Chapter 5 Section 5 Counting Techniques

Chapter 5 – Section 5 Learning objectives Solving counting problems using the Multiplication Rule Solving counting problems using permutations Solving counting problems using combinations Solving counting problems involving permutations with nondistinct items Compute probabilities involving permutations and combinations 1 2 3 5 4

Chapter 5 – Section 5 Learning objectives Solving counting problems using the Multiplication Rule Solving counting problems using permutations Solving counting problems using combinations Solving counting problems involving permutations with nondistinct items Compute probabilities involving permutations and combinations 1 2 3 5 4

Chapter 5 – Section 5 The classical method, when all outcomes are equally likely, involves counting the number of ways something can occur This section includes techniques for counting the number of results in a series of choices, under several different scenarios

Chapter 5 – Section 5 Example If there are 3 different colors of paint (red, blue, green) that can be used to paint 2 different types of toy cars (race car, police car), then how many different toys can there be? 3 colors … 2 cars … 3 • 2 = 6 different toys This can be shown in a table or in a tree diagram Example If there are 3 different colors of paint (red, blue, green) that can be used to paint 2 different types of toy cars (race car, police car), then how many different toys can there be?

Chapter 5 – Section 5 A table of the different possibilities This is a rectangle with 2 rows and 3 columns … 2 • 3 = 6 entries Green Police Car Blue Police Car Red Police Car Police Car Green Race Car Blue Race Car Red Race Car Race Car Green Blue Red

Chapter 5 – Section 5 A tree diagram of the different possibilities This also shows that there are 6 possibilities Paint Car Race Police Blue Race Car Blue Police Car Green Race Car Green Police Car Red Race Car Red Police Car Red Blue Green

Chapter 5 – Section 5 The Multiplication Rule of Counting applies to this type of situation If a task consists of a sequence of choices With p selections for the first choice With q selections for the second choice With r selections for the third choice … Then the number of different tasks is p • q • r • …

Chapter 5 – Section 5 Example Part A A child is coloring a picture of a shirt and pants There are 5 different colors of markers How many ways can this be colored? By the multiplication rule 5 • 5 = 25 Example Part A A child is coloring a picture of a shirt and pants There are 5 different colors of markers How many ways can this be colored? Example Part A A child is coloring a picture of a shirt and pants There are 5 different colors of markers Example Part A A child is coloring a picture of a shirt and pants

Chapter 5 – Section 5 Example Part B A child is coloring a picture of a shirt and pants There are 5 different colors of markers The child wants to use 2 different colors How many ways can this be colored? By the multiplication rule 5 • 4 = 20 Example Part B A child is coloring a picture of a shirt and pants There are 5 different colors of markers The child wants to use 2 different colors How many ways can this be colored? Example Part B A child is coloring a picture of a shirt and pants There are 5 different colors of markers Example Part B A child is coloring a picture of a shirt and pants Example Part B A child is coloring a picture of a shirt and pants There are 5 different colors of markers The child wants to use 2 different colors

Chapter 5 – Section 5 Example continued Allowing the same marker to be used twice 5 • 5 = 25 Requiring that there be two different markers 5 • 4 = 20 There are 5 selections for the first choice for both Part A and Part B of this example But they differ for the second choice … there are only 4 selections for Part B Example continued Allowing the same marker to be used twice 5 • 5 = 25 Requiring that there be two different markers 5 • 4 = 20

Chapter 5 – Section 5 Example continued Part A, allowing the same marker to be used twice, is called counting with repetition and has formulas such as 5 • 5 • 5 • … Part B, requiring that there be two different markers, is called counting without repetition and has formulas such as 5 • 4 • 3 • … Example continued Part A, allowing the same marker to be used twice, is called counting with repetition and has formulas such as 5 • 5 • 5 • …

Chapter 5 – Section 5 One way to help write these products is using the factorial symbol n! n! = n • (n-1) • (n-2) • … • 2 • 1 We start off by saying that 0! = 1 and 1! = 1 For example 5! = 5 • 4 • 3 • 2 • 1 = 120 Notice how 5! looks like the 5 • 4 • 3 from the previous example One way to help write these products is using the factorial symbol n! n! = n • (n-1) • (n-2) • … • 2 • 1 One way to help write these products is using the factorial symbol n! n! = n • (n-1) • (n-2) • … • 2 • 1 We start off by saying that 0! = 1 and 1! = 1 For example 5! = 5 • 4 • 3 • 2 • 1 = 120

Chapter 5 – Section 5 Learning objectives Solving counting problems using the Multiplication Rule Solving counting problems using permutations Solving counting problems using combinations Solving counting problems involving permutations with nondistinct items Compute probabilities involving permutations and combinations 1 2 3 5 4

Chapter 5 – Section 5 The problem of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is an example of a permutation A permutation is an ordered arrangement, in which r different objects are chosen out of n different objects with repetition not allowed The number of ways is written nPr The problem of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is an example of a permutation The problem of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is an example of a permutation A permutation is an ordered arrangement, in which r different objects are chosen out of n different objects with repetition not allowed

