Korey Nishimoto. Nature of Mathematics

Transcription

1 Korey Nishimoto Nature of Mathematics

2 Math 100 Nature of Mathematics, Smith 13th ed Notes Korey Nishimoto Math Department, Kapiolani Community College August 13, 2018

3 Page 3 Contents Contents Nature of Logic Deductive Reasoning Conjunction and Disjunction Negation Homework ruth ables and the Conditional he Conditional Converse, Inverse, and Contrapositive ruth ables Homework Operators and Laws of Logic he Bi-Conditional Operations (De Morgan s Law) Homework he Nature of Proof Direct Proof Proof By Contrapositive Proof By Contradiction Homework he Nature of Sets Sets, Subsets, and Venn Diagrams How to Write Sets Venn Diagrams and the Universal and Empty Set Subsets and Proper Subsets How many Subsets are there? vs { } Homework Operations with Sets Intersections and Unions Cardinality est Your Knowledge Homework Application of Sets Given Numerals and a Picture Draw Venn Diagrams with Appropriate Data Homework Infinite and inite Sets Countable and Uncountable Homework he Nature of Counting Permutations and Counting undamental Counting Principle Permutations Distinguishable Permutations Homework Combinations Combinations Vs Permutations Binomial heorem Counting low Chart Homework Counting Without Counting Random Counting Problems 66

4 Page 4 Contents Homework he Nature of Probability Introduction to Probability heoretical Probability Empirical Probability Homework Mathematical Expectations Expected Value Homework Probability Models Complement Probability Conditional Probabilities Homework Calculating Probabilities Probabilities Using Counting Principles Or Probabilities And Probabilities General Probability Instructions Homework he Nature of Statistics requency Distributions Stem and Leaf Plot Bar and Line Graph Pie Graph Combine it All Homework Descriptive Statistics Measures of Central endency Box Plots Homework he Normal Distribution Determining And Using he Normal Distribution Standard Scores and Percentiles Normal Distribution Instructions Homework he Nature of inancial Mathematics Interest Growth and Decay actor ypes of Interest Different Uses of Interest Homework Month Payments, Loans and Savings Monthly Payments or Savings Monthly Payments or Loans Homework inancial Project

5 Nature of Logic Page 5 3 Nature of Logic 3.1 Deductive Reasoning Definition 3.1.1: A statement is a declarative sentence that is either true or false. It cannot be both true and false. Example: 3.1.1: Some examples of statements and sentences that are not statements. 1) I have a car. 2) I hate you! 3) Hawaii is located on the moon. 4) How are you today? 5) Chickens and cats are animals. 6) his statement is false. Notice number 5 in example uses the word and. We will now discuss the connecting words of two or more statements Conjunction and Disjunction Conjunctions and disjunctions are a very important part of logic. hey allow us to create more common and useful statements. We will first discuss the compound statement conjunction, then disjunction. Definition 3.1.2: he word and is called a conjunction and is used to combine statements. Let p and q be statements, then p and q is written as p q. Note: Remembering truth values is difficult if you try to remember the entire truth table. Instead remember the special case for each type of statement discussed and you will have less to remember. he truth table below gives us an understanding of when this compound statement is true () or false (). p q p q Notice that the only time that the compound conjunction is true is when both the left and right hand sides are true. It is easier to remember this rule than each line separately. All you need to do is check if both sides are true. If it is not, the entire conjunctions is automatically false. We will see some examples further down in example Example: 3.1.2: Your parents tell you they will give you a car and $1,000,000 if you get straight A s. You get straight A s. If your parents only give you a car, or only $1,000,000, did they lie to you? he answer is yes. Regardless of how happy you are to get a car or $1,000,000 they still lied. he only time that they would not have lied is if they had given you both. It is also clear that they lied if they gave you neither. Definition 3.1.3: he word or is called a disjunction and is used to combine statements. Let p and q be statements, then p or q is written p q. Similarly the truth table will help us understand disjunctions.

6 Nature of Logic In each statement we should determine the type of statement which is made. or example number 7, (p r) (p q) is an or statement or a disjunction. he main connector is highlighted in red. What is the main connector for 8)? p q Page 6 p q Notice in this case the only way a disjunction is false is when both the left and right hand side are false. Same as example 3.1.2, but with statement, If you get straight A s they will give you a car or $1,000,000. Now the only time that your parents would have lied to you is if they gave you neither. One thing to note, in mathematical logic, or is inclusive. his means that we may have one the other or both, to make the statement true. Because of this it is easiest to remember when disjunctions are false. You only need to check to see if both mini statements are false. If they are not then the statement is automatically true. Example: 3.1.3: Determine the truth value of the given statements. Assume that p and r are true, while q is false. 1) p q 2) p q 3) r (p q) ( ) 4) r p 5) (p q) r ( ) 6) p q r 7) (p r) (p q) ( ) ( ) 8) (p r) (q (r q)) ( ) ( ( )) ( ) Negation We commonly use words like; not, don t, false, and not true. hese are all words that give us a negation. Consider the statement I like to eat apples, the negation could be I do not like to eat apples or It is not true that I like to eat apples. he negation of statements including compound statements using disjunctions and conjunctions is very common in our everyday speech. In example 3.1.4, 3 we keep the statement p q and negate the entire statement instead of the common error of negating each part separately. Definition 3.1.4: he negation of a statement changes the truth value of that statement. It is written as. It should be clear that the negation of a negation gives back the original. his is because the truth value switches two times. or. Example: 3.1.4: Let p represent the statement Nursing informatics is a growing field and let q represent Critical care will always be in demand. ranslate the following. 1. p q Nursing informatics is not a growing field and critical care will always be in demand.

7 Nature of Logic Page 7 2. (p q) It is not the case (It is not true) that nursing informatics is a growing field or critical care will always be in demand. 3. (p q) It is not the case (It is not true) that nursing informatics is a growing field and critical care will always be in demand. Notice that in each case we work similar to the acronym PEMDAS that you learned previously. his time there are no exponents, multiplication, division, addition, or subtraction. However you do work from the inside of the parenthesis out, by evaluating each step one at a time. Now that we have seen an example of how to convert the english statements into its negations, lets consider the symbolic version. We want to be able to take a compound statement where we know the truth value of each mini-statement, and apply the negation in the correct order. his is similar to the example 3.1.4, except we will plug in and, then apply the negations as necessary. Example: 3.1.5: Let p represent a false statement and let q represent a true statement. ind the truth value of the given compound statement. 1) ( p) ( ) 2) (p q) ( ) 3) (p q) ( ) ( ) 4) ( p q) ( ) ( ) 5) [ p ( q p)] [ ( )] [ ( )] [ ] 6) [( p q) q] [( ) ] [( ) ] [ ]

8 Page Homework 3.1 Problem 1. Answer the following questions. 1) What is the special case for an or statement? What is every other case? 2) What is the special case for an and statement? What is every other case? 3) What is a negation? 4) What is a statement? 5) Write the truth table for a conjunction. 6) Write the truth table for a disjunction. Problem 2. Answer the following questions. Justify your answer. 1) If q is false, what must be the truth value of the statement (p q) q? 2) If q is true, what must be the truth value of the statement q (q p)?

9 Page 9 Problem 3. Let p represent a false statement and let q represent a true statement. ind the truth value of the given compound statement. 1) (p q) 2) ( p q) 3) [ p ( q p)] 4) [( p q) q] Problem 4. Let p represent a true statement, and let q and r represent false statements. ind the truth value of the given compound statements. 1) [ q (r p)] 2) (p q) (p q) Problem 5. rue or alse. Give an explanation. 1) A statement p q is only false when both p and q are false. 2) A statement p q is only true when both p and q are true. 3) he negation of a statement p is always false. 4) A double negation will always give you the original statement back.

