Combinatorial Proofs

Save this PDF as:

Size: px
Start display at page:

Transcription

1 Combinatorial Proofs

2 Two Counting Principles Some proofs concerning finite sets involve counting the number of elements of the sets, so we will look at the basics of counting. Addition Principle: If A and B are disjoint finite sets with A =n and B = m, then A B = n + m. Multiplication Principle: If a procedure can be broken into m stages and if there are r outcomes in the first stage,s outcomes in the second stage independent of the choice in the first stage, t outcomes in the third stage again independent of the previous choices then the total procedure has rst... composite outcomes.

3 Examples If I teach two classes and one class has 30 students and the other has 12 students, how many students do I teach? There is not enough information to answer the question!!! If there is no overlap, then the addition principle gives 42, but if there is an overlap the addition principle does not apply. However, if we know that there are 2 students in both classes then we can use the addition principle as follows: #in first class only + #in second class only + #in both = 40.

4 Examples Toss two dice. How many ways can the outcome not be doubles? Think of the toss as a process with two stages, the first stage is the toss of one die and the second is the toss of the other. 1 st method: There are 6 outcomes for the first stage, and to not get doubles, there are 5 outcomes for the second stage. By the product rule there are 30 possibilities. [Note that the number of outcomes for the second stage does not depend on which outcome occurred in the first stage.] 2 nd method: Total number of outcomes (including doubles) is, by the product rule 6 6 = 36. Of these, 6 are doubles, so the number that are not doubles is 36 6 = 30.

5 Examples There are 15 different apples and 10 different pears. How many ways are there for Jack to pick an apple or a pear and then for Jill to pick an apple and a pear? There are two natural stages, Jack's pick followed by Jill's pick. Jack chooses one out of 25 fruit, so there are 25 outcomes in this stage. To calculate the number of outcomes for Jill's pick we must know what Jack picked: If Jack picked an apple, then Jill has 14(10) = 140 choices. If Jack picked a pear, then Jill has 15(9) = 135 choices. The product rule does not apply directly, but we can use the addition rule to break up the problem into subproblems in which it does apply. Jack picks an apple + Jack picks a pear 15(140) + 10(135) = = 3450

6 Lists (ordered arrangements) In general when creating lists, if there are n 1 selections for first place, n 2 for second,... n r for the r-th place (each selection independent of the preceding) then the number of lists that can be created equals n 1 n 2...n r (product rule). If the selections are taken from an n-set and repeats (replacement) are allowed then the number of lists is given by n n n... n = n r. If no repeats are allowed then the number of lists is given by P(n,r) = n(n-1)(n-2)...(n-r+1) In the special case of complete lists (permutations) we have P(n,n) = n!

7 Permutation Problems How many ways are there to pick 2 successive cards from a standard deck of 52 such that: a. The first card is an Ace and the second is not a Queen? b. The first is a spade and the second is not a Queen? a) We are creating a list of two things. There are 4 choices for the first item and (51 4) = 47 choices for the second, so there are 4(47) = 188 ways to do this. b) This time there are 13 choices for the first item, but the number of choices for the second depends on what was chosen first. If the Queen of Spades was picked first, there are (51 3) = 48 choices for the second card. If a different spade had been selected then there are 51-4 = 47 choices for the second card. So, we get 1(48) + 12(47) ways to do this.

8 Permutation Problems There are 50 cards numbered 1 to 50. Two different cards are chosen at random. How many ways can this be done? In how many of these picks will one card be twice the other? In this case we are not forming a list (there is no first card of the two selected), but we can still use the list idea to solve the problem. If we were forming a list, the number of possible lists would be 50(49) since the cards are different. But, two cards A and B would appear twice in these lists, once as A,B and again as B,A. So, to answer the question we should divide by 2 to get ½(50)(49) = 25(49). To answer the second question, we note that the two cards are now distinguished, there is a larger one and a smaller one. If we select the larger one first there are 25 choices (it must be even), and then the smaller card is determined, so there are no additional choices.

