Idea Transcript
LESSON
19.3
Name
Combinations and Probability
Class
19.3 Combinations and Probability Essential Question: What is the difference between a permutaion and a combination?
Resource Locker
Common Core Math Standards
A combination is a selection of objects from a group in which order is unimportant. For example, if 3 letters are chosen from the group of letters A, B, C, and D, there are 4 different combinations.
S-CP.9(+) Use permutations and combinations to compute probabilities of compound events and solve problems.
ABC
MP.7 Using Structure Give an example of a combination to a partner and explain how you know it is a combination.
ENGAGE
View the Engage section online. Discuss the photograph. Ask students to guess the location where the photo was taken and describe what is happening there. Then preview the Lesson Performance Task.
ACD
BCD
Appetizers
Language Objective
Nachos
Chicken Wings
Chicken Quesadilla
Vegetarian Egg Rolls
Potato Skins
Soft Pretzels
Beef Chili
Guacamole Dip
© Houghton Mifflin Harcourt Publishing Company • ©Tom Henderson/The Food Passionates/Corbis
PREVIEW: LESSON PERFORMANCE TASK
ABD
A restaurant has 8 different appetizers on the menu, as shown in the table. They also offer an appetizer sampler, which contains any 3 of the appetizers served on a single plate. How many different appetizer samplers can be created? The order in which the appetizers are selected does not matter.
Mathematical Practices
You can choose a number of objects in such a way that the order matters, in which case you choose a permutation, or you can choose in such a way that order does not matter, in which case you choose a combination.
Finding the Number of Combinations
Explore
The student is expected to:
Essential Question: What is the difference between a permutation and a combination?
Date
Find the number appetizer samplers that are possible if the order of selection does matter. This is the number of permutations of 8 objects taken 3 at a time.
8! 8! P 3 = __ = _ = 336 5! 8 - 3 !
(
8
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Find the number of different ways to select a particular group of appetizers. This is the number of permutations of 3 objects.
3! 3! P 3 = __ = _ = 6 0! 3 - 3 !
3
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HARDCOVER PAGES 707714 Turn to these pages to find this lesson in the hardcover student edition.
Appetizers
Nachos
Chicken Wings Egg Rolls Vegetarian ls Soft Pretze Dip Guacamole
dilla
Chicken Quesa
. does matter of selection the order possible if ers that are at a time. zer sampl s taken 3 s of 8 object number appeti Find the permutation number of This is the 8! 8! _ = 336 = _ _ = 5! 8P 3 This is the ! appetizers. 8 - 3 group of a particular ways to select different s. number of Find the s of 3 object permutation number of 3! 3! _= 6 = __ = 0! 3P 3 ! 3 - 3
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Lesson 3 973 Module 19
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To find the number of possible appetizer samplers if the order of selection does not matter, divide the answer to part A by the answer to part B. 336 So the number of appetizer samplers that can be created is _ = 56 . 6
EXPLORE Finding the Number of Combinations
Reflect
1.
Explain why the answer to Part A was divided by the answer to Part B. Since the order of selection does not matter, the answer to Part A contained duplications
INTEGRATE TECHNOLOGY
of each possible sampler. Dividing by the answer to Part B removed the duplicates of each
Students have the option of doing the Explore activity either in the book or online.
sampler. 2.
On Mondays and Tuesdays, the restaurant offers an appetizer sampler that contains any 4 of the appetizers listed. How many different appetizer samplers can be created? 1680 8P 4 = ____ = 70. The number of appetizer samplers that can be created is ___ 24 P 4
3.
QUESTIONING STRATEGIES
4
In general, are there more ways or fewer ways to select objects when the order does not matter? Why? There are fewer ways to select objects when the order does not matter. This is because
Why is it important to distinguish between a permutation and a combination when counting? They may have a different number of possible arrangements.
multiple selections are counted as the same combination.
Explain 1
Finding a Probability Using Combinations
How can you tell if an arrangement is a combination? An arrangement is a combination if the order of the items does not matter.
The results of the Explore can be generalized to give a formula for combinations. In the Explore, the number of combinations of the 8 objects taken 3 at a time is
_
3! 8! 0! = _ 8! 8! 1! ÷ ·_ ·_ P ÷ 3P3 = _ =_ (8 - 3)! (8 - 3)! 3! (8 - 3)! 3! (3 - 3)! This can be generalized as follows. 8 3
Combinations The number of combinations of n objects taken r at a time is given by
Example 1
n! Cr = _ r!(n - r)!
Find each probability.