10 Nature of Logic Page ruth ables and the Conditional We have already seen some truth tables in the previous section and we know that truth tables help us understand the truth value of a given compound statement given any combination of truth values of the mini statements. In this section we will learn how to construct these truth tables along with a new connecting word he Conditional Let us first discuss the new connecting logical symbol so that we may use it in our construction of truth tables. Definition 3.2.1: An if-then statement is called a conditional statement. Let p and q be statements. hen the conditional if p, then q is written as p q. he first statement which is attached to the if is called the antecedent / hypothesis, and the statement attached to the then or after is called the consequence / conclusion. Example: 3.2.1: Some examples of the conditional. 1. If Hawaii wasn t so expensive, the island would be completely full. 2. If I don t exercise, then I will gain weight. he conditional is a little weird with respects to the truth value. I will explain after the truth table below. p q p q Notice that the only time that a conditional statement is false is when. Many people think that if the first statement is false, then the conditional statement must also be false. his is the example I like to think about that helped me understand. Example: 3.2.2: If it rains today, then the road will be wet. Let p be it rains today, and q be the road will be wet. Here are the cases. 1) It rained today so the road is wet. 2) It rained today but the road is not wet. 3) It didn t rain today but the road is wet. 4) It didn t rain today so the road is not wet. Which one of these doesn t make sense? I think it is pretty clear that 1 and 4 make sense. Most people will agree that 2 also makes sense. he debate usually happens in 3. he question is, it didn t rain so is it possible that the road is wet? Yes. Someone could have spilled water, snow could have melted, or a fire hydrant could have been broken. his statement can happen. You may also consider the statement, If you win the lottery, then you will have a million dollars. One thing to note, conditional statements do not give us cause and effect. An example of this may be If a unicorn is born, then I will have a new iphone or If the sun rises, then monkeys are mammals. Both of these statements are true. he first statement is true since the antecedent is false. his automatically makes the conditional true. he second

11 Nature of Logic Page 11 conditional is true since both the antecedent and the consequence are true. I think it should be clear that these two statements have to cause and effect relation ship. he conditional statement is one of the most commonly used statements and thus has some special versions Converse, Inverse, and Contrapositive Note: his procedure works for any conditional statement. If you follow the rules, then the converse, inverse, and contrapositive is easy to find. he converse of the contrapositive is the inverse. he converse says to take the left and the right of the contrapositive and flip them. So q p converts to q p which is the inverse. his section gives people a lot of trouble, and is actually used incorrectly a lot because people believe they are using it correctly. We will discuss this after we state the converse, inverse, and contrapositive. Definition 3.2.2: Given the conditional p q, we define: 1. Converse: q p he converse says to take the conditional s left hand side and put it on the right, then take the right hand side and put in on the left. his is the converse. 2. Inverse: p q he inverse says to take the conditional s left hand side and put negation on it, then take the right hand side and put a negation on it. his is the inverse. 3. Contrapositive: q p he Contrapositive says to take the conditional s left hand side and put it on the left, then negate it. hen take the right hand side and put in on the left and negate it. his is the contrapositive. Example: 3.2.3: Lets find the converse, inverse, and contrapositive statements of the following conditionals. 1. p (q p) (a) he converse says to flip the the left and the right hand sides of the original. (q p) p (b) he inverse says to negate both the left and the right hand side of the original. p (q p) (c) he contrapositive says to flip the left and the right hand sides of the original, then negate both of them. (q p) p 2. (p q) (p q) (a) he converse says to flip the left and the right hand sides of the original. (p q) (p q) (b) he inverse says to negate both the left and the right hand side of the original. (p q) (p q) (c) he contrapositive says to flip the left and the right hand sides of the original, then negate both of them. (p q) (p q) 3. (p q) (p q)

12 Nature of Logic Page 12 (a) he converse says to flip the left and the right hand sides of the original. (p q) (p q) (b) he inverse says to negate both the left and the right hand side of the original. (p q) (p q) (p q) (p q) (c) he contrapositive says to flip the left and the right hand sides of the original, then negate both of them. (p q) (p q) (p q) (p q) Which of these are the same? Let the class discuss with each other and use examples. Many people believe that the inverse is the same as the conditional. Lets reconsider the statement If it rains today then the road will be wet. Can we infer that If it does not rain, then the road will not be wet. In order to show which of these are the same we will need to construct a truth table ruth ables We have seen truth tables before, and we know that the truth table gives us the truth value of a given statement for all possible combinations of truth values of the mini-statements. Let s discuss the actual method to construct these tables. Instructions 1. Determine the amount of rows needed by calculating 2n, where n is the number of unique statements. If we have statements p and q, then there are 2 unique statements, and there will be 22 4 rows. Whereas if there are statements p, q, and r, then there will be 23 8 rows. 2. Draw the table with the given amount of rows, with columns for each mini statement and the possibly large compound statement. 3. ill in the truth value for the mini statements. he first column will have the top half and the bottom half. Now take the amount of and divide by 2. his is the amount of in the second column before alternating to, then alternating back to until the entire column is full. Continue this process until all mini statement s columns are full. 4. Now evaluate the truth table and circle the column of your final answer. Construct a truth table for the following. Example: 3.2.4: Construct a truth table for the following. 1. or the converse, inverse, and contrapositive of the conditional p q. p q p q q p p q

13 Nature of Logic q Page 13 p Notice that the conditional and the contrapositive have the same truth table, while the inverse and the converse have the same truth table. his means that the end value of the compound statement is the same, or the blues are the same. his tells us that the converse and inverse as well as the conditional and the contrapositive are equivalent statements. his is the only way to determine if two statements are equivalents. You cannot determine equivalence using 1 case of the truth table. All the different cases must be the same in order for the statements to be equivalent. 2. (p q) ( p q) p q (p q) ( p q) his is an example of a special compound statement called a autology. his is where the compound statement is true regardless of the truth value of the mini statements. Let s now combine everything we learned using conditional, disjunctions, and conjunctions to determine the truth value of the statement with given truth values. Example: 3.2.5: Determine the truth value of the statement when p is false, q and r is true 1) (p q) r ( ) 2) (p r) (r q) ( ) ( ) 3) (q p) (p r) ( ) ( ) 4) (p r) (q p) ( ) ( ) 5) (p q) (q r) ( ) ( ) 6) [(p q) r] (p [q r]) [( ) ] ( [ ]) [ ] ( [ ]) ( )

14 Page Homework 3.2 Problem 1. rue or alse. Provide explanation. 1) Let p and q be statements and consider the conditional p q. he contrapositive of the converse statement is the inverse of p q. 2) Given that p is true and q is false, the conditional p q is true. 3) Given that p is false and q is false, the conditional p q is true. 4) ) If 3+1 7, then ) If , then Problem 2. ind the truth value of the given statement, given that p is true and q and r are false. 1) [p (p q)] ( q r) 2) [(p q) (p r)] (p r) 3) [(p r) (q r)] 4) [ (p r) (q p)] (q r)

15 Page 15 Problem 3. Write the converse, inverse, and contrapositive statements for the following conditional statements. 1) ) If 3+1 7, then ) (p q) (p q) 4) p q Problem 4. Give a real life example of a conditional statement not used in class. hen convert the condition statement you created to its converse, inverse and contrapositive statements. Your answer should be in English.

16 Page 16 Problem 5. Create a truth table for the following compound statements. Label in the following order, p, q, r. 1) ( q p) q 2) (p q) (p q) 3) [( p q) (p q)] 4) [ (p q) (q p)] [(r q) ( p r)]

17 Nature of Logic Page Operators and Laws of Logic Sometimes it is useful to be able to do operations in logic. We may want to simplify before finding the truth value. Simplify does exactly what is sounds like, it simplifies the expression. In this section we will discuss the bi-conditional as well as operations in logic he Bi-Conditional We will first discuss the related logical statement the bi-conditional. his is exactly as it sounds 2-conditionals. Definition 3.3.1: An if and only if statement, also written as iff, is called a bi-conditional statement. Let p and q be statements. hen the bi-conditional is written p q. It is a little easier to understand this as (p q) (q p). Meaning, both must imply each other. Lets think about this for a second. he statements must imply each other, and conditional statements are false only when. So we get truth table p p q q In this section p. 98, the book says that there is a difference between and as well as and. hey say that the single lined arrow is used of the logic symbol while the double lined arrow is used for logical equivalence. or us, we will always use and, and if we want to say logically equivalent, then we must first show logical equivalence using a truth table, and then just simply state they are logically equivalent. Lets do this using a rule of logic called De Morgan s law Operations (De Morgan s Law) heorem 3.3.1: De Morgan s law gives us a way to algebraically manipulate the negation of a conjunction or disjunction. (p q) p q (p q) p q We will prove the first part of De Morgan s law p q (p q) p q he truth tables are the same, hence (p q) p q Next lets discuss the negation of conditional and bi-conditional statements. heorem 3.3.2: he negation of a conditional and bi-conditional are given as, (p q) p q,

18 Nature of Logic Page 18 and (p q) [(p q) (q p)] (p q) (q p) (p q) (q p). Example: 3.3.1: Use the rules above to find the simplified negation of the following statements. 1) p (q r) [p (q r)] p (q r) p q r 2) (p q) (r p) [(p q) (r p)] (p q) (r p) (p q) (r p) 3) (p q) (r p) [(p q) (r p)] (p q) (r p) (p q) (r p) 4) (p r) (q [r p]) [(p r) (q [r p])] (p r) (q [r p]) ( p r) ( q [r p]) ( p r) ( q r p]) 5) p q (p q) [(p q) (q p)] (p q) (q p) (p q) (q p) 6) (p q) (r p) [(p q) (r p)] [(p q) (r p)] [(r p) (p q)] [(p q) ( r p)] [(r p) ( p q)]

19 Page Homework 3.3 Problem 1. ind the simplified negation of each statement. If applicable, use De Morgan s laws. 1) (p q) (p r) 2) (m n) (k j) 3) (p r) (p q) 4) [(p q) (q r)] (p q) Problem 2. Create the truth table for the following statements. 1) (p q) (q p) 2) [ (p q) ([p q] p)]

20 Page 20 Problem 3. rue or alse. Must provide explanation. 1) he negation of a conjunction is equivalent to a disjunction. 2) he negation of the contrapositive q p is the converse q p. 3) Let p and q be true. hen p q is true and p q is true. his means that they are logically equivalent. 4) he Bi-conditional statement (p q) ( p q) is a tautology.