9 Permutation Problems How many integers greater than 6600 have distinct digits, not including the digits 7,8,or 9? Since only 7 digits are allowed, we can not have any integers with more than 7 decimal places. We also can not have any with less than 5 decimal places and still be greater than We look at the cases of 5, 6 and 7 decimal places separately. For a 5 place integer, there are only 6 choices for the first digit since we can't use 0 there. There are also 6 choices for the second digit since we can't use the digit used in the first place, but we are now allowed to use 0. The total number is thus 6(6)(5)(4)(3). Similarly, for a 6 place number we get 6(6)(5)(4)(3)(2) and a 7 place number give 6(6)(5)(4)(3)(2)(1). So, our answer is: 6(6)(5)(4)(3) + 6(6)(5)(4)(3)(2) + 6(6)(5)(4)(3)(2)(1) = 10,800

10 Permutation problems Use the prime decomposition to show that every square integer has an odd number of divisors (including 1 and itself), and every non-square integer has an even number of divisors. Observe that the divisors of an integer are made up of the same primes that make up the integer. We can count the number of divisors (factors) by listing, for each prime divisor, the number of times that prime appears in the factor (including the possibility that it doesn't appear at all). Thus, looking at the prime decomposition of an integer n we can say: p 1 a 1 p2 a 2 pk a k has a1 1 a 2 1 a k 1 factors. If n is a square, all the exponents are even, so the number of factors is a product of odd numbers and so is odd. If n is not a square, then at least one exponent is odd, so the number of factors has an even integer divisor and is even.

11 r-sets of an n-set Subsets of size r of a set of size n are called combinations of n elements taken r at a time. The number of such subsets is given by the binomial coefficient C(n,r), also written as n r and read as "n choose r". We determine a formula for C(n,r) by using an obvious, but important counting principle: If you count the elements of a set in two different ways - YOU GET THE SAME ANSWER! Duh!!

12 Combinations Example: We have already counted the number of lists of r elements taken from a set of n elements, P(n,r). Now we count these in a different way. First select the r elements (this can be done in C(n,r) ways) then arrange the r elements in lists in all possible ways (P(r,r) ways to do this). So,... P(n,r) = C(n,r) P(r,r) C n, r = n r P n, r = P r, r = n n 1 n 2 n r 1 r! = n n 1 n r 1 n r n r r! n r! = n! r! n r!.

13 Combinations Given a set of n elements, there is only one subset that has 0 elements, i.e., the empty set. So, C(n,0) = 1 n N. Also, there is only one subset that contains n elements. Thus C(n,n) =1 n N. Our formula for computing C(n,r) would, in these two cases, give: n 0 n! = n n 0! n! = n! n!0! In order to have these formulas make sense, we must define 0! = 1. It is also convenient to define C(n,r) = 0 if r < 0 or r > n.

14 Poker Hands A poker hand consists of 5 cards from a deck of 52 cards. A four of a kind hand has 4 cards of the same denomination. There are =13 48 =624 four of a kind hands. 1 1 A full house consists of 3 cards of one denomination and 2 cards of another denomination. There are = =3,744 full house hands. 1 1 A flush consists of 5 cards of the same suit but not consecutive. There are 1 [ ] =5,108flush hands.

15 Committee Selections Suppose that there are 10 Republicans, 8 Democrats and 2 Independent legislators who are eligible for committee membership. There are C(20,5) 5 member committees that can be formed. How many 5 member committees have at least 2 Republicans? C(10,2) C(18,3) How many 6 member committees have as many Republicans as Democrats? C(10,2) C(8,2) C(2,2) + C(10,3) C(8,3) How many 5 member committees have no more than 2 Republicans? C(10,5) + C(10,1) C(10,4) + C(10,2) C(10,3)

16 The Binomial Theorem A binomial is an algebraic expression with two terms, like x + y. When we multiply out the powers of a binomial we can call the result a binomial expansion. Of course, multiplying out an expression is just a matter of using the distributive laws of arithmetic, a(b+c) = ab + ac and (a + b)c = ac + bc. In the examples below, I have multiplied out a few binomial expansions, without combining like terms or even writing products as powers. To make things easier to see, I have used a different color for the terms in each factor, so that you can see where they came from.

17 The Binomial Theorem

18 Binomial Theorem Notice that each term of the binomial expansions is a product composed of one item of each color. This suggests another way to look at this process of multiplication. Think of each factor in the product as a box containing two items, an x and a y (the colors are used here to keep track of which box an x or a y came from). We form one of the terms in the binomial expansion by picking either an x or a y from each of the boxes and multiplying our selections together. Since there are 2 choices for what we can pick from each of the boxes, the total number of possible different selections is = 2 n. Now look at the examples again, and notice that the number of terms in the binomial expansions is exactly the total number of possiblities, i.e., for the cube there are 8 terms, for the 4th power there are 16 terms, etc. This means that every possible way of choosing one item from each box gives a term in the binomial expansion.