There are 4 boys and 8 girls on the debate team. The coach randomly chooses 3 of the students to participate in a competition. What is the probability that the coach chooses all girls? The sample space S consists of combinations of 3 students taken from the group of 12 students. 12! = 220 n(S) = 12C 3 = _ 3!9! Event A consists of combinations of 3 girls taken from the set of 8 girls. 8! = 56 n(A) = 8C 3 = _ 3!5! n(A) 56 = _ 14 . The probability that the coach chooses all girls is P(A) = _ = _ 55 220 n(S)
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EXPLAIN 1
© Houghton Mifflin Harcourt Publishing Company
n
Finding a Probability Using Combinations INTEGRATE TECHNOLOGY
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Math Background
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Graphing calculators have a built-in function that calculates combinations. The nC r function is available from the MATH PRB menu. Have students enter data in the same way they enter n and r when applying the permutation function. For example, to find 6C 4, first enter 6. Then press MATH and use the arrow keys to choose the PRB menu. Select 3:nCr and press ENTER. Now enter 4 and press ENTER to see that 6C 4 = 15.
In this lesson students are introduced to combinations. A combination of n n! . Pascal’s triangle is a objects taken r at a time is given by the rule nC r = ________ r!(n - r)! number triangle with rows arranged according to the combination formula, starting with n = 0 and continuing indefinitely. Each successive row can be found by adding elements from the row above. It has many fascinating properties, and it can be used as a shortcut in algebra when factoring polynomials.
Combinations and Probability 974
B
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Check that students understand how the
There are 52 cards in a standard deck, 13 in each of 4 suits: clubs, diamonds, hearts, and spades. Five cards are randomly drawn from the deck. What is the probability that all five cards are diamonds? The sample space S consists of combinations of n(S) =
formula for the number of combinations of n objects taken r at a time is related to the formula for the number of permutations of n objects taken r at a time. The number of combinations is equal to the number of permutations divided by r!.
52
C
5
52! = _ = 2,598,960 5!47!
Event A consists of combinations of 5 cards drawn from the 13 diamonds. n(A) =
13
C
5
13! =_= 5!8!
1287
The probability of randomly selecting
5 cards that are diamonds is
1287 33 n(A) P(A) = _ = __ = _. ( ) nS 2,598,960 66, 640
QUESTIONING STRATEGIES The formula for the number of combinations is a ratio. Can a combination ever be a fraction less than 1 ? Explain. No; as with the formula for permutations, although the formula is a ratio, it always simplifies to a whole number because it is a method for counting arrangements.
Your Turn
4.
A coin is tossed 4 times. What is the probability of getting exactly 3 heads?
The number of outcomes in the sample space S is found by using the Fundamental Counting Principle since each flip can result in heads or tails. n(S) = 2 · 2 · 2 · 2 = 2 4 = 16 Event A consists of combinations of 3 heads taken from the set of 4 coin flips, so 4! n(A) = 4C 3 = ___ =4 3!1!
The probability of getting exactly 3 heads is © Houghton Mifflin Harcourt Publishing Company
Can the number of combinations ever be equal to the number of permutations? Explain. only when the combination is taken 1 at a time, when r = 1, which is the same as a permutation
5 cards drawn from 52 cards.
n(A)
4 1 P(A) = ____ = __ =_ 4 16 n(S)
5.
A standard deck of cards is divided in half, with the red cards (diamonds and hearts) separated from the black cards (spades and clubs). Four cards are randomly drawn from the red half. is the probability they are all diamonds? The sample space S consists of combinations of 4 cards drawn from the 26 red cards, so 26! n(S) = 26C 4 = ____ = 14, 950. 4!22!
Event A consists of combinations of 4 cards drawn from the 13 diamonds, so 13! n(A) = 13C 4 = ___ = 715. 4!9!
The probability of getting all diamonds is n(A)
715 11 P(A) = ____ = _____ = ___ . 14, 950 230 n(S)
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Small Group Activity Have students create two situations in which 2 out of 16 objects are selected, with order important in one of the situations and not important in the other. Ask students to use the situations to compare and contrast permutations and combinations, explaining why the number of selections is greater in one than in the other, and describing the relationship between the two. Tell students it may be necessary to create several situations in which order is and is not important to be able to describe the relationship between permutations and combinations.
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Explain 2
Finding a Probability Using Combinations and Addition
AVOID COMMON ERRORS Students may have difficulty using combinations to find probability. Have students break down the probability problem into parts. First find the size of the sample space. Then find the number of outcomes associated with the event. Finally, write the ratio.
Sometimes, counting problems involve the phrases “at least” or “at most.” For these problems, combinations must be added. For example, suppose a coin is flipped 3 times. The coin could show heads 0, 1, 2, or 3 times. To find the number of combinations with at least 2 heads, add the number of combinations with 2 heads and the number of combinations with 3 heads ( 3C 2 + 3C 3). Example 2
Find each probability.
A coin is flipped 5 times. What is the probability that the result is heads at least 4 of the 5 times?