21 Nature of Logic Page he Nature of Proof In this section we will be discussing different methods that are commonly used in mathematics to prove statements in mathematics. We will first begin with direct reasoning or direct proof Direct Proof his uses a straight reasoning of the conditional statement, without any modifications. We must first assume the antecedent (Why is this?) before we start. And use the antecedents information to try to conclude the concequence. Example: 3.4.1: Prove that the sum of 2 even numbers is even. (Did you know that every even number can be written as 2k, where k is an integer. Integers{... 3, 2, 1, 0, 1, 2, 3...}.) his can be rewritten as a conditional as such if we add two even numbers, then the sum is odd. Now first assume that the antecedent (we add two even numbers) is true. his is the same as 2a + 2b where a and b are integers. actor a 2 to get 2(a + b). Since a + b is an integer we are of the form 2 times an integer which makes the sum even Proof By Contrapositive he next type of proof is called indirect reasoning/contrapositive proof. his uses the contrapositive statement q p. We do this because sometimes the negation of q has more information than p itself. Lets do two examples like this. Example: 3.4.2: Prove the following using a contrapositive proof. 1. Prove that if you multiply two integers together and get an even number, then at least one must be even. Did you also know that if a number is not even then it is odd and is written as 2k+1 again where k is an integer. We want to prove the contrapositive and the contrapositive give us, If none of the numbers are even, then when you multiply them together the result is not even/odd. Now we know about even number so assume the antecedent of the contrapositive is true (None of the numbers are even). hen they must be of the form 2a + 1 and 2b + 1 where a and b are integers. Multiplying them together we get (2a + 1)(2b + 1) 4ab + 2a + 2b + 1 2(2ab + a + b) + 1. Similarly as in example 3.4.1, ab is an integer and adding three integers is an integer, hence we have 2 times an integer plus 1. his is odd or not even. 2. If x2 is even, then x is even. Contrapositive: If x is not even, then x2 is not even. Assume x is not even,

22 Nature of Logic Page 22 hence x 2a + 1 where a is an integer. his gives us x2 (2a + 1)2 4a2 + 4a + 1 2(2a2 + 2a) + 1. Similarly in ex part 1, this is of the form 2 times an integer plus 1, hence not even or odd. his works because p q q p, shown in 3.2, example Proof By Contradiction Lastly we will discuss the negation/proof by contradiction. his is not in the book, but it is one of the fundamental proof methods in math. he negation of p q is p q. If we show that p q is false, then what must p q be? It is true because the negation is false. If we negate a true statement then we get a false statement. Our goal is to try to find a contradiction. Example: 3.4.3: Prove by contradiction. 1. Let n be an integer. If n2 is odd, then n is odd. he negation give us n2 is odd and n is even. We can assume both of these to come up with a contradiction. Since n is even n 2a where a is an integer. Hence n2 (2a)2 4a2 2(2a2 ). his is of the form of 2 times an integer which makes this even. However we first assumed that n2 is odd. hese two conflict and give us a contradiction. 2. Prove if a and b are integers, then a2 4b 6 2. he negation gives us a and b are integers and a2 4b 2. Notice if we bring b to the right hand side we get, a b 2(1 + 2b). his again is the form of an even number so a2 is even. We proved in example part 2, that if a2 is even then so is a. hus a 2c where c is an integer. Substituting back we get (2c)2 4b 2. We get 4c2 4b 2 which gives us 4(c2 b) 2. Now divide both sides by to to get 2(c2 b) 1. Notice the left hand side is of the form 2 times an integer, which makes it even, but 1 is not even. his is a contradiction.

23 Page Homework 3.4 Problem 1. Prove by contradiction. 1. Let n be an integer. If n is odd, then n 2 is odd. 2. Let n be an integer. If n is even, then n 2 is even. Problem 2. Give a contrapositive proof. 1. Let n be an integer. If 3n + 1 is even, then n is odd.

24 Page 24 bonus Let m and n be integers. If n 2 + m 2 is even, then n + m is even. Problem 3. Give a direct proof for the following. 1. Prove that the sum of two odd numbers is even. 2. Prove that the sum of three odd numbers is odd.

25 he Nature of Sets Page 25 2 he Nature of Sets 2.1 Sets, Subsets, and Venn Diagrams Note: When learning sets, the key is to remember the definition and how to read them. he reason why people cannot do math is because they cannot read math. Practice reading the symbols in a way that you understand it. If you understand it a different way than is being taugh, discuss it with you teacher to ensure that it is correct. In this section, we will have a brief introduction to a mathematics called set theory. Definition 2.1.1: A set is a collection of distinct objects called elements. Definition 2.1.2: he complement of a set is the set of all elements not in the set. his is denoted a A0 where A is the set and A0 (your book uses A ) is the complement of set A. When writing a set, mathematicians have certain conventions. or example, a set is always denoted by a capital letter, an element is usually denoted by a lower case letter, and there is a difference between and, which we will discuss latter How to Write Sets irst lets discuss some sets using the different methods of writing sets. 1. Listing method (Your book calls this roster method): Example: 2.1.1: Let try to construct the set of integers between and including 5 and 10 (x Z and 5 x 10). Can we see this includes numbers 5,6,7,8,9, and 10? o write this in set notation all we have to do is give the set a name, call the set of integers between and including 5 and 10 A. hen write A {5, 6, 7, 8, 9, 10}. he listing method is usually used when we are writing either finite amounts of elements, or there is a pattern to the infinite number of elements. Example: 2.1.2: x 10. Let B be the numbers larger or equal to 10. B {10, 11, 12,...}. Lets look back at A. What are the elements of A? hey are 5,6,7,8,9, and 10. o write this symbolically, we write 5 A. his is read as 5 is in set A. his is a very common notation and we have already used it once to describe the integers. Here are some more common sets in math. (a) he set of integersz {..., 3, 2, 1, 0, 1, 2, 3,...} (b) he set of natural numbersn {1, 2, 3, 4, 5,...} (c) he set of whole numbersw {0, 1, 2, 3, 4, 5,...} (d) he set of rational numbersq { ab a, b Z and b 6 0} (e) he set of irrational numbersq0 {x x 6 Q} (f) he set of real numbersr {x x is on a number line}. Notice that d, e, and f use a different notation which we will now discuss. 2. Set Builder Notation:

26 he Nature of Sets Page 26 Example: 2.1.3: Let is write A above in set builder notation. A {x x Z and 5 x 10}. his reads A is equal to the set of x such that x is an integer and is between or including 5 and 10. his type of set is usually used when the set is to cumbersome to write as a list. he difficulty of this type of set is being able to read the symbols. Math is a language and you must be able to read it. wo sets are equal when they have the same elements. Order does not matter. his means A {1, 2, 3} and B {3, 1, 2} are the same and A B Venn Diagrams and the Universal and Empty Set Sets are difficult to understand without a picture. he representation of a set as a picture is called a Venn Diagram. Before we can draw these we need two more definition. Definition 2.1.3: he universal set is the set of all possible elements in the given situation. It is denoted as U. 1. or the common sets above U R. 2. We did an experiment testing a new drug. Does it make sense that the universal set is all people? No. It is all people in the experiment. 3. hree people in this class got A s on the exam. What is the Universal set? All the people in this class. Note: he empty set is not the same as { }. he empty set is a set which contains nothing (No Elements). However the set { } has one element in it, namely the empty set. Remember to read the set notation {} as the set containing. Definition 2.1.4: he empty set is the set that contains no elements. It is denoted as. his is why you cannot write 0 as or {}. Example: 2.1.4: Let us draw a Venn Diagram for the previous set A {5, 6, 7, 8, 9, 10} UR 5,6,7,8,9,10 his tells us that the blue circle only has the elements 5,6,7,8,9, and 10 in it. while the inside the box but outside of the blue circle has the rest of R. One thing to note, Venn diagrams do not need to be circles. Most people use circles as a convention, but it is not necessary. You may have used the shape of a horse to represent set A, and the answer would still be correct provided that 5, 6, 7, 8, 9, and 10 are in A and you label U.

27 he Nature of Sets Page Subsets and Proper Subsets he last thing to discuss in this section is subsets and proper subsets. Definition 2.1.5: A subset B of A is define as such. Every element in B is also in A. his is written as B A and is read B is a subset of A. Definition 2.1.6: A proper subset B of A is define as such. Every element in B is also in A, but A 6 B. We will use the symbol and is read as in. his is written as B A and is read B is a proper subset of A. You can think of this as B is inside of A. Definition 2.1.7: wo sets A and B are disjoint if they share no elements in common. hat is for every x A, x 6 B, and for every x B, x 6 A. he only difference between subset and proper subset is that subset allows the possibility of equality where as proper subset does not. When in doubt subset will always work assuming that B is contained in A. Notice that in definition we use. his is not the same as. We have shown before that two sets are equal when they have the exact same elements. x A, says that the number x is an element of A, not that x and A are the same. Let us now draw some Venn diagrams which show the different cases that can happen. 1. B and A have nothing in common or are disjoint. U A 2. B is a proper subset of A or, B A. B

28 Page 28 he Nature of Sets U A B 3. A B, which can also be written as A B and B A. U B A One thing to note is the definition of the empty set and subset. he empty set is the set which contains no elements, while the subset definition states a set is a subset of a second set if all of its elements are contained in that second set. What can we sat about the empty set, and its relationship to the subset definition? he empty set is a subset of any set. All its elements (which are none) are contained in any other possible set. Going back to the common math sets. Which of these sets are subsets of each other? Give me some examples.