19 Binomial Theorem

20 Binomial Theorem We would like to be able to say exactly how many terms are in each group (this would be the coefficient of the term if it was written out in the normal way). We can do this by counting because we know how the terms are formed. To get a term with, say, 3 x's and 1 y, all we have to do is decide which box we are going to pick the y from and then pick the x's from all the other boxes. Thus, the number of terms of this type is the number of ways we can pick 1 (box) from a set of 4 (boxes). This is the number of subsets of size 1 in a set of size 4, i.e. C(4,1). Looking at the group of terms having 2 x's and 2 y's, we know that we get them by picking 2 of the 4 boxes to take the y's from, so there are C(4,2) = 6 such terms. Now, consider the first and last terms of this example. In order to get xxxx, you don't pick any boxes to take y's from, and you can only do this in one way, i.e., C(4,0) = 1. Similarly, to get yyyy, you must pick all the boxes to take y's from, and again there is only one way to do this since C(4,4) = 1. These two statements are true for any number of boxes, since C(n,0) = C(n,n) = 1 for any n.

21 Binomial Theorem We can now drop the colors and write things normally to get: The general case can be written as: Because the numbers C(n,k) appear as the coefficients of the terms in a binomial expansion, they are called binomial coefficents.

22 Counting Proofs Using the formula for C(n,r) it is easy to see that C(n, r) = C(n, n-r). n n r = n! n r! n n r! = n! n r! r! = n r. However, it is also possible to see this relationship without resorting to the formulas. To every subset of size r in an n set, there corresponds a unique subset of size n-r, namely, the complement of the r set. Thus, the total number of size r subsets must equal the total number of size n-r subsets. Reasoning this way is supplies a combinatorial proof as opposed to the algebraic proof of manipulating formulas.

23 Combinatorial Proofs C(n,m) C(m,k) = C(n,k) C(n-k, m-k) To give a combinatorial proof of this binomial identity, we need to find a counting problem for which one side or the other is the answer and then find another way to do the count. Let S be a set with n elements. Consider the ordered pairs (M,K) where M S with m elements and K M with k elements. The number of these ordered pairs is, C(n,m) C(m,k) using the product rule and first counting the number of choices for M and then the number of choices for K. Now we count the ordered pairs in a different way. First choose a K, this can be done in C(n,k) ways, and next count the number of M's that contain this K. Since we must add m-k elements to K to get an M, and these are in the complement of K, of size n-k, we have C(n-k,m-k) ways to do this. So, C(n,m) C(m,k) = C(n,k) C(n-k, m-k) since these represent two ways to count the same thing.

24 An old favorite Thm: If a set S has n elements then its power set has 2 n elements. Pf: We count the number of subsets by size. There are C(n,0) of size 0, C(n,1) of size 1,..., C(n,r) of size r, etc. Thus, the total number of subsets of S is the sum: 0 n 1 n n n n = i=0 n i. On the other hand, the binomial theorem gives, by setting x = y = 1: n x y n = n 2 n = 1 1 n = i=0 i=0 n i 1i n i xi y n i n 1 n i = i=0 n i.

25 Combinatorial Proofs C(n,r) = C(n-1,r-1) + C(n-1,r). C(n,r) counts the number of r-sets in an n-set. We now count these in a different way. Single out one element of the n-set and think of it as being colored red. Every r-set either contains the red element or not, so C(n,r) = # r-sets with red element + # r-sets without red element. To get a subset without the red element, we must select r elements out of n-1 (i.e., those that are not the red element) and there are C(n-1,r) ways to do this. To get a subset with the red element, we must select r-1 additional elements from the n-1 non-red elements to add to the red element. This can be done in C(n-1, r-1) ways. So, C(n, r) = C(n-1, r-1) + C(n-1, r).

