMENT ii. MASOAMBETA iii. PROFESSOR

1.3 Lesson 3: Combination

Suppose we have a collection of n objects. A combination of these n objects taken r at a time, is any selection of r objects where order does not matter. For example: How many ways can one choose two boys from 5?

In the example, the issue of putting boys on two fixed position is missing. Recall our solution from “first principle” (without using any formula/technique).

Let A, B, C, D, E be our boys. Then, we have 10 pairs: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. The required answer is 10.

The notation 5 C 2 means number of ways of choosing 2 boys from 5. At this time we do not know the formula for 5 C 2. However, we know 5 P 2.

Recall: For any choice of 2 boys, AB, we have 2 P 2 = 2 × 1 = 2 possibilities.Therefore,

(^5) C 2 × (^2) P 2 = 5 P 2 ∴ 5 C 2 =

5 P 2

2 P 2

Notation

Combination is denoted, nCr or (^) nCr or Crn to mean number of ways of choosing r items from n.

NB: If the chosen r are to be placed into r fixed positions, then rPr = r! possible permutations. Therefore,

nPr = nCr ×r (^) Pr

1.3 Lesson 3: Combination 1 UNIT 1 : COUNTING TECHNIQUES

nPr = nCr × r! ∴ nCr =

nPr r! =^

n! (n − r)!/r! ∴ nCr = (^) r!(nn −! r)!

Theorem 1.5 Combination

i. nCr = (^) r!(nn−!r)!

ii. nCr ≡ nCn−r

Proof

i. nCr = (^) r!(nn−!r)! already proved.

ii. Number of ways of choosing r items from n is equivalent to choosing n − r from n to DISCARD. Therefore, nCr ≡ nCn−r

Example 1.

a) A class has 6 girls and four boys. How many ways can we choose 2 girls and 2 boys from this class?

Solution Number of ways of choosing 2 girls from 6 is 6 C 2. Number of ways of choosing 2 boys from 4 is 4 C 2. By product rule, the number of ways of choosing 2 girls and 2 boys is 6 C 2 ×^4 C 2 = 90

b) Five (5) points lie on a circumference of a circle. How many chords can we have by joining any two points? Solution

Each chord requires two points. Therefore, number of chords are 5 C 2 = 10 chords.

c) Find the total number of triangles that can be drawn using points in (b) above as the corners (vertices) of a triangle? Solution

For us to draw a triangle, there is need of joining three points. Thus, we are choosing three points out of five. On the other hand, we are choosing 2 points out of five to discard. Therefore, the total number of triangles are: 5 C 3 =^6 C 2 = 10 triangles.

1.4 Lesson 4: Binomial Coefficients 1 UNIT 1 : COUNTING TECHNIQUES

a) How many strictly increasing sequences of 3 numbers can you form from the set { 3 , 1 , 2 , 9 , 7 }. For instance, the choice of { 3 , 1 , 3 } gives 1, 2 , 3 as a strictly increasing sequence.

Solution

In here, we observe that any choice of 3 different numbers give one unique strictly increas- ing sequences. Therefore, the number of sequences required is the same as number of ways of choosing 3 numbers from 5 numbers which is also equivalent to choosing 2 numbers from 5 to discard. Thus, (^5) C 3 ≡ 5 C 2 = 10

b) Figure1 shows roads intersecting at right angle. How many shortest routes are there from A to B?

Figure 1: Crossing Roads

Solution

Let x be the horizontal movement and y be a vertical movement as labeled above. Then, every route from A to B is uniquely expressed as the sequence of 3x’s and 2y’s. Therefore, the number of possible shortest routes is equivalent to number of sequences of 3x’s and 2y’s. Thus we have: (^5) C 3 = 10 = (^5) C 2

Recall: b + b = 2b. This implies that (x + y) + (x + y) = 2(x + y) = 2x + 2y. Thus, we can now generalize for c = constat that c(x + y) = cx + cy. Suppose we are multiplying three brackets, (x 1 +x 2 )(y 1 +y 2 )(z 1 +z 2 ). We can choose to multiply out the first and the second brackets and finish with the third bracket as:

(x 1 + x 2 )(y 1 + y 2 )(z 1 + z 2 ) = [(x 1 + x 2 )y 1 + (x 1 + x 2 )y 2 ](z 1 + z 2 ) = [x 1 y 1 + x 2 y 1 + x 1 y 2 + x 2 y 2 ](z 1 + z 2 ) = x 1 y 1 z 1 + x 2 y 1 z 1 + x 1 y 2 z 1 + x 2 y 2 z 1 + x 1 y 1 z 2 + x 2 y 1 z 2 + x 1 y 2 z 2 + x 2 y 2 z 2

Each term has a member of first second and third brackets. Therefore, expansion is a choice of one member from each bracket without repeating a choice.

1.4 Lesson 4: Binomial Coefficients 1 UNIT 1 : COUNTING TECHNIQUES

For instance, the coefficient of x^3 y^2 in the expansion of (x + 5)^5 will be 5 C 2 ≡ 5 C 2 = 10. Thus, (x + 5)^5 is a product of five brackets.

(x + y)^5 = (x + y)(x + y)(x + y)(x + y)(x + y) y^2 ⇒ y’s came from choices of y from 2 out of 5 brackets x^3 ⇒ x’s came from choices of x from 3 out of 5 brackets ∴ Number of ways of getting x^3 y^2 is 5 C 2 ≡ 5 C 2 = 10

In other words (x + 5)^5 has 5 C 2 x^3 y^2 = 10x^3 y^2 as one term and x^3 y^2 has a coefficient of 10.