Chapter 5 – Section 5 The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20 In general, since there are n choices for the first selection, n-1 choices for the second, etc. nPr = n • (n-1) • (n-2) • … • (n-(r-1))‏ The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20 In general, since there are n choices for the first selection, n-1 choices for the second, etc. nPr = n • (n-1) • (n-2) • … • (n-(r-1))‏ is the formula for permutations The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20 In general, since there are n choices for the first selection, n-1 choices for the second nPr = n • (n-1) • (n-2) • … • (n-(r-1))‏ The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20 The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20 In general, since there are n choices for the first selection nPr = n • (n-1) • (n-2) • … • (n-(r-1))‏ Third choice = (n – 2)‏ rth choice = (n – (r-1))‏ Second choice = (n – 1)‏ First choice = (n – 0)‏

Chapter 5 – Section 5 A mathematical way to write the formula for the number of permutations is This is a very convenient mathematical way to write a formula for nPr, but it is not a particularly efficient way to actually compute it In particular, n! gets rapidly gets very large

Chapter 5 – Section 5 Learning objectives Solving counting problems using the Multiplication Rule Solving counting problems using permutations Solving counting problems using combinations Solving counting problems involving permutations with nondistinct items Compute probabilities involving permutations and combinations 1 2 3 5 4

Chapter 5 – Section 5 For some problems, the order of choice does not matter Order matters example Choosing one person to be the president of a club and another to be the vice-president Two different roles Order does not matter example Choosing two people to go to a meeting The same role For some problems, the order of choice does not matter For some problems, the order of choice does not matter Order matters example Choosing one person to be the president of a club and another to be the vice-president Two different roles

Chapter 5 – Section 5 When order does not matter, this is called a combination A combination is an unordered arrangement, in which r different objects are chosen out of n different objects with repetition not allowed The number of ways is written nCr When order does not matter, this is called a combination When order does not matter, this is called a combination A combination is an unordered arrangement, in which r different objects are chosen out of n different objects with repetition not allowed

Chapter 5 – Section 5 Comparing the description of a permutation with the description of a combination The only difference is whether order matters No repetition Out of n objects Choose r objects Order does not matter Order matters Combination Permutation

Chapter 5 – Section 5 To list a combination of r items Any of the r items can be written first Any of the remaining (r-1) items can be written second Etc. Each of these r! lists is the same combination but a different permutation Thus each combination corresponds to r! permutations To list a combination of r items Any of the r items can be written first Any of the remaining (r-1) items can be written second Etc. To list a combination of r items Any of the r items can be written first Any of the remaining (r-1) items can be written second Etc. Each of these r! lists is the same combination but a different permutation

Chapter 5 – Section 5 Example If there are 8 researchers and 3 of them are to be chosen to go to a meeting A combination since order does not matter There are 56 different ways that this can be done Example If there are 8 researchers and 3 of them are to be chosen to go to a meeting Example If there are 8 researchers and 3 of them are to be chosen to go to a meeting A combination since order does not matter

Chapter 5 – Section 5 Because each combination corresponds to r! permutations, the formula nCr for the number of combinations is

Chapter 5 – Section 5 Is a problem a permutation or a combination? One way to tell Write down one possible solution (i.e. Roger, Rick, Randy)‏ Switch the order of two of the elements (i.e. Rick, Roger, Randy)‏ Is this the same result? If no – this is a permutation – order matters If yes – this is a combination – order does not matter Is a problem a permutation or a combination? Is a problem a permutation or a combination? One way to tell Write down one possible solution (i.e. Roger, Rick, Randy)‏ Switch the order of two of the elements (i.e. Rick, Roger, Randy)‏

Chapter 5 – Section 5 Learning objectives Solving counting problems using the Multiplication Rule Solving counting problems using permutations Solving counting problems using combinations Solving counting problems involving permutations with nondistinct items Compute probabilities involving permutations and combinations 1 2 3 5 4

Chapter 5 – Section 5 Our permutation and combination problems so far assume that all n total items are different Sometimes we have a permutations but not all of the n items are different This is a more complicated problem How many ways are there?