29 he Nature of Sets Page 29 Irrationals Real numbers Rational numbers Integers Whole numbers Natural numbers How many Subsets are there? Let me pose a question. How many subsets are there of the set A {5, 6, 7, 8, 9, 10}? here happen to be 64 subsets in total. Lets demonstrate this using a couple slightly smaller sets. Example: 2.1.5: How many subsets are there for the following sets? 1) A {1, 2, 3} 2) A {cat, dog} {cat}, {dog}, {cat, dog}, {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3}, here are 4 subsets in total. here are 8 subsets in total. 3) here is 1 subset. 4) A {2, 4, 6, 8} {2},{4},{6},{8}, {2,4},{2,6},{2,8},{4,6},{4,8},{6,8}, {2,4,6},{2,4,8},{2,6,8},{4,6,8}, {2,4,6,8}, here are 16 subsets in total. Can we see a pattern? Notice that we originally started with 1 subset when the set was the empty set. hen we moved to 4 when the set had 2 elements, 8 when the set had 3 elements, and lastly 16 when the set had 4 elements. What is the pattern here? heorem 2.1.1: he total number of subsets is 2n where n is the number of elements in the set. his will help us know if there is a missing subset. Many times we forget the two simplest ones, the empty set and the entire set.

30 he Nature of Sets Page 30 Based on this how many proper subsets do we have? Proper subset is a set contained in another, but is not equal to it. So we have 2 n 1 proper subsets. here is only one subset which is equal vs { } An interesting thing to consider is the difference between and { }. What are the elements of and what are the elements of { }? he empty set, or contains no element, while { } contains one element, namely. his is a little confusing, but { } is a set which contains a set as its elements. You can think of this as a group of groups. or examples A { Dogs, Cats, Mammals, Bugs, ish}. Each of these categories has a lot of different elements. or example; golden retriever, labradors, chihuahua, pit-bull, and shiba inu, are all different types of dogs. his makes A a set of sets.

31 Page Homework 2.1 Problem 1. Insert or in each blank so that the resulting statement is true. 1. {2, 3, 100, 55} {x x is an even integer.} 2. {7, 33, 1 2 } Q 3. {x x is an odd letter in the alphabet.} {x x a type of elephant.} 4. {Boston, Orlando, Honolulu} he set of states. Problem 2. Let U {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. ind the compliment of each set. 1) A {2, 4, 6, 8, 10} 2) A {10, 8, 6, 9, 7, 3, 5, 2, 4, 1} 3) A {1, 3, 4, 5, 8, 1, 4, 5} 4) A Problem 3. Draw one venn diagram with all of the following properties. U is the universal set, A B, C B but no elements of C are in A, and no elements of a set D is in A, B, or C.

32 Page 32 Problem 4. ind the number of subsets and the number of proper subsets of each set. or 1, also write all the different subsets. 1) {a, b, c, d} 2) he set containing the days of the week. 3) {x x is an odd integer between -4 and 6.} 4) {x x is a whole number less than 4} Problem 5. Draw one venn diagram with all of the following properties. U is the universal set, A B C, A D,and D B and D C Problem 6. rue or alse. Provide explanation. 1. Assume that A is not the universal set. rue alse A is never a subset of A. 2. If A B, then A B. rue alse 3. If A B and B C, then A C. rue alse

33 Page 33 Problem Give two real life examples not done in class of sets A and B so that A B. 2. Draw a Venn Diagram for both. 3. Justify in words that the examples you created have the property A B.

34 he Nature of Sets Page Operations with Sets In this section we will discuss different types of operations involving sets. his includes intersections (and), union (or), and cardinality (count). We have already discussed and as well as or in the logic section. his is the exact same thing Intersections and Unions Note: You can read A B as the elements or things, in both the sets A and B. As a picture, this reads as the overlap of the sets A and B. irst lets discuss intersections. Definition 2.2.1: he intersection of two sets A and B is the set of elements that are in both A and B. It is written as A B. he Venn diagram is draw as U A Note: You can read A B as the elements or things, in set A, set B, or in both A and B (In one, the other, or both). As a picture, this reads as the circle A, circle B, or the overlap of A and B. A B B Definition 2.2.2: he union of two sets A and B is the set of elements that are in either A or B. Remember, or is inclusive. his means one the other or both. It is written as A B. he Venn diagram is draw as U A A B B A Bblue or the union, notice that A B includes A, B, and A B. his is the same as saying in A, B, or in both. he union and intersection are similar to the logic version. We had a rule for negations of and and or statements called De Morgan s law. his still holds for sets. heorem 2.2.1: Let A and B be sets. We have the following properties. (A B)0 A0 B 0 and (A B)0 A0 B 0.

35 he Nature of Sets Page 35 Example: 2.2.1: ind the following sets. 1) {x, y, z, a, b, c} {xy, yz, ab, bc} his is empty or the. Just because there is an x in both does not mean that x is an element of both. Look carefully in the second set. he element is actually xy not x. 3) { cat, dog, chicken} { dog, ant, bird} A B {cat, dog, chicken, ant, bird} Don t forget two things. One we want the elements in one, the other, or both, and two, writing elements twice does nothing, so we do not need to write dog again. 2) { 2, 3, 5} { 2, 2, 1 5} hey both share 2 and 5. Hence A B { 2, 5} 4) (Q Q0 )0 (Q Q0 )0 Q0 Q00 Q0 Q his reads as in the rationals, in the irrationals, or in both. What is every rational number combined with every irrational number? his is the entire number line. his A B R Cardinality he last thing that we need to discuss is cardinality. Definition 2.2.3: Cardinality is the number of elements in the set. or example let A {2, 4, 6}. he cardinality of A of n(a) 3. Your book uses A, but I think that this is confusing since most people know as the absolute value sign. his, n(a), reads as n of A, can be thought of, n, or number of elements of A. heorem 2.2.2: he cardinality of the union of two sets A and B is n(a B) n(a) + n(b) n(a B). If A and B are disjoint, then n(a B) n(a) + n(b). his is because n(a B), hence n(a B) 0. We get this formula because of the way we count sets. hink about this in terms of a Venn Diagram. U A A B B

36 he Nature of Sets Page 36 Why is A B maroon colored? Its because the blue part and the red part overlap each other. You can think of the shading as the counting. hat is n(a)blue shade and n(b)red shade. If this is the case, notice we have counted A B twice. Once with the ble shade and once with the red shade. his is the reason why we need to take away (subtract) the overlap A B once. Lets do the following examples where the sets are given and we must find the cardinality of each of the following. Example: 2.2.2: Let A {x 0 x 10 and x Z}, B {2, 4, 6, 8}, and C {3, 9, 27}. Solve for the following. 1) n(a) 11 2) n(b) 4 3) n(c) 3 4) n(a B) n(a) + n(b) n(a B) Since B is a subset of A, n(a B) n(a). 5) n(a C) n(a) + n(c) n(a C) ) n(c B) n(c) + n(b) n(c B) ) n(a B) n(a) + n(b) n(a B) his makes sense as well since B A, A B B 8) n(b C) n(b) + n(c) n(b C) his makes sense because B and C are disjoint. hus their intersection is empty. In this example we wil be given the cardinality of certain sets. We must find the cardinality of the following. Example: 2.2.3: Let n(a) 17, n(b) 13, n(c) 12, n(a B) 10, and n(b C) 15. ind the following. 1) n(a B) n(a) + n(b) n(a B) ) n(b C) n(b) + n(c) n(b C) est Your Knowledge Example: 2.2.4: Are the following sometimes true, always true, or false? 1. A B A B A B is some times a subset of A B. his happens when A B. We then have A BA A A and A B A A A. hus A A. 2. A B A B

37 he Nature of Sets Page 37 his is always true by definition of A B. A B is the set of all elements in A, B, or both (A B). 3. n(a B) n(b A). We know that unions and intersections are commutative. hat is A B B A. Hence n(a B) n(b A). Now that we have a understanding of cardinality, lets draw a Venn diagram. Example: 2.2.5: Draw a Venn diagram with the following properties. A B C 6, A B, and A 6 C. Also shade the following. (A B) C 0. U B A C

38 Page Homework 2.2 Problem 1. Match each term in Group 1 to the appropriate description in group ) he complement of A. A. he set of elements that are in A or in B or in both A and B. 2) he union of A and B. B. he set of elements common to both A and B. 3) he intersection of A and B. C. he set of elements in the universal set that are not in A. Problem 2. Use the listing method to describe the following sets. U {a, b, c, d, e, f, g}, X {a, c, e, g}, Y {a, b, c}, Z {b, c, d, e, f}. 1) X Y 2) Y Z 3) X Y 4) X (Y Z) 5) Y (X Z) 6) (Y Z ) X 7) (X Y ) Z 8) (Z X ) Y 9) (Y X ) Z 10) (X Y ) (Y X )

39 Page 39 Problem 3. Let A and B be two sets. Is the following statements always true or sometimes true. Give an explanation. 1) A (A B) 2) A (A B) 3) (A B) A 4) (A B) A 5) n(a B) n(a) + n(b) 6) n(a B) n(a) + n(b) n(a B) Problem 4. Draw one venn diagram that represents the following operations. A B, A C and C B. Shade in (A B ) C.