In College Algebra, you learnt the Binomial Expansion as the expansion of the form (x+y)n^ = nC 0 x (^0) yn (^) +n (^) C 1 x (^1) yn− (^1) +n (^) C 2 x (^2) yn− (^2) + · · · + nCnxny (^0) from which the following formula is derived.

Theorem 1.6 Binomial Theorem

(x + y)n^ =

∑^ n p=

nCpxpyn−p

Notation

The notation nCp ≡

(n p

(x + y)n^ =

∑^ n p=

n p

xpyn−p

To find the coefficient of the expansion, it does not matter the letter you use.

Example 1.

Find the coefficient of x^3 y^2 in the expansion of (2x + y^1 /^2 )^10

Solution

We would observe that y^2 = y^1 /^2 × y^1 /^2 × y^1 /^2 × y^1 /^2 comes from 4 brackets. 2 x must have been chosen from 6 brackets to give x^3. Therefore, the corresponding term in the expansion is 10 C 6 (2x)^6 (y^1 /^2 )^4 = 210(2x)^6 (y^1 /^2 )^4. Therefore, the coefficient of x^3 y^2 in the expansion of (2x + y^1 /^2 )^10 is 210.

Activity 1.

Q.1 The term and its coefficient that contains x^5 in the expansion of

2 + x)^9.

Q.2 Find the coefficient of the term that contains x^8 y^18 in the expansion of (3x + 2y^2 )^17.

Q.3 The middle term in the expansion of (2x − 3 y)^50. What is its coefficient?

Q.4 Find the coefficient of x^10 in the expansion of (2 − x^25 )^200.

1.5 Lesson 5: Multinomial Theorem 1 UNIT 1 : COUNTING TECHNIQUES

Alternatively, regard the two splitters as pencils, to have 10 + 2 = 12 pencils and choose any two pencils to be splitters. Each choice of 2 pencils from 12 gives a unique solution to x + y + z. The total number of non-negative integer solutions is 12 C 2 = (^) 2!10!12! = 66

c) How many non-negative integer solutions has the equation w + x + y + z = 10 got?

Solution

The solution to this follows the reasoning applied in (b) above. Just add three spitters to 10 pencils and consider them as pencils altogether. Thus, we will have 13 C 3 = (^) 3!10!13! = 286

The last two parts in Example1.8 gives us the generalization stated in this Theorem about non-negative integer solutions.

Theorem 1.8 Non-negative Integer Solutions

i. Number of non-negative integer solutions to x 1 + x 2 + x 3 + · · · + xp is n+p−^1 Cp− 1 ii. Number of ways of placing n− indistinguishable items in k boxes is n+k−^1 Ck− 1

Example 1.

How many non-negative integer solutions has the equation w + x 2 + x 3 + x 4 = 50 got if w ≥ 2?

Solution

If we let w = 1, x 2 = 1, x 3 = 1 and x 4 = 48, then it is not a solution since w 6 ≥ 2. Thus, if w = 2 + x 1 , where x 1 ∈ { 0 } ∪ N will be a solution.Therefore,

w + x 2 + x 3 + x 4 = 50 ⇒ 2 + x 1 + x 2 + x 3 + x 4 = 50 ⇒ x 1 + x 2 + x 3 + x 4 = 50 − 2 ⇒ x ︸ 1 + x (^2) ︷︷+ x 3 + x (^4) ︸ non-negative integers

Therefore, applying Theorem1.8, the solution is 48+3C 3 = 53 C 3 = 20825

Activity 1.

Q.1 Calculate the coefficient of x^2 c^5 d^3 in the expansions of:

i. (x^2 − d + c)^10 ii. (x^2 + d − c + 2)^12 iii. (2 + xd^1.^5 + c^2.^5 )^10

Q.2 Find the total number of integer solutions of u + v + w + x = 40.

Q.3 Find the total number of integer solutions of u + v + w + x = 40 if u ≥ 15 or x ≥ 10

1.6 Lesson 6: Sum Rule 1 UNIT 1 : COUNTING TECHNIQUES

1.6 Lesson 6: Sum Rule

Suppose Box 1 has 10 Mangoes and Box 2 has 15 Oranges. A person is offered to pick 2 fruits of the same type. How many ways can he/she do that? It is true that Mangoes and Oranges are altogether different fruits. Thus, there is no intermediate fruit between them. Let x be number of all possible ways of picking a pair of Mangoes from Box 1 and y be number of possible ways of picking a pair of Oranges from Box 2. Then the total number of ways of picking a pair of fruits is x + y = 10 C 2 +^15 C 2 = 150. In general, the following theorem applies:

Theorem 1.9 Sum Rule

Let P and Q be two disjoint sets.

Then n(P ∪ Q) = n(P ) + n(Q). Example 1.

a) Suppose there are 10 football coaches and 7 netball coaches participating in certificate of fitness test under AFCON trials at Bingu International Stadium. If you are part of the trainers and are asked to choose the best three coaches, any coach can be chosen in 10 C 3 + (^7) C 3 = 155 ways.

b) If a bookshelf in the departmental library has 5 calculus texts, 3 algebra texts, 6 differential equations texts and 4 statistics texts. Find the number of ways a student can select one of the texts. Here again, using the sum rule, a text book can be chosen in 5 + 3 + 6 + 4 + 9 = 18 ways.

Nevertheless, using same principals, the rule can be extended to three or more disjoint events as shown in Example1.10(b). Theorem 1.10 nCr = n−^1 Cr− 1 (Pascal’s Triangle )

The proof to the theorem is simple. Let x be an item among the n available items. To choose r items from n, we have to either i. Include x among the r− chosen items OR

ii. Exclude x among the r−chosen items.