Chapter 5 – Section 5 Example How many ways to put 3 A’s, 2 N’s, and 2 T’s to try to make a seven letter sequence? ____ ____ ____ ____ ____ ____ ____ Each of the blanks can be filled in with either an A or a N or a T The three A’s are the same … the two N’s are the same … the two T’s are the same Example How many ways to put 3 A’s, 2 N’s, and 2 T’s to try to make a seven letter sequence? ____ ____ ____ ____ ____ ____ ____ Example How many ways to put 3 A’s, 2 N’s, and 2 T’s to try to make a seven letter sequence? ____ ____ ____ ____ ____ ____ ____ Each of the blanks can be filled in with either an A or a N or a T

Chapter 5 – Section 5 Example continued Where can the A’s go? There are 7 possible places Any 3 of them are possible Order does not matter So 7C3 different ways to put in the A’s

Chapter 5 – Section 5 Example continued Where can the N’s go? There are 4 possible places (since 3 of the 7 have been taken by the A’s already)‏ Any 2 of them are possible Order does not matter So 4C2 different ways to put in the N’s And there are 2C2 different ways to put in the T’s

Chapter 5 – Section 5 Example continued Altogether there are 7C3 • 4C2 • 2C2 different ways This is Notice that the denominator is 3, 2, 2 … the numbers of each letter Example continued Altogether there are 7C3 • 4C2 • 2C2 different ways Example continued Altogether there are 7C3 • 4C2 • 2C2 different ways This is

Chapter 5 – Section 5 The general formula for the number of permutations of n total objects where there are n1 of the first kind n2 objects of the second kind … and nk of the kth kind is

Chapter 5 – Section 5 Learning objectives Solving counting problems using the Multiplication Rule Solving counting problems using permutations Solving counting problems using combinations Solving counting problems involving permutations with nondistinct items Compute probabilities involving permutations and combinations 1 2 3 5 4

Chapter 5 – Section 5 Probabilities using the classical method involve counting the number of possibilities Often the number of possibilities is some permutation or some combination The permutation / combination formulas can be used to calculate probabilities

Chapter 5 – Section 5 A permutation example In a horse racing “Trifecta”, a gambler must pick which horse comes in first, which second, and which third If there are 8 horses in the race, and every order of finish is equally likely, what is the chance that any ticket is a winning ticket? Order matters, so this is a permutations problem A permutation example In a horse racing “Trifecta”, a gambler must pick which horse comes in first, which second, and which third If there are 8 horses in the race, and every order of finish is equally likely, what is the chance that any ticket is a winning ticket?

Chapter 5 – Section 5 A permutation example continued There are 8P3 permutations of the order of finish of the horses The probability that any one ticket is a winning ticket is 1 out of 8P3, or 1 out of 56

Chapter 5 – Section 5 A combination example The Powerball lottery consists of choosing 5 numbers out of 55 and then 1 number out of 42 The grand prize is given out when all 6 numbers are correct What is the chance of getting the grand prize? Order does not matter, so this is a combinations problem (for the 5 balls)‏ A combination example The Powerball lottery consists of choosing 5 numbers out of 55 and then 1 number out of 42 The grand prize is given out when all 6 numbers are correct What is the chance of getting the grand prize?

Chapter 5 – Section 5 A combination example continued There are 55C5 combinations of the 5 numbers There are 42 possibilities for the last ball, so the probability of the grand prize is 1 out of which is pretty small A combination example continued There are 55C5 combinations of the 5 numbers

Summary: Chapter 5 – Section 5 The Multiplication Rule counts the number of possible sequences of items Permutations and combinations count the number of ways of arranging items, with permutations when the order matters and combinations when the order does not matter Permutations and combinations are used to compute probabilities in the classical method

Example Phil Black, a former Navy SEAL instructor, has developed a game called the Push Up Game (PUG). The game consists of a drawing a card from the FitDeck, a pack of 56 cards, and doing the exercise listed on the card (many of these are variations on push-ups.) Black claims the FitDeck cards can be shuffled to produce more than 3 trillion workouts. (Source: Ryckman, L. Scripps Howard News Service, Push Upmanship, Post Register. December 5, 2005.) Assuming all the cards are unique, how many different workouts can be created by shuffling the deck and then doing the exercise on each card in the order they are drawn? Is Black’s claim accurate?

(56! ≈ 7.11x1074; yes)

Example The dial on a combination lock shows the integers 0 through 39. Turning the dial to three specific numbers (in order) opens the lock. a. How many different locks can a manufacturer produce before they have to reuse combinations, if numbers cannot be repeated in a combination (i.e., 25-17-25 is not allowed)? b. How many different locks can a manufacturer produce before they have to reuse the combination if numbers can be repeated in a combination?

(40P3 = 59,280) (403=64,000)

Example A bride is trying to design her wedding ring. Based on her tastes and the groom’s finances, she has limited her choices to: five different bands, three diamonds, and four metals (platinum, gold, white gold, and silver). How many different ring combinations can she choose? A student places four books in her backpack. How many ways can she order these books?

(5·3·4 = 60 ring combinations) (4! = 24)

Example A statistics professor is preparing an exam from a test bank. If the test bank consists of 30 problems and the professor will include 10 of them, how many different ways can the professor select the problems? (The order of the questions on the exam is not important.)

(30C10 = 30,045,015)

Example A statistics professor has written an exam with 10 questions. To discourage cheating, this professor changes the order of the questions and creates several versions of the exam. In how many ways can these questions be arranged?

(10! = 3,628,800)