26 Pascal's Triangle This last binomial identity is the formation rule for what is known as Pascal's Triangle: By numbering the rows (top down) starting at 0, and the entries in a row (left to right) starting at 0, then the entry in the r th position of the n th row is C(n,r). C(n,r) = C(n-1,r-1) + C(n-1,r)

27 Pascal's Triangle Pascal's Triangle has many properties: Triangular Numbers:

28 Pascal's Triangle Pascal's Triangle has many properties: Tetrahedral Numbers:

29 Pascal's Triangle Fibonacci numbers hidden in the triangle

30 Pascal's Triangle Illustrates the binomial identity: C(n,0) + C(n+1,1) C(n+k,k) = C(n+k+1,k)

31 Pascal's Triangle We call the triangle Pascal's Triangle in honor of the French mathematician Blaise Pascal ( ) who in 1653 wrote Traité du triangle arithmétique (Treatise on the Arithmetic Triangle). The properties we have mentioned, and many more, are collected in this work, many of them first discovered by Pascal. HOWEVER,...

32 Pascal's (?) Triangle

33 Pascal's (?) Triangle Even this is not the earliest reference we have to the triangle: ~1100 Omar Khayyam mentions the triangle and does not claim credit for it. So he probably obtained it from earlier Chinese or Indian sources.

34 One More Counting Proof 2 n n = 0 2 n n 1 2 n n 2 Let a set S consist of n red balls and n blue balls. We want to count how many ways there are to select n balls (regardless of color) from S. The answer to this problem is clearly C(2n, n). We now count in a different way. Each selection of n balls contains a certain number, say x, of red balls and also n-x blue balls. Clearly, x ranges from 0 to n. For any particular value of x, we can choose the red balls in C(n,x) ways and the blue balls in C(n, n-x) ways. So the number of ways to get an n set with x red balls is C(n,x) C(n, n-x) = C(n,x) 2 since we know that C(n,x) = C(n, n-x). So, counting the n-sets by how many red balls they contain, gives: C(2n,n) = C(n,0) 2 + C(n,1) C(n,n) 2

4. Binomial Expansions

4. Binomial Expansions 4.. Pascal's Triangle The expansion of (a + x) 2 is (a + x) 2 = a 2 + 2ax + x 2 Hence, (a + x) 3 = (a + x)(a + x) 2 = (a + x)(a 2 + 2ax + x 2 ) = a 3 + ( + 2)a 2 x + (2 + )ax 2 +

Math 2001 Homework #10 Solutions

Math 00 Homework #0 Solutions. Section.: ab. For each map below, determine the number of southerly paths from point to point. Solution: We just have to use the same process as we did in building Pascal

NOTES ON COUNTING KARL PETERSEN

NOTES ON COUNTING KARL PETERSEN It is important to be able to count exactly the number of elements in any finite set. We will see many applications of counting as we proceed (number of Enigma plugboard

0018 DATA ANALYSIS, PROBABILITY and STATISTICS

008 DATA ANALYSIS, PROBABILITY and STATISTICS A permutation tells us the number of ways we can combine a set where {a, b, c} is different from {c, b, a} and without repetition. r is the size of of the

Module 6: Basic Counting

Module 6: Basic Counting Theme 1: Basic Counting Principle We start with two basic counting principles, namely, the sum rule and the multiplication rule. The Sum Rule: If there are n 1 different objects

Discrete Mathematics & Mathematical Reasoning Chapter 6: Counting

Discrete Mathematics & Mathematical Reasoning Chapter 6: Counting Colin Stirling Informatics Slides originally by Kousha Etessami Colin Stirling (Informatics) Discrete Mathematics (Chapter 6) Today 1 /

Continued fractions and good approximations.

Continued fractions and good approximations We will study how to find good approximations for important real life constants A good approximation must be both accurate and easy to use For instance, our

Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important

Combinations and Permutations

Combinations and Permutations What's the Difference? In English we use the word "combination" loosely, without thinking if the order of things is important. In other words: "My fruit salad is a combination

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a

MATHEMATICAL THEORY FOR SOCIAL SCIENTISTS THE BINOMIAL THEOREM. (p + q) 0 =1,

THE BINOMIAL THEOREM Pascal s Triangle and the Binomial Expansion Consider the following binomial expansions: (p + q) 0 1, (p+q) 1 p+q, (p + q) p +pq + q, (p + q) 3 p 3 +3p q+3pq + q 3, (p + q) 4 p 4 +4p

Spring 2007 Math 510 Hints for practice problems

Spring 2007 Math 510 Hints for practice problems Section 1 Imagine a prison consisting of 4 cells arranged like the squares of an -chessboard There are doors between all adjacent cells A prisoner in one

SOLUTIONS, Homework 6

SOLUTIONS, Homework 6 1. Testing out the identities from Section 6.4. (a) Write out the numbers in the first nine rows of Pascal s triangle. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

Chapter 3: The basic concepts of probability

Chapter 3: The basic concepts of probability Experiment: a measurement process that produces quantifiable results (e.g. throwing two dice, dealing cards, at poker, measuring heights of people, recording

Discrete Mathematics Lecture 5. Harper Langston New York University

Discrete Mathematics Lecture 5 Harper Langston New York University Empty Set S = {x R, x 2 = -1} X = {1, 3}, Y = {2, 4}, C = X Y (X and Y are disjoint) Empty set has no elements Empty set is a subset of

WHERE DOES THE 10% CONDITION COME FROM?

