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Thinking Mathematically. Chapter 11: Counting Methods and Probability Theory. Thinking Mathematically. Section 1: The Fundamental Counting Principle. The Fundamental Counting Principle.

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## Thinking Mathematically

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**Thinking Mathematically**Chapter 11: Counting Methods andProbability Theory**Thinking Mathematically**Section 1: The Fundamental Counting Principle**The Fundamental Counting Principle**If you can choose one item from a group of M items and a second item from a group of N items, then the total number of two-item choices is M N. You the numbers! MULTIPLY**cereal**eggs pancakes OJ apple OJ apple OJ apple 1 2 3 4 5 6 The Fundamental Counting Principle At breakfast, you can have eggs, pancakes or cereal. You get a free juice with your meal: either OJ or apple juice. How many different breakfasts are possible?**Example: Applying the Fundamental Counting Principle**• The Greasy Spoon Restaurant offers 6 appetizers and 14 main courses. How many different meals can be created by selecting one appetizer and one main course? • Using the fundamental counting principle, there are 14 6 = 84 different ways a person can order a two-course meal.**Example: Applying the Fundamental Counting Principle**• This is the semester that you decide to take your required psychology and social science courses. • Because you decide to register early, there are 15 sections of psychology from which you can choose. Furthermore, there are 9 sections of social science that are available at times that do not conflict with those for psychology. In how many ways can you create two-course schedules that satisfy the psychology-social science requirement?**Solution**The number of ways that you can satisfy the requirement is found by multiplying the number of choices for each course. You can choose your psychology course from 15 sections and your social science course from 9 sections. For both courses you have: 15 9, or 135 choices.**The Fundamental Counting Principle**The number of ways a series of successive things can occur is found by multiplying the number of ways in which each thing can occur.**Example: Options in Planning a Course Schedule**Next semester you are planning to take three courses - math, English, and humanities. Based on time blocks and highly recommended professors, there are 8 sections of math, 5 of English, and 4 of humanities that you find suitable. Assuming no scheduling conflicts, there are: 8 5 4 = 160 different three course schedules.**Example**Car manufacturers are now experimenting with lightweight three-wheeled cars, designed for a driver and one passenger, and considered ideal for city driving. Suppose you could order such a car with a choice of 9 possible colors, with or without air-conditioning, with or without a removable roof, and with or without an onboard computer. In how many ways can this car be ordered in terms of options?**Solution**This situation involves making choices with four groups of items. color - air-conditioning - removable roof - computer 9 2 2 2 = 72 Thus the car can be ordered in 72 different ways.**Example: A Multiple Choice Test**You are taking a multiple-choice test that has ten questions. Each of the questions has four choices, with one correct choice per question. If you select one of these options per question and leave nothing blank, in how many ways can you answer the questions?**Solution**We DON’T blindly multiply the first two numbers we see. The answer is not 10 4 = 40. We use the Fundamental Counting Principle to determine the number of ways you can answer the test. Multiply the number of choices, 4, for each of the ten questions 4 4 4 4 4 4 4 4 4 4 =1,048,576**Example: Telephone Numbers in the United States**Telephone numbers in the United States begin with three-digit area codes followed by seven-digit local telephone numbers. Area codes and local telephone numbers cannot begin with 0 or 1. How many different telephone numbers are possible?**Solution**We use the Fundamental Counting Principle to determine the number of different telephone numbers that are possible. 8 10 10 8 10 10 10 10 10 10 =6,400,000,000**Thinking Mathematically**Section 2: Permutations**Permutations**• A permutation is an arrangement of objects. • No item is used more than once. • The order of arrangement makes a difference.**Example: Counting Permutations**Based on their long-standing contribution to rock music, you decide that the Rolling Stones should be the last group to perform at the four-group Offspring, Pink Floyd, Sublime, Rolling Stones concert. Given this decision, in how many ways can you put together the concert?**Solution**We use the Fundamental Counting Principle to find the number of ways you can put together the concert. Multiply the choices: 3 2 1 1 = 6 Thus, there are six different ways to arrange the concert if the Rolling Stones are the final group to perform. 3 choices 2 choices 1 choice 1 choice whichever of the two remaining offspring pink floyd sublime only one remaining stones**Example: Counting Permutations**You need to arrange seven of your favorite books along a small shelf. How many different ways can you arrange the books, assuming that the order of the books makes a difference to you?**Solution**You may choose any of the seven books for the first position on the shelf. This leaves six choices for second position. After the first two positions are filled, there are five books to choose from for the third position, four choices left for the fourth position, three choices left for the fifth position, then two choices for the sixth position, and only one choice left for the last position. 7 6 5 4 3 2 1 = 5040 There are 5040 different possible permutations.**Factorial Notation**If n is a positive integer, the notation n! is the product of all positive integers from n down through 1. n! = n(n-1)(n-2)…(3)(2)(1) note that 0!, by definition, is 1. 0!=1**Permutations of n Things Taken r at a Time**The number of permutations possible if ritems are taken from n items: n! nPr= = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1) (n – r)! n! = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1) (n - r) (n - r- 1) . . . (2)(1) (n – r)! = (n - r) (n - r- 1) . . . (2)(1)**Permutations of n Things Taken r at a Time**The number of permutations possible if ritems are taken from n items: nPr: starting at n, write down r numbers going down by one: nPr = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1) 1 2 3 4 r**Problem**A math club has eight members, and it must choose 5 officers --- president, vice-president, secretary, treasurer and student government representative. Assuming that each office is to be held by one person and no person can hold more than one office, in how many ways can those five positions be filled? We are arranging 5 out of 8 people into the five distinct offices. Any of the eight can be president. Once selected, any of the remaining seven can be vice-president. Clearly this is an arrangement, or permutation, problem. 