EXPLAIN 2
The number of outcomes in the sample space S can be found by using the Fundamental Counting Principle since each flip can result in heads or tails. n(S) = 2 · 2 · 2 · 2 · 2 = 2 5 = 32
Finding a Probability Using Combinations and Addition
Let A be the event that the coin shows heads at least 4 times. This is the sum of 2 events, the coin showing heads 4 times and the coin showing heads 5 times. Find the sum of the combinations with 4 heads from 5 coins and with 5 heads from 5 coins. 5! + _ 5! = 5 + 1 = 6 n(A) = 5C 4 + 5C 5 = _ 4!1! 5!0! n(A) 6 =_ 3. The probability that the coin shows at least 4 heads is P(A) = _ = _ 32 16 n(S)
INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Students may have difficulty recognizing how
Three number cubes are rolled and the result is recorded. What is the probability that at least 2 of the number cubes show 6?
n(S) = 6
3
© Houghton Mifflin Harcourt Publishing Company • ©Eldad Carin/Shutterstock
The number of outcomes in the sample space S can be found by using the Fundamental Counting Principle since each roll can result in 1, 2, 3, 4, 5, or 6. = 216
Let A be the event that at least 2 number cubes show 6. This is the sum of 2 events, 2 number cubes showing 6 or 3 number cubes showing 6. The event of getting 6 on 2 number cubes occurs 5 times since there are 5 possibilities for the other number cube. n(A) =
3! 3! ___ ___ 5 · 3C 2 + 3C 3 = 5 · 2!1! + 3!0! = 15 + 1 = 16
16 2 n(A) The probability of getting a 6 at least twice in 3 rolls is P(A) = _ = _ = _. n(S) 27 216
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to use addition with combinations, and understanding why addition is used. Review how to find the probability of simple events that involve addition when rolling a number cube, such as rolling at least a 3 or at most a 4.
QUESTIONING STRATEGIES When you find a probability using combinations, will both parts of the probability ratio necessarily be combinations? Explain. No, the method used for counting each part of the ratio depends on the problem.
Lesson 3
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Graphic Organizers
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Have the class work together to create a graphic organizer that summarizes and demonstrates the rules for counting: the Fundamental Counting Principle, the permutation rule, and the combination rule. For each rule, have students include the formula and an example. Suggest that students highlight words in the examples that indicate why and how the rule is applied.
Combinations and Probability 976
Your Turn
ELABORATE
6.
AVOID COMMON ERRORS
A math department has a large database of true-false questions, half of which are true and half of which are false, that are used to create future exams. A new test is created by randomly selecting 6 questions from the database. What is the probability the new test contains at most 2 questions where the correct answer is “true”?
The number of outcomes in the sample space S can be found by using the Fundamental
Students sometimes compute a combination for n(S) and then choose the numerator of the probability ratio carelessly. Emphasize the importance of accurately identifying both parts of the probability ratio.
Counting Principle since each question is either true or false. n(S) = 2 · 2 · 2 · 2 · 2 · 2 = 2 6 = 64 Let A be the event that at most 2 questions are true. This is the sum of 3 events: 2 true questions, 1 true question, or no true questions. 6! 6! 6! n(A) = 6C 2 + 6C 1 + 6C 0 = ___ + ___ + ___ = 15 + 6 + 1 = 22 2!4! 1!5! 0!6!
Because the questions are equally likely to be true or false, the probability that the test contains at most 2 true questions is
SUMMARIZE THE LESSON
n(A)
22 11 P(A) = ____ = __ = __ 64 32 n(S)
What are combinations and how can you use them to calculate probabilities? A combination is a grouping of objects in which order does not matter. You can use combinations to find the number of outcomes in a sample space or in an event.
7.
There are equally many boys and girls in the senior class. If 5 seniors are randomly selected to form the student council, what is the probability the council will contain at least 3 girls?
The number of outcomes in the sample space S an be found by using the Fundamental Counting Principle since each selection is either a boy or a girl. n(S) = 2 · 2 · 2 · 2 · 2 = 2 5 = 32 Let A be the event that at least 3 girls are selected. This is the sum of 3 events: selecting 3 girls, 4 girls, or 5 girls. 5! 5! 5! n(A) = 5C 3 + 5C 4 + 5C 5 = ___ + ___ + ___ = 10 + 5 + 1 = 16 3!2! 4!1! 5!0!
Because a senior is equally likely to be a boy or a girl, the probability that the council will
© Houghton Mifflin Harcourt Publishing Company
contain at least 3 girls is n(A)
16 1 P(A) = ____ = __ =_ 32 2 n(S)
Elaborate 8.
Discussion A coin is flipped 5 times, and the result of heads or tails is recorded. To find the probability of getting tails at least once, the events of 1, 2, 3, 4, or 5 tails can be added together. Is there a faster way to calculate this probability? The sum of the probabilities of all possible outcomes is equal to 1. Determine the probability of getting no tails (or 5 heads) and subtract this value from 1.