40 Page 40 Problem 5. Give an example of two real life sets, not used in class, A and B such that A B and A B and B A. (Read as not equal to. Anytime there is a line through a symbol, read it as not a. What is?) 1. Draw a Venn diagram and label. 2. Justify that your example represents the properties asked in the question. 3. What is A B as a sentence using your real life examples.

41 he Nature of Sets Page Application of Sets Application of sets is a common idea which we have used everyday for almost all our lives. We discuss things like how many people work full time and go to school, or how many native Hawaiian students are in a S..E.M degree. It is easiest to jump straight into this topic with examples Given Numerals and a Picture Example: 2.3.1: Use the numerals representing cardinalities in the Venn diagrams to give the cardinality of each set specified. U 8 A 0 5 B 2 1) A B n(a B)5 2) A B n(a B)7 3) A B 0 n(a B 0 )0 4) A0 B n(a0 B)2 5) A0 B 0 n(a0 B 0 )8 Now lets try an example which has three sets instead of two. he general idea of how to find the cardinality of sets is still the same. Example: 2.3.2: Use the given data to find the cardinality of the given sets. U 8 3 A C 2 B 2 0

42 he Nature of Sets 1) A B C n(a B C) 1 2) A B C 0 n(a B C 0 ) 3 3) A B 0 C n(a B 0 C) 4 4) A0 B C n(a0 B C) 0 5) A0 B 0 C n(a0 B 0 C) 2 6) A B 0 C 0 n(a B 0 C 0 ) 8 7) A0 B C 0 n(a0 B C 0 ) 2 8) A0 B 0 C 0 n(a0 B 0 C 0 ) 8 Page 42 As previously stated, the most important part of this is the ability to read the problem. or example, in example part 3, we want to to the number of elements that are in A and C, but outside of B. his is the most difficult thing to do in math, but the most important Draw Venn Diagrams with Appropriate Data We originaly started with the Venn diagram and evaluated the cardinality of each given set. Now we will go backwards. We will be given the cardinality of each set, then draw the Venn Diagram. Example: 2.3.3: Draw a Venn diagram and use the given information to fill in the number of elements in each region. n(a) 57, n(a B) 35, n(a B) 81, n(a B C) 15, n(a C) 21, n(b C) 25, n(c) 49, n(b 0 ) 52 U A B C 18 Example: 2.3.4: ind the value of n(b) if n(a) 20, n(a B) 6 and n(a B) 30. Now draw a Venn diagram for this data. n(a B) n(a) + n(b) n(a B) n(b) n(a B) n(a) + n(a B), so n(b)

43 he Nature of Sets Page 43 U A B Example: 2.3.5: A middle school counselor, attempting to correlate school performance with leisure interests, found the following information for a group of students. 1) 34 had seen Despicable Me 2) 29 had seen Epic 3) 26 had seen urbo 4) 16 had seen Despicable Me and Epic 5) 12 had seen Despicable Me and urbo 6) 10 had seen Epic and urbo 7) 4 had seen all three of these films 8) 5 had seen none of the three films 1. How many students have seen urbo only? irst we find how many people have seen all three. here are 4 people who saw all three and this is given by (7). We also know that n(e ) 10 so the number of people who have seen Epic and urbo but not Despicable me is 6. Do this again for Despicable me and Epic to get that 8 people have seen Despicable me and urbo, but not Epic. As a Venn diagram we have U D E Notice to get the number of people who have seen urbo, we add x 26 by (3). his tells us that the number of people who have seen urbo only is 8. We can also think of this as n(urbo only)n( ) n(d ) n(e ) + n( D E) How many have seen exactly two of the films? We already have solve two of the parts we need in 1). We only need to find the number of people who have seen Despicable me and Epic, but not urbo.

44 he Nature of Sets Page 44 Using the same strategy as in 1) we get In order to get the number of people who have seen exactly two films we need, n(despicable me and Epic only)+n(despicable me and urbo only)+n(urbo and Epic only) How many students were surveyed? So far we have U 12 D 8 4 E 6 8 We can solve for the number of people who have seen Despicable me only and urbo only the same way we solved for the number of people who saw urbo only in 1). his gives us n(despicable me only) and n(epic only) he only thing left is to calculate the number of people who saw none. But this is already given by (8). We add all these numbers up to get Draw a Venn diagram. U 5 12 D E Example: 2.3.6: A survey was conducted among 75 patients admitted to a hospital cardiac unit during a two-week period. Let B he set of patients with high blood pressure, C he set of patients with high cholesterol levels, and S he set of patients who some cigarettes. he survey produced the following data. 1) n(b) 47 2) n(b S) 33 3) n(c) 46 4) n(b C) 31 5) n(s) 52 6) n(b C S) 21

45 he Nature of Sets Page 45 7) n[(b C) (B S) (C S)] 51 ind the number of these patients who, 1. Had either high blood pressure or high cholesterol levels, but not both. By theorem 2.2.2, n(b C) n(b) + n(c) n(b C). However we don t want the number of people with high blood pressure or high cholesterol levels. We want n(b C But not both). his says to take away the people with both, hence n(b C But not both) n(b C) n(b C) n(b) + n(c) n(b C) n(b C) (31) Had fewer than two of the indications listed. his statement asks us to find the number of people that have only one or no indications. o make this problem easier, we must first find the only missing intersection n(c S). We will use 7 to do this. As a Venn diagram, 7) give us the following shaded area has 51 people in it. U B C S his is n[(b C) (B S) (C S)] n(b S)+n(B C)+n(S C) 2n(B S C) n(s C) 2(21). Hence 5122+n(S C) n(s C) 29. Using this and the method used in the second explanation of part a) in the example above, n(b only) n(b) n(b C) n(b S) + n(b S C) n(c only) n(c) n(c B) n(c S) + n(c B C) and n(s only) n(s) n(s C) n(s B) + n(s C C) Lastly we need to find the number of people with no indications. We can do this by the following. n(number of people with no indications) otal number of particpants [n(b) + n(c) + n(s) n(b S) n(c S) n(b C)] 75 [ ] 2. Adding these all up we get the number of people that had fewer than two of the indications listed is Were smokers but had neither high blood pressure nor high cholesterol levels. his question is asking for the number of people who were smokers only. his is given by n(s) n(b S) n(s C) + n(b C S) Did not have exactly two of the indications listed. his is easier if you solve for the number of people that have exactly two and then deduct that number from the total number of participants. n(b and C only) n(b C) n(b C S)

46 he Nature of Sets Page 46 n(b and S only) n(b S) n(b C S) n(s and C only) n(s C) n(b C S) Hence the number of people that did not have exactly two of the indications listed, is 75-( ) Draw a Venn diagram U 2 10 B S 11 C 7 8

47 Page Homework 2.3 Problem 1. At a southern university, half of the 48 mathematics majors were receiving federal financial aid. 1) 5 had Pell Grants 2) 14 participated in the College Work Study Program 3) 4 had OPS scholarships 4) 2 had OPS scholarships and participated in Work Study 5) hose with Pell grants had no other federal aid. Create a Venn diagram and answer how many of the 48 math majors had: 1) No federal aid? 2) More than one of these three forms of aid? 3) ederal aid other than these three forms? 4) A OPS scholarship or Work Study? 5) Exactly one of these three forms of aid. 6) No more than one of these three forms of aid?

48 Page 48 Problem U.S. adults were surveyed. Let A he set of respondents who believe in astrology, R he set of respondents who believe on reincarnation, and Y he set of respondents who believe in the spirituality of yoga. he survey revealed the following information: 1) n(a) 35 2) n(r) 36 3) n(y ) 32 4) n(a R) 19 5) n(r Y ) 8 6) n(a Y ) 10 7) n(a R Y ) 6 Create a Venn diagram and answer how many of the respondents believe in: 1) Astrology but not reincarnation? 2) At least one of these three things? 3) Reincarnation but neither of the others? 4) Exactly two of these three things? 5) None of the three?