8P5 = 8!/(8-5)! = 8!/3! = 8 · 7 · 6 · 5 · 4 = 6720**Permutations with duplicates.**• In how many ways can you arrange the letters of the word minty? • That's 5 letters that have to be arranged, so the answer is 5P5 = 5! = 120 • But how many ways can you arrange the letters of the word messes? • You would think 6!, but you'd be wrong!**messes**here are six permutations of messes m e s s e s 1 m e s s e s well, all 3! arrangements of the s's look the same to me!!!! This is true for any arrangement of the six letters in messes, so every six permutations should count only once. The same applies for the 2! arrangement of the e's 2 m e s s e s m e s s e s 3 m e s s e s 4 m e s s e s 5 6**Permutations with duplicates.**• How many ways can you arrange the letters of the word messes? • The problem is that there are three s's and 2 e's. It doesn't matter in which order the s's are placed, because they all look the same! • This is called permutations with duplicates.**Permutations with duplicates.**• Since there are 3! = 6 ways to arrange the s's, there are 6 permutations that should count as one. Same with the e's. There are 2! = 2 permutations of them that should count as 1. • So we divide 6! by 3! and also by 2! • There are 6!/3!2! = 720/12 = 60 ways to arrange the word messes.**Permutations with duplicates.**• In general if we want to arrange n items, of which m1, m2, .... are identical, the number of permutations is**Problem**A signal can be formed by running different colored flags up a pole, one above the other. Find the number of different signals consisting of 6 flags that can be made if 3 of the flags are white, 2 are red, and 1 is blue 6!/3!2!1! = 720/(6)(2)(1) = 720/12 = 60**Thinking Mathematically**Section 3: Combinations**Combination: definition**A combination of items occurs when: • The item are selected from the same group. • No item is used more than once. • The order of the items makes no difference.**How to know when the problem is a permutation problem or a**combination problem • Permutation: • arrangement, arrange • order matters • Combination • selection, select • order does not matter.**Example: Distinguishing between Permutations and**Combinations • For each of the following problems, explain if the problem is one involving permutations or combinations. • Six students are running for student government president, vice-president, and treasurer. The student with the greatest number of votes becomes the president, the second biggest vote-getter becomes vice-president, and the student who gets the third largest number of votes will be student government treasurer. How many different outcomes are possible for these three positions?**Solution**• Students are choosing three student government officers from six candidates. The order in which the officers are chosen makes a difference because each of the offices (president, vice-president, treasurer) is different. Order matters. This is a problem involving permutations.**Example: Distinguishing between Permutations and**Combinations • Six people are on the volunteer board of supervisors for your neighborhood park. A three-person committee is needed to study the possibility of expanding the park. How many different committees could be formed from the six people on the board of supervisors?**Solution**• A three-person committee is to be formed from the six-person board of supervisors. The order in which the three people are selected does not matter because they are not filling different roles on the committee. Because order makes no difference, this is a problem involving combinations.**Example: Distinguishing between Permutations and**Combinations • Baskin-Robbins offers 31 different flavors of ice cream. One of their items is a bowl consisting of three scoops of ice cream, each a different flavor. How many such bowls are possible?**Solution**• A three-scoop bowl of three different flavors is to be formed from Baskin-Robbin’s 31 flavors. The order in which the three scoops of ice cream are put into the bowl is irrelevant. A bowl with chocolate, vanilla, and strawberry is exactly the same as a bowl with vanilla, strawberry, and chocolate. Different orderings do not change things, and so this problem is combinations.**Combinations of n Things Taken rat a Time**n n! = nCr = r!(n – r)! r Note that the sum of the two numbers on the bottom (denominator) should add up to the number on the top (numerator).**Computing Combinations**• Suppose we need to compute 9C3 • r = 3, n – r = 6 • The denominator is the factorial of smaller of the two: 3!**Computing Combinations**• Suppose we need to compute 9C3 • r = 3, n – r = 6 • In the numerator write (the product of) all the numbers from 9 down to n - r + 1 = 6 + 1 = 7: • There should be the same number of terms in the numerator and denominator: 9 8 7**Computing Combinations**• If called upon, there's a fairly easy way to compute combinations. • Given nCr , decide which is bigger: r or n – r. • Take the smaller of the two and write out the factorial (of the number you picked) as a product. • Draw a line over the expression you just wrote.**Computing Combinations**• If called upon, there's a fairly easy way to compute combinations. • Now, put n directly above the line and directly above the leftmost number below. • Eliminate common factors in the numerator and denominator. • Do the remaining multiplications. • You're done!**Computing Combinations**• Suppose we need to compute 9C3 . • n – r= 6, and thesmaller of 3 and 6 is 3. 3 4 9 8 7 = 3 4 7 = 84 3 2 1 1 1**Problem**• A three-person committee is to be formed from the eight-person board of supervisors. We saw that this is a combination problem 2 = 8 7 = 56**A deck of cards**suits ranks aces face cards**Problem**• How many poker hands (five cards) are possible using a standard deck of 52 cards? • We need to pick any five cards from the deck. Therefore we are selecting 5 out of 52 cards.**Problem**• A poker hand of afull house consists of three cards of the same rank (e.g. three 8's or three kings or three aces.) and two cards of another rank. How many full houses are possible? • First we need to select 1 out of the 13 possible ranks. 13 • Then we need to select 3 out of the 4 cards of that rank. 4C3 = 4 • By the fundamental counting principle, there are 13 4 = 52 ways to do this part of the problem.

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