9.
If nC a = nC b , what is the relationship between a and b? Explain your answer. n! n! = _______ . This will occur when a = b or a + b = n. The equation is true if _______ a!(n - a)!
b!(n - b)!
10. Essential Question Check-In How do you determine whether choosing a group of objects involves combinations? Combinations are used when the order of selection does not matter. Module 19
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Connect Vocabulary Have students make a chart to summarize what they know about combinations. Sample:
Combination Formula
A grouping of objects in which order does not matter n! _________ nC r = r!(n - r)!
Example
Combinations of 2 letters from A, B, and C: AB AC BC
Definition
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Evaluate: Homework and Practice 1.
A cat has a litter of 6 kittens. You plan to adopt 2 of the kittens. In how many ways can you choose 2 of the kittens from the litter?
C2 =
6
2.
EVALUATE • Online Homework • Hints and Help • Extra Practice
720 6! 6! __ = _ = _ = 15 ways 2!(6 - 2)!
48
2!4!
ASSIGNMENT GUIDE
An amusement park has 11 roller coasters. In how many ways can you choose 4 of the roller coasters to ride during your visit to the park? 39,916,800 11! 11! = = = 330 ways 11C 4 = 120,960 4!7! 4!(11 - 4)!
__ _ __
3.
Four students from 30-member math club will be selected to organize a fundraiser. How many groups of 4 students are possible?
C4 =
30
4.
30! 30! __ = _ = 27,405 groups 4!(30 - 4)!
A school has 5 Spanish teachers and 4 French teachers. The school’s principal randomly chooses 2 of the teachers to attend a conference. What is the probability that the principal chooses 2 Spanish teachers? The sample space S consists of combinations of 2 teachers chosen from the 9 teachers, and event A consists of combinations of 2 teachers chosen from the 5 Spanish teachers. 362,880 9! 9! = = = 36 n(S) = 9C 2 = 10,080 2!7! 2!(9 - 2)!
n(A) = 5C 2 =
Explore Finding the Number of Combinations
Exercises 1–3, 22–23, 25–26
Example 1 Finding a Probability Using Combinations
Exercises 4 –7, 12–21
Example 2 Finding a Probability Using Combinations and Addition
Exercises 8–11
COMMUNICATING MATH
5! 5! 120 _ = _ = _ = 10
After students have completed some of the exercises, ask them to discuss how they are able to distinguish combinations and permutations. Then have them explain how they were able to identify n and r in various examples. Ask them to tell whether they find one easier to work with than the other, and why.
2!(5 - 2)!
2!3!
12
n(A) _ 5 10 _ _ = =
18 36 n(S) There are 6 fiction books and 8 nonfiction books on a reading list. Your teacher randomly assigns you 4 books to read over the summer. What is the probability that you are assigned all nonfiction books? The sample space S consists of combinations of 4 books chosen from the 14 books, and event A consists of combinations of 4 books chosen from the 8 nonfiction books. n(S) = 14C 4 =
14! 14! __ = _ = 1001
n(A) = 8C 4 =
8! 8! _ = _ = 70
P(A) =
Practice
_ _ _
© Houghton Mifflin Harcourt Publishing Company
P(A) =
5.
4!26!
Concepts and Skills
4!(14 - 4)!
4!(8 - 4)!
4!10!
4!4!
n(A) _ 10 70 _ = =_ n(S)
1001
143
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Depth of Knowledge (D.O.K.)
Mathematical Practices
1–12
1 Recall of Information
MP.2 Reasoning
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2 Skills/Concepts
MP.4 Modeling
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2 Skills/Concepts
MP.1 Problem Solving
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2 Skills/Concepts
MP.4 Modeling
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2 Skills/Concepts
MP.2 Reasoning
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MP.3 Logic
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Combinations and Probability 978
A bag contains 26 tiles, each with a different letter of the alphabet written on it. You choose 3 tiles from the bag without looking. What is the probability that you choose the tiles with the letters A, B, and C?
6.
INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Ask students to share their rationales for
Let S be the sample space, which consists of combinations of 3 tiles chosen from 26 tiles, and let A be the event that you choose the tiles with the letters A, B, and C.
solving probability problems involving combinations. In particular, have them explain how they computed the values for n(S) and n(A) to find the probability.
n(S) = 26C 3 =
26! 26! __ = _ = 2600
n(A) = 3C 3 =
n(A) 3! 3! 1 __ = _ = 1 ⇢ P(A) = _ = _
3!(26 - 3)!
3!(3 - 3)!
3!23!
n(S)
3!0!
2600
You are randomly assigned a password consisting of 6 different characters chosen from the digits 0 to 9 and the letters A to Z. As a percent, what