49 Page 49 Problem 3. rue or alse. Give explanation. 1. [(A B) (B C)] (A B ) (B C ) 2. If A is disjoint from B and B is disjoint from C then A is disjoint from C. Problem 4. Use the numerals representing cardinalities in the Venn diagram to give the cardinality of each set specified. 1) A B C 2) A B C U 3 3) A B C 4) A B C A B 6 5) A B C 6) A B C C 5 7) A B C 8) A B C 9) A B C Problem 5. Draw a Venn diagram and use the given information to fill in the number of elements in each region. 1) n(a) 15 2) n(a B C) 5 3) n(a C) 13 4) n(a B ) 9 5) n(b C) 8 6) n(a B C ) 21 7) n(b C ) 3 8) n(b C) 32

50 he Nature of Sets Page Infinite and inite Sets We will only discuss this section briefly. here will be no solving problems from this section, but there may be problems that test your understanding of the material. his section is very interesting and worth discussing. Definition 2.4.1: A set A is said to be finite if n(a) x N of n(a) 0. Definition 2.4.2: A set is said to be infinite if it can be placed into a 1-1 correspondence with a proper subset of itself. his asks us to pair elements together. hat is one element in a set A is paired with one element from set B where B A. Notice that this definition eliminates the possibility that a finite set is infinite. his is because no finite set can be paired up 1-1 with a proper subset of its self. Example: 2.4.1: Let A {1, 2, 3}. Are there any proper subsets of A that have 1-1 correspondence with A? here are none. he proper subsets of A are {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3},. We know that there a total of proper subsets of A, which we have. Notice that all of these sets have cardinality less than n(a). Hence it is impossible to have 1-1 correspondence. Example: 2.4.2: Is the set N finite or infinite? We know that N {1, 2, 3, 4,...}. Consider the set of even natural numbers B {2, 4, 6, 8...}. Notice that the pairs can be given by the following N{ B{ 1, l 2, 2, l 4, 3, l 6, 4, l 8,...}...} his is a weird property. Even though every element of B is inside of N, we can still make a 1-1 correspondence because of the infinite amount of elements Countable and Uncountable heorem 2.4.1: If you can make a 1-1 correspondence with N the set is a special infinite, called countable, or countably infinite. his is denoted by ℵ0 (aleph-naught). Example: 2.4.3: Give some examples of infinite sets using the common math sets and determine if they are countable or not. Some examples are Z, N, Q, Q0, and R. We have already shown that N is countable. What about the rest? he integers is pretty easy to show, since it is exactly like the natural numbers. N{ Z{ 1, l 0, 2, l -1, 3, l 1, 4, l -2, 5, l 2,...}...} he tricky one is Q. What would happen if we made a 1-1 correspondence as such N{ { 1, l 1 1, 2, l 1 2, 3, l 1 3, 4, l 1 4, 5, l 1 5,...}...}

51 he Nature of Sets Page 51 Notice that we would never be able to change the numerator. his causes a problem, that can be easily be fixed with some creativity. We can solve this by write the set Q out and moving in a diagonal fashion. We do this as such, his allows us to move out of the first row. Each red line represents one element of the set N and gives us the 1-1 correspondence. What about Q0 and R? We won t show it, but they are not countable. However they are infinite. If they are not countably infinite, what could they be? hese sets are called uncountably infinite or uncountable.since it is not countable, the cardinality is not ℵ0. It happens to be the case that there are different ℵ numbers that represent cardinality. his also means that there are different sized infinities. Do you think that Q and R are the same size? How about Q0 and R? What about the set of all subsets of R and R itself?

52 Page Homework 2.4 Problem 1. Can you provide a real life example of an infinite set? Justify your answer. Problem 2. rue or alse. Provide explanations. 1. If A B and B is an infinite set, then A is infinite. rue alse 2. Every subset of a finite set is also finite. rue alse 3. Since N is an infinite set and R is an infinite set, n(n) n(r). rue alse 4. he intersection of infinite sets is infinite. rue alse 5. he union of infinite sets is infinite. rue alse 6. he complement of is infinite. rue alse

53 he Nature of Counting Page he Nature of Counting 12.1 Permutations and Counting he goal of this section is to give us the tools to be able to handle the probability section. We will learn two different different ways to count. he first is called the fundamental counting principle undamental Counting Principle heorem : he fundamental counting principle gives the number of ways to perform a multiple number of tasks. If task A can be performed in m different ways and B can be performed in n different ways, then performing task A then task B can be done in m n different ways. his principle holds for more than 2 tasks. Example: : Let A be a group with 3 people, B be a group with 3 people, and C be a group with 2 people. I want to find all the possible ways to group people from group A, B, and C. What if I only want to know the number of ways to group people from A,B, and C? Label the people in A as 1,2, and 3, the people in B as a,b, and c, and the people in C as α and β. he number of ways to make groups is given below. A B a 1 b c a 2 b c a 3 b c C α β α β α β α β α β α β α β α β α β Each red path designates a different way to group people from group A, B, and C. or example, if we follow the path on the top, we would have 1 from A, a from C, and α from C, forming the group (1,a,α). his is a relatively small sample which could be done through this tree graph. However it is much simpler to answer the question about the number of different groups by using theorem By theorem , we have the total of different groups is given by multiplying the number of elements in each set. Hence, we get If you were to count all the possible paths along the red lines above, you would come out to this same number. Notice, in each column of red lines, we are essentially adding a number of paths that is the same as the number of elements in that set, but we are doing this the same number of

54 he Nature of Counting Page 54 times as the number of elements in the previous set. he first column adds 3 paths, then the second column adds three paths three times, and lastly the 3 column adds 2 paths 3 times for each group of a, b and c. Example: : ind the number of ways to form groups, where the groups are taken from A, B, then C. 1) n(a) 4, n(b) 6, and n(c) 20 We will have different groups. 2) n(a) 10, n(b) 10, and n(c) 10 We will have different groups. 3) A, B {1, 2, 3, 4, 5, 6}, and C {dog, cat, pig} We will have different groups. 4) A N, B {1}, and C {2} Since N is an infinite set, we will have an infinite number of different groups. his fundamental principle leads to the next topic called permutations Permutations Permutations in essence is the fundamental counting principle. We will first define permutations and its notation. heorem : A permutation of r elements from a set A with n elements is the number of different arrangements of those r elements selected without repetition. Order matters as well. his says that group (1,2) is different than group (2,1). Permutations is denoted at n Pr. n! where n! n(n 1)(n 2)... (3)(2)(1) n Pr (n r)! 0! 1 hink about the fundamental counting principle. It says to take the number of elements in the first group and multiply it to the number of elements in the second and so on. Well, in permutations consider taking the set A {1, 2, 3, 4, 5}, and we want to select a group of 2 where we do not allow repetition. How many total groups of 2 are there? Lets use the fundamental counting principle to see how this works. he first selection is from A with 5 elements, but the second choice is from A minus the element we chose first. his means that in the second choice we can only choose from 4 elements. By theorem , we have groups of 2. Notice however that we will get the same answer using 5 P2 5! (5 2)! 3 2 Essentially the top n! gives you all possible choices. hen by dividing by (n r)!, we remove all the repeated versions where order mattered. Example: : In a math class, the professor says that he will only give out 3 A s which will be distributed at random to a class of 30 students. he first person picked gets an A+, the second gets an A, and the third gets an A-. Regardless of how you do in the class the only students who get A s will be these three students. How many different ways can he give out these A s? We first need to determine if replacement are happens. Can the professor pick an A then put it back into the running for the next person? Why or why not? He cannot

55 he Nature of Counting Page 55 do this because he said that he would give out only one A. Next determine if order matters. his one is a little clearer, because the first person picked gets as different grade than the other two. Because we satisfy these conditions, we use permutations. 30 P3 30! , 360. (30 3)! here are 24,360 ways for the professor to give out the A s. Lets think about another situation. What would happen if we wanted to pick a group of size zero? How many different ways can we do this? Example: : We have a group of 5 dogs staying at a kennel. he owner of the kennel want to know how many ways there are to group 0 dogs, assuming that the order matters. What is the answer? One way to think of this is, to say we want to have a group of zero dogs and there is only one way to do this. hat is get a group of zero dogs. Another way to show this is mathematically. We know that we cannot pick a dog twice and order matters, hence we use permutations. By theorem , 5 P0 5! 5! 1. (5 0)! 5! here is only one way to group 0 dogs Distinguishable Permutations What if we want to find the number of ways to permute the letters of Hawaii. Can you tell the difference between the arrangement waiiah and waiiah? Well one had waiiah while the other waiiah. Notice we cannot determine the difference between these two arrangements. We need a new theorem to find these. heorem : he number of distinguishable permutations of n objects in which n1 is the number of objects of one kind, n2 is the number of objects of another kind,..., and nk is the number of objects of a further kind, is given by n!. n1! n2!... nk! Example: : How many different permutations of the letters of the given words are there? 1) Hawaii Notice a and i are repeated two times. hus n1 2 and n2 2 and we have ! 180 2!2! 4 here are 180 distinguishable permutations of the letters of Hawaii. 2) Maui No letters are repeated so the denominator is 1. 4! ! his is the same as 4 P4. here are 24 distinguishable permutations of the letters of Maui.