1 WHERE DOES THE 10% CONDITION COME FROM? The text has mentioned The 10% Condition (at least) twice so far: p. 407 Bernoulli trials must be independent. If that assumption is violated, it is still okay

Question: What is the probability that a five-card poker hand contains a flush, that is, five cards of the same suit?

ECS20 Discrete Mathematics Quarter: Spring 2007 Instructor: John Steinberger Assistant: Sophie Engle (prepared by Sophie Engle) Homework 8 Hints Due Wednesday June 6 th 2007 Section 6.1 #16 What is the

1.3 Polynomials and Factoring

1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable.

A1. Basic Reviews PERMUTATIONS and COMBINATIONS... or HOW TO COUNT

A1. Basic Reviews Appendix / A1. Basic Reviews / Perms & Combos-1 PERMUTATIONS and COMBINATIONS... or HOW TO COUNT Question 1: Suppose we wish to arrange n 5 people {a, b, c, d, e}, standing side by side,

Math Workshop October 2010 Fractions and Repeating Decimals

Math Workshop October 2010 Fractions and Repeating Decimals This evening we will investigate the patterns that arise when converting fractions to decimals. As an example of what we will be looking at,

Just the Factors, Ma am

1 Introduction Just the Factors, Ma am The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive

Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

Pigeonhole Principle Solutions

Pigeonhole Principle Solutions 1. Show that if we take n + 1 numbers from the set {1, 2,..., 2n}, then some pair of numbers will have no factors in common. Solution: Note that consecutive numbers (such

Definition and Calculus of Probability

In experiments with multivariate outcome variable, knowledge of the value of one variable may help predict another. For now, the word prediction will mean update the probabilities of events regarding the

k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

Contemporary Mathematics- MAT 130. Probability. a) What is the probability of obtaining a number less than 4?

Contemporary Mathematics- MAT 30 Solve the following problems:. A fair die is tossed. What is the probability of obtaining a number less than 4? What is the probability of obtaining a number less than

Squares and Square Roots

SQUARES AND SQUARE ROOTS 89 Squares and Square Roots CHAPTER 6 6.1 Introduction You know that the area of a square = side side (where side means the length of a side ). Study the following table. Side

A set is a Many that allows itself to be thought of as a One. (Georg Cantor)

Chapter 4 Set Theory A set is a Many that allows itself to be thought of as a One. (Georg Cantor) In the previous chapters, we have often encountered sets, for example, prime numbers form a set, domains

8.3 Probability Applications of Counting Principles

8. Probability Applications of Counting Principles In this section, we will see how we can apply the counting principles from the previous two sections in solving probability problems. Many of the probability

Fractions and Decimals

Fractions and Decimals Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles December 1, 2005 1 Introduction If you divide 1 by 81, you will find that 1/81 =.012345679012345679... The first

Lecture 1 Introduction Properties of Probability Methods of Enumeration Asrat Temesgen Stockholm University

Lecture 1 Introduction Properties of Probability Methods of Enumeration Asrat Temesgen Stockholm University 1 Chapter 1 Probability 1.1 Basic Concepts In the study of statistics, we consider experiments

Jan 17 Homework Solutions Math 151, Winter 2012. Chapter 2 Problems (pages 50-54)

Jan 17 Homework Solutions Math 11, Winter 01 Chapter Problems (pages 0- Problem In an experiment, a die is rolled continually until a 6 appears, at which point the experiment stops. What is the sample

An Introduction to Combinatorics and Graph Theory. David Guichard

An Introduction to Combinatorics and Graph Theory David Guichard This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/

What is the probability of throwing a fair die and receiving a six? Introduction to Probability. Basic Concepts

Basic Concepts Introduction to Probability A probability experiment is any experiment whose outcomes relies purely on chance (e.g. throwing a die). It has several possible outcomes, collectively called

Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00

18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list

Stanford Math Circle: Sunday, May 9, 2010 Square-Triangular Numbers, Pell s Equation, and Continued Fractions

Stanford Math Circle: Sunday, May 9, 00 Square-Triangular Numbers, Pell s Equation, and Continued Fractions Recall that triangular numbers are numbers of the form T m = numbers that can be arranged in

Discrete Mathematics: Homework 6 Due:

Discrete Mathematics: Homework 6 Due: 2011.05.20 1. (3%) How many bit strings are there of length six or less? We use the sum rule, adding the number of bit strings of each length to 6. If we include the

Basics of Counting. The product rule. Product rule example. 22C:19, Chapter 6 Hantao Zhang. Sample question. Total is 18 * 325 = 5850

Basics of Counting 22C:19, Chapter 6 Hantao Zhang 1 The product rule Also called the multiplication rule If there are n 1 ways to do task 1, and n 2 ways to do task 2 Then there are n 1 n 2 ways to do

Section 6-5 Sample Spaces and Probability

492 6 SEQUENCES, SERIES, AND PROBABILITY 52. How many committees of 4 people are possible from a group of 9 people if (A) There are no restrictions? (B) Both Juan and Mary must be on the committee? (C)

Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note 11

CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note Conditional Probability A pharmaceutical company is marketing a new test for a certain medical condition. According

1. By how much does 1 3 of 5 2 exceed 1 2 of 1 3? 2. What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3.

1 By how much does 1 3 of 5 exceed 1 of 1 3? What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3 A ticket fee was \$10, but then it was reduced The number of customers

ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 +

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

Homework 3 (due Tuesday, October 13)

Homework (due Tuesday, October 1 Problem 1. Consider an experiment that consists of determining the type of job either blue-collar or white-collar and the political affiliation Republican, Democratic,

Coach Monks s MathCounts Playbook! Secret facts that every Mathlete should know in order to win!

Coach Monks s MathCounts Playbook! Secret facts that every Mathlete should know in order to win! Learn the items marked with a first. Then once you have mastered them try to learn the other topics.. Squares

Fibonacci Numbers and Greatest Common Divisors. The Finonacci numbers are the numbers in the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,...

Fibonacci Numbers and Greatest Common Divisors The Finonacci numbers are the numbers in the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,.... After starting with two 1s, we get each Fibonacci number

Generating Functions Count

2 Generating Functions Count 2.1 Counting from Polynomials to Power Series Consider the outcomes when a pair of dice are thrown and our interest is the sum of the numbers showing. One way to model the

There are 8000 registered voters in Brownsville, and 3 8. of these voters live in

Politics and the political process affect everyone in some way. In local, state or national elections, registered voters make decisions about who will represent them and make choices about various ballot

4. Joint Distributions

Virtual Laboratories > 2. Distributions > 1 2 3 4 5 6 7 8 4. Joint Distributions Basic Theory As usual, we start with a random experiment with probability measure P on an underlying sample space. Suppose

6.3 Conditional Probability and Independence

222 CHAPTER 6. PROBABILITY 6.3 Conditional Probability and Independence Conditional Probability Two cubical dice each have a triangle painted on one side, a circle painted on two sides and a square painted

3.1. RATIONAL EXPRESSIONS

3.1. RATIONAL EXPRESSIONS RATIONAL NUMBERS In previous courses you have learned how to operate (do addition, subtraction, multiplication, and division) on rational numbers (fractions). Rational numbers

I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

EXPONENTS. To the applicant: KEY WORDS AND CONVERTING WORDS TO EQUATIONS

To the applicant: The following information will help you review math that is included in the Paraprofessional written examination for the Conejo Valley Unified School District. The Education Code requires

Topic 1 Probability spaces

CSE 103: Probability and statistics Fall 2010 Topic 1 Probability spaces 1.1 Definition In order to properly understand a statement like the chance of getting a flush in five-card poker is about 0.2%,

Elements of probability theory

2 Elements of probability theory Probability theory provides mathematical models for random phenomena, that is, phenomena which under repeated observations yield di erent outcomes that cannot be predicted

MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

10 k + pm pm. 10 n p q = 2n 5 n p 2 a 5 b q = p

Week 7 Summary Lecture 13 Suppose that p and q are integers with gcd(p, q) = 1 (so that the fraction p/q is in its lowest terms) and 0 < p < q (so that 0 < p/q < 1), and suppose that q is not divisible