56 he Nature of Counting 3) Kauai he only repeat is the a, which is repeated 2 times. Hence 5! ! 2 here are 120 distinguishable permutations of the letters of Kauai. Page 56 4) Humuhumunukunukua pua a his one has a lot of letters repeated letters. Humuhumunukunukua pua a here are 2 h, 9 u, 2 m, 2 n, 2 k, 3 a, and 1 p. Hence the formula gives us 21! 1, 466, 593, 128, 000 2!9!2!2!2!3! here are 1,466,593,128,000 distinguishable permutations of the letters of Humuhumunukunukua pua a.

57 Page Homework 12.1 Problem 1. Create a tree diagram showing all the possible outcomes of tossing three fair coins. hen list the ways of getting each result. 1) More than three tails. 2) ewer than 2 tails. 3) At least 2 tails. 4) No more than 2 tails. Problem 2. Evaluate the expression without a calculator. 1) 16! 14! 2) 5! (5 2)! 3) 8! 6!(8 6)! 4) 100! 99!(100 99)! Problem 3. ind the number of distinguishable arrangements of the letters of each word or phrase. 1) GOOGOL 2) BANANA

58 Page 58 Problem 4. How many five digit numbers are there in our system of counting numbers. Problem 5. rue or alse. Provide explanation. 1. (5 2)! 3! 5! 0 2. he amount of ways to arrange the letters Lallay is 6! 2!2!. 3. n! (n 1)! is always n. Problem 6. Evaluate each expression. 1) 14 P 9 2) 11 P 3 3) 12 P 9 4) 9 P 3

59 Page 59 Problem 7. irst, second, and third prizes are to be awarded to three different people. If there are ten eligible candidates, how many outcomes are possible. Problem 8. At a wedding reception, the bride, the groom, and four attendants will form a reception line. How many ways can they be arranged in each of the following cases? 1. Any order will do. 2. he bride and groom must be the last two in line. 3. he groom must be the last in line with the bride next to him.

60 he Nature of Counting Page Combinations he book states that the permutations are without repetitions. hey also hint toward the idea that combinations is opposite or with repetition. his is not the case. Combinations also must be without repetition. Combinations is discussing the amount aways to group thing, given that the order does not matter. or example, making a team of James and Cathy is the same as making a team of Cathy and James Combinations Vs Permutations heorem : A combination of r elements from a set A with n elements is the number of different arrangements of those r elements selected without repetition with the property that order does matter. his says that group (1,2) is the same as group (2,1). Combinations is denoted at n Cr. n n! with r n. C n r r!(n r)! r Example: : You are the president of the math club. he club has members Cathy, James, Rob, Mat, and Halley. We want to form a three person committee. How many possible ways can we do this? 1. Is there repeats? No, we would not have a three person committee if James was selected twice. 2. Does order matter? No order does not matter because a committee of James, Cathy, and Rob is the same as Cathy, James, and Rob or Rob Cathy, and James. hese are the questions that need to be asked every time there is a counting problem. We will give a flow chart later. With this, we will use combination to get 5 5! 5! C3 3 3!(5 3)! 3!2! here are 10 possible three person committees. What if the question was stated a little differently. We want to chose a secretary, treasurer, and a vice president. How many ways can we do this? Notice that the answer to question 1 is still the same. However, the order of which we pick members will be important. his tells us to use permutations. Hence, there are 5 P3 5! 5! (5 3)! 2! 2 possible ways to make a three person committee consisting of a secretary, treasurer, and a vice president. Does this make sense? he permutations is larger than the combinations, since order matter, we should have more possibilities. Now our previous example of James, Cathy, and Rob versus Cathy, James, and Rob or Rob Cathy, and James are all different. In combinations these would be worth one, but in permutations this is worth 3. Some of the most commonly found examples involve cards.

61 he Nature of Counting Page 61 Example: : How many 5 card hands can you make from a standard deck of cards. 1. Is there repeats? No, we cannot have 2 kings of hearts with a standard deck of cards. 2. Does order matter? No, because a 5 card hand 2, 3, 4, 5, and 6 is the same as 6, 5, 4, 3, and 2. Hence we use combinations and get 52! ! 2, 598, C5 5!(52 5)! 3!47! We have 2,598,960 possible 5 card hands. Example: : How many ways can you get a full house of three tens and two queens? Notice that this is a problem that requires both the fundamental counting principle as well as combinations. irst we think about the different events discussed. Let AGet three tens and BGet two queens. By theorem we get n(a) n(b) Ways to get three tens Ways to get 2 queens. Now each of these parts is a counting problem in its own right. But they are solved the same way. or the first ways to get three tens, 1. Is there repeats? No, similarly as example Does order matter? No, similarly as example So we have 4! ! !(4 3)! 2!(4 2)! here are a total of 24 ways to get a full house of 3 tens and 2 queens Binomial heorem I won t discuss this to much, but if you would like to read more about it, you can find information in your book p or htm. he binomial theorem can be used to count, but we will be using it in a different way. irst we will state the theorem and then give an example of its application. heorem : or any positive integer n, n Z, n n 0 n n 1 n n 2 2 n n 0 n n n 1 (a + b) a b + a b+ a b ab + a b n 1 n Example: : What is 37?

62 he Nature of Counting Page 62 Notice 37 can be rewritten as (1 + 2)7. his is of the form of the right hand side of theorem Hence, (1 + 2) (1)(17 )(1) + (7)(16 )(2) + 21(15 )(4) + 35(14 )(8) + 35(13 )(16) + 21(12 )(32) + 7(11 )(64) + 1(10 )(128) his can be used to expand really large powers quickly. In previous math classes you have expanded problems like (x + 2)2. What if the problem was (x + 2)7? Evaluating (x + 2)(x + 2)(x + 2)(x + 2)(x + 2)(x + 2)(x + 2) would be very tedious. Where as evaluating (x+2) x 2 + x 2 + x 2 + x 2 + x 2 + x 2 + x 2 + x (1)(x7 )(1)+(7)(x6 )(2)+21(x5 )(4)+35(x4 )(8)+35(x3 )(16)+21(x2 )(32)+7(x1 )(64)+1(x0 )(128) is much easier.

63 Page Counting low Chart Straight to undamental Counting Principle Are items selected with replacement? Yes Use undamental Counting Principle No Does Order Matter? Yes Are Some Items Identical? Yes Use n! n 1! n 2!... n k! No No Use Combination Rule Permutations n! r!(n r)! n! (n r)!

64 Page Homework 12.2 Problem 1. Evaluate each expression. 1) 14 C 9 2) 10 C 8 3) 11 C 7 4) 12 C 9 Problem 2. irst, second, and third prizes are to be awarded to three different people. If there are ten eligible candidates, how many outcomes are possible. Problem 3. Your boss has decided to start a project of building parks in inner cities. She hopes to bring youth programs to these areas in hopes of creating a safe space for children to be. She needs a team of 5 people to run the project and has tasked you to create the team. here are 7 people in your section and you need to make the team from your colleagues. How many possible teams are there?

65 Page 65 Problem 4. yler is to build six homes on a block in a new subdivision, using two different models: standard and deluxe. (All standard model homes are the same, and all deluxe models are the same.) 1) How many different choices does yler have in positioning the six houses if he decides to build three standard and three deluxe models. 2) If yler builds two deluxe and four standards, how many different positioning can he use. Problem 5. In how many ways could eight people be divided into two groups of three people and a group of two people? Problem 6. Use the Binomial heorem to evaluate 2 5. Problem 7. Use the Binomial heorem to evaluate (x + 1) 5.

66 he Nature of Counting Page Counting Without Counting In this section we will be doing a multitude of problems where it is unknown which method of counting we will use. You will need to use section to figure out if you need to use the fundamental counting principle, combinations, or permutations. he most difficult part about the upcoming probability section is counting. If you can count well, then the probability section will be much easier Random Counting Problems Example: : How many different Hawaii state license plates are there? 1. Is there repeats? Yes, so we use the fundamental counting principle. Notice that Hawaii has a license plate ordered by 3 letters then 3 numbers. here are 26 letters in the alphabet and 10 possible numbers. By theorem , we have , 567, 000. here are 17,567,000 different license plates. Example: : How many different Hawaii license plates are there if repetition is not allowed? We will do this in two different ways. irst we will do this using theorem he first task is to pick a letter which has 26 choices, the second task is to pick a letter which has 25, and the third task is to pick a letter which has 24, assuming no repetition. Also the first number or fourth task has 10 choices, the second number of fifth task has 9 choices, and the third number or sixth task has 8 choices. By theorem , we have , 232, 000. We may also do this using permutations since order of the letters matters and order of the numbers matters. We know that it is permutations since 1. Are there repeats? No. We cannot draw the same number or letter twice. 2. Does order matter? Yes. his is true since license plate ABC123 is not the same as BAC123. We would do this as Number of ways to get 3 letters he number of ways to get three numbers 26 P3 10 P3 26! 10! 26! 10! , 232, 000 (26 3)! (10 3)! 23! 7! We conclude there are 11,232,000 possible license plates without repetition. Example: : How many different ticket numbers are there in the Powerball? he instructions for the Powerball are as follows. hey draw 5 white balls from 69 different numbers and 1 red ball from 26 different balls numbered You need to guess the right white numbers regardless of the order and the the red number

67 he Nature of Counting Page 67 correctly to win the jackpot. 1. Are there repeats? here are no repeats because the balls are taken out of the box, and they are not put back in. 2. Does order matter? No order does not matter. his is stated in the way that the Powerball is played. Hence we use combinations to get, 69 C ! , 201, !(69 5)! he total number of ways to get a Powerball ticket is 292,201,338. Example: : Lets say that the committee that runs the Powerball has found that to many people are winning. What is an easy way for them to fix this problem using what we know about counting? Instead of using combinations, where order does not matter, we would use permutations. his would increase the number of possibilities. 69 P ! (69 5)! , 064, 160, 560. he new Powerball system has a total of 35,064,160,560 different tickets. Another way to do this is to increase the cost of the Powerball tickets. his changes the expected value of the game, something that we will learn in the probability section. All this means is that the amount of money expected to go to the people running the game will increase. Example: : How many ways are there to arrange the numbers 1,2,1, and Are there repeats? In other words even though there are two ones, the first one cannot be repeated twice. Nor could we get four ones. 2. Order matters, and some items are identical, hence we have a permutations problem where we need cannot distinguish between the arrangements 1142 and his says to use theorem ! ! 2 here are 12 different arrangements of the numbers 1,2,1, and 4. Example: : In a race with 30 runners where 8 trophies will be given to the top 8 runners (the trophies are distinct: first place, second place, etc), how many ways can this be done? 1. Is there repeats? No, since two runners cannot get a 1st place trophie. 2. Order matters, because the trophies are for different places.

68 he Nature of Counting Page 68 We are working with permutations, and we have 30 P8 30! , 989, 936, 000. (30 8)! here are 235,989,936,000 different ways for 8 runners to win trophies with 30 total runners. Example: : How many arrangements of the letters are there of the word repetition. We have the letters e,t, and i repeated twice. Using theorem , we have 10! , 600 2!2!2! 2 4 here is a total of 453,600 arrangements of the letters of the word repetition. Example: : An election ballot asks voters to select three city commissioners from a group of six candidates. In how many ways can this be done? 1. Is there repeats? Same person cannot be the first and second commissioner. His/her name cannot be placed back into the hat. 2. Order does not matter. his is just a group of three commissioners with no distinction between them. We use combinations to get ! !(6 3)! here are 20 different three person commissioner teams. he following problems were found online at 0/9/8/ /permutationscombinationshw.pdf

69 Page Homework 12.3 Problem 1. In a race in which six automobiles are entered and there are no ties, in how many ways can the first four finishers come in? Problem 2. he model of the car you are thinking of buying is available in nine different colors and three different styles (hatchback, sedan, or station wagon). In how many ways can you order the car? Problem 3. A book club offers a choice of 8 books from a list of 40. In how many ways can a member make a collection? Problem 4. A medical researcher needs 6 people to test the effectiveness of an experimental drug. If 13 people have volunteered for the test, in how many ways can 6 people be selected? Problem 5. rom a club of 20 people, in how many ways can a group of three members be selected?

70 Page 70 Problem 6. rom the 30 pictures I have of my daughter s first birthday, my digital picture frame will only hold 3 at a time. How many different groups of 3 pictures can I put on the frame? Problem 7. A popular brand of pen is available in three colors (red, green or blue) and four tips (bold, medium, fine, or micro). How many different choices of pens do you have with this brand? Problem 8. A corporation has ten members on its board of directors. In how many ways can it elect a president, vice-president, secretary and treasurer? Problem 9. or a segment of a radio show, a disc jockey can play 7 songs. If there are 12 songs to select from, in how many ways can the program for this segment be arranged? Problem 10. How many different ways can a director select 4 actors from a group of 20 actors to attend a workshop on performing in rock musicals?

71 he Nature of Probability Page he Nature of Probability 13.1 Introduction to Probability Probability, as stated earlier, is essentially counting. We will explain probability using things that we have already learned in the counting section (section 12) as well as the set theory section (section 2). We first will need some definitions. Definition : An experiment is an observation of any physical occurrence. Definition : he sample space is the set of all possible outcomes or events of an experiment. You can think of this as the universal set U. Remember U changes depending on what we are talking about (definition 2.1.3). Definition : An event or outcome of an experiment are the possible things that can happen. Events are denoted using capitol letters. Example: : ind the sample space and some of the possible events. 1. lipping a coin. he sample space is S{Getting a heads, Getting a tails}. Each of these elements are events and we denote it as Hgetting a heads and getting a tails. 2. Rolling a die. he sample space is S{Roll a 1, Roll a 2, Roll a 3, Roll a 4, Roll a 5, Roll a 6,}. he events are ARoll a 1, BRoll a 2, CRoll a 3, DRoll a 4, ERoll a 5, and Roll a 6. Now since events are subsets of S, they may include more than one element. or example, in example part 2, another event could be GRoll a 4,5, or a heoretical Probability irst, when we talk about probability it is likely that I may forget to state fair, standard, or unbiased when discussing this topic. You can assume that the items that we are using are fair or unbiased unless stated otherwise. heoretical probability is well theoretical. Every one knows that the probability of getting a heads when flipping a coin is 21. Does this necessarily mean that if I flip two coins, that one must be a head and the other must be a tail? Of course not. Probability is the concept of likeliness, not what must happen. Definition : Let E be an event. he theoretical probability of E if the ratio of ways to get into event E divided by the number of outcomes. Assuming that all the events have the same likelihood of happening. P (E) Ways to get into event E n(e) he number of outcomes n(s) heorem : Properties of probability: Let E be an event within the sample space S. hat is E is a subset of S. hen the following properties hold P (E) 1 he probability of an event is a number from 0 through P ( ) 0 he probability of an impossible event is 0.

72 he Nature of Probability Page P (S) 1 he probability of a certain event is P (E) + P (E 0 ) 1. Instructions 1. Write down the event(s) (What you want) and S physically what you are doing. 2. Write explicitly the ways to get into event E, or use section 12 to determine the ways to get into event E. 3. Write explicitly the ways to get into event S, or use section 12 to determine the ways to get into event S. 4. Write the formula. P (E) Ways to get into event E he number of outcomes n(e) n(s). 5. Plug in and simplify. Example: : ind the probability of the following events. 1. What is the probability of getting a 4 when rolling a fair 6 sided dice? (a) Let E Roll a 4. (b) here is only one way to do this i.e. roll a 4. So n(e) 1. (c) here are six outcomes. Roll a 1,2,3,4,5 or 6. So n(s) 6 (d) P (E) is 16. Ways to get into event E he number of outcomes n(e) n(s) 16. he probability of rolling a 4 2. What is the probability of getting 1 head and 1 tail when flipping two fair coins? (a) Let E lip 1 head and 1 tail. (b) here are two ways to do this i.e. lip a H or H. So n(e) 2. (c) here are 4 possible outcomes. By theorem , the number of outcomes is We can also list them as HH, H, H, or. So n(s) 4 n(e) to get into event E 2 1 (d) P (E) Ways he number of outcomes n(s) 4 2. he probability of flipping two coins and getting 1 head and 1 tail is What is the probability of getting a queen when drawing a random card from a standard 52 card deck? (a) Let E Draw a queen. (b) here are four ways to do this i.e. Q Q Q Q Q Q Q Q Note: he events and the sample space relate. he sample space is physically what you did in the event. or example, if E is roll a 6, then S is roll a dice. However if E is roll a 6 and get a heads in coin flip, then S is rolling a dice and flipping a coin. n(e) 4. (c) here are 52 possible outcomes. his is just by the number of cards. So n(s) 52

73 to get into event E (d) P (E) Ways he number of outcomes 1. drawing a queen is 13 n(e) n(s) he probability of 4. What is the probability of getting a queen or a king when drawing a random card from a standard 52 card deck? (a) Let E Draw a queen or a king. (b) Don t forget that or is inclusive. Meaning we can draw a king, queen, or both at the same time. he last one is impossible, we only care about drawing a king or a queen separately. here are 8 ways to do this i.e. K Q Q K K KQ Q K KQ Q KQ KQ n(e) 8. (c) here are still 52 possible outcomes. So n(s) 52 to get into event E (d) P (E) Ways he number of outcomes 2 drawing a queen or a king is 13. n(e) n(s) he probability of 5. What is the probability of drawing a red 4? (a) Let E Draw a red four. (b) his is asking us to draw a card that is both red and 4. here are 2 ways to do this i.e he Nature of Probability Page 73 n(e) 2. (c) here are still 52 possible outcomes. So n(s) 52 to get into event E (d) P (E) Ways he number of outcomes 1 drawing a red four is 26. n(e) n(s) he probability of he following example is a very famous example which represents theorem part 4. After reading this example, think about how this would play out in a game of deal or no deal. What should you do? Example: : Monty Hall Problem: Suppose you re on a game show, and you re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what s behind the doors, opens