PYTHAGOREAN TRIPLES KEITH CONRAD

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

. 0 1 10 2 100 11 1000 3 20 1 2 3 4 5 6 7 8 9

Introduction The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive integer We say d is a

RECURSIVE ENUMERATION OF PYTHAGOREAN TRIPLES

RECURSIVE ENUMERATION OF PYTHAGOREAN TRIPLES DARRYL MCCULLOUGH AND ELIZABETH WADE In [9], P. W. Wade and W. R. Wade (no relation to the second author gave a recursion formula that produces Pythagorean

PEN A Problems. a 2 +b 2 ab+1. 0 < a 2 +b 2 abc c, x 2 +y 2 +1 xy

Divisibility Theory 1 Show that if x,y,z are positive integers, then (xy + 1)(yz + 1)(zx + 1) is a perfect square if and only if xy +1, yz +1, zx+1 are all perfect squares. 2 Find infinitely many triples

E3: PROBABILITY AND STATISTICS lecture notes

E3: PROBABILITY AND STATISTICS lecture notes 2 Contents 1 PROBABILITY THEORY 7 1.1 Experiments and random events............................ 7 1.2 Certain event. Impossible event............................

Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.

Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

Elementary Algebra. Section 0.4 Factors

Section 0.4 Contents: Definitions: Multiplication Primes and Composites Rules of Composite Prime Factorization Answers Focus Exercises THE MULTIPLICATION TABLE x 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5

PROBABILITY 14.3. section. The Probability of an Event

4.3 Probability (4-3) 727 4.3 PROBABILITY In this section In the two preceding sections we were concerned with counting the number of different outcomes to an experiment. We now use those counting techniques

SPECIAL PRODUCTS AND FACTORS

CHAPTER 442 11 CHAPTER TABLE OF CONTENTS 11-1 Factors and Factoring 11-2 Common Monomial Factors 11-3 The Square of a Monomial 11-4 Multiplying the Sum and the Difference of Two Terms 11-5 Factoring the

Probabilistic Strategies: Solutions

Probability Victor Xu Probabilistic Strategies: Solutions Western PA ARML Practice April 3, 2016 1 Problems 1. You roll two 6-sided dice. What s the probability of rolling at least one 6? There is a 1

MAT 1000. Mathematics in Today's World

MAT 1000 Mathematics in Today's World We talked about Cryptography Last Time We will talk about probability. Today There are four rules that govern probabilities. One good way to analyze simple probabilities

Math 115 Spring 2011 Written Homework 5 Solutions

. Evaluate each series. a) 4 7 0... 55 Math 5 Spring 0 Written Homework 5 Solutions Solution: We note that the associated sequence, 4, 7, 0,..., 55 appears to be an arithmetic sequence. If the sequence

Chapter 3. Distribution Problems. 3.1 The idea of a distribution. 3.1.1 The twenty-fold way

Chapter 3 Distribution Problems 3.1 The idea of a distribution Many of the problems we solved in Chapter 1 may be thought of as problems of distributing objects (such as pieces of fruit or ping-pong balls)

Math 319 Problem Set #3 Solution 21 February 2002

Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod

a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

Integers and division

CS 441 Discrete Mathematics for CS Lecture 12 Integers and division Milos Hauskrecht milos@cs.pitt.edu 5329 Sennott Square Symmetric matrix Definition: A square matrix A is called symmetric if A = A T.

The Advantage Testing Foundation Solutions

The Advantage Testing Foundation 013 Problem 1 The figure below shows two equilateral triangles each with area 1. The intersection of the two triangles is a regular hexagon. What is the area of the union

Graduate Management Admission Test (GMAT) Quantitative Section

Graduate Management Admission Test (GMAT) Quantitative Section In the math section, you will have 75 minutes to answer 37 questions: A of these question are experimental and would not be counted toward

It is not immediately obvious that this should even give an integer. Since 1 < 1 5

Math 163 - Introductory Seminar Lehigh University Spring 8 Notes on Fibonacci numbers, binomial coefficients and mathematical induction These are mostly notes from a previous class and thus include some

Lemma 5.2. Let S be a set. (1) Let f and g be two permutations of S. Then the composition of f and g is a permutation of S.

Definition 51 Let S be a set bijection f : S S 5 Permutation groups A permutation of S is simply a Lemma 52 Let S be a set (1) Let f and g be two permutations of S Then the composition of f and g is a

Statistics 100A Homework 2 Solutions

Statistics Homework Solutions Ryan Rosario Chapter 9. retail establishment accepts either the merican Express or the VIS credit card. total of percent of its customers carry an merican Express card, 6

Factoring Polynomials

UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can

15 Prime and Composite Numbers

15 Prime and Composite Numbers Divides, Divisors, Factors, Multiples In section 13, we considered the division algorithm: If a and b are whole numbers with b 0 then there exist unique numbers q and r such

Probability definitions

Probability definitions 1. Probability of an event = chance that the event will occur. 2. Experiment = any action or process that generates observations. In some contexts, we speak of a data-generating

UNIT TWO POLYNOMIALS MATH 421A 22 HOURS. Revised May 2, 00

UNIT TWO POLYNOMIALS MATH 421A 22 HOURS Revised May 2, 00 38 UNIT 2: POLYNOMIALS Previous Knowledge: With the implementation of APEF Mathematics at the intermediate level, students should be able to: -

MATH REVIEW KIT. Reproduced with permission of the Certified General Accountant Association of Canada.

MATH REVIEW KIT Reproduced with permission of the Certified General Accountant Association of Canada. Copyright 00 by the Certified General Accountant Association of Canada and the UBC Real Estate Division.

I. WHAT IS PROBABILITY?

C HAPTER 3 PROBABILITY Random Experiments I. WHAT IS PROBABILITY? The weatherman on 0 o clock news program states that there is a 20% chance that it will snow tomorrow, a 65% chance that it will rain and

MTH6109: Combinatorics. Professor R. A. Wilson

MTH6109: Combinatorics Professor R. A. Wilson Autumn Semester 2011 2 Lecture 1, 26/09/11 Combinatorics is the branch of mathematics that looks at combining objects according to certain rules (for example:

Math 55: Discrete Mathematics

Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 7, due Wedneday, March 14 Happy Pi Day! (If any errors are spotted, please email them to morrison at math dot berkeley dot edu..5.10 A croissant

Homework until Test #2

MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

A fairly quick tempo of solutions discussions can be kept during the arithmetic problems.

Distributivity and related number tricks Notes: No calculators are to be used Each group of exercises is preceded by a short discussion of the concepts involved and one or two examples to be worked out

Introduction to Probability

3 Introduction to Probability Given a fair coin, what can we expect to be the frequency of tails in a sequence of 10 coin tosses? Tossing a coin is an example of a chance experiment, namely a process which

3 Some Integer Functions

3 Some Integer Functions A Pair of Fundamental Integer Functions The integer function that is the heart of this section is the modulo function. However, before getting to it, let us look at some very simple

5544 = 2 2772 = 2 2 1386 = 2 2 2 693. Now we have to find a divisor of 693. We can try 3, and 693 = 3 231,and we keep dividing by 3 to get: 1

MATH 13150: Freshman Seminar Unit 8 1. Prime numbers 1.1. Primes. A number bigger than 1 is called prime if its only divisors are 1 and itself. For example, 3 is prime because the only numbers dividing

number of favorable outcomes total number of outcomes number of times event E occurred number of times the experiment was performed.

12 Probability 12.1 Basic Concepts Start with some Definitions: Experiment: Any observation of measurement of a random phenomenon is an experiment. Outcomes: Any result of an experiment is called an outcome.

Factoring (pp. 1 of 4)

Factoring (pp. 1 of 4) Algebra Review Try these items from middle school math. A) What numbers are the factors of 4? B) Write down the prime factorization of 7. C) 6 Simplify 48 using the greatest common

1.2. Successive Differences

1. An Application of Inductive Reasoning: Number Patterns In the previous section we introduced inductive reasoning, and we showed how it can be applied in predicting what comes next in a list of numbers

Section 6.4: Counting Subsets of a Set: Combinations

Section 6.4: Counting Subsets of a Set: Combinations In section 6.2, we learnt how to count the number of r-permutations from an n-element set (recall that an r-permutation is an ordered selection of r

Chapter 6. 1. What is the probability that a card chosen from an ordinary deck of 52 cards is an ace? Ans: 4/52.

Chapter 6 1. What is the probability that a card chosen from an ordinary deck of 52 cards is an ace? 4/52. 2. What is the probability that a randomly selected integer chosen from the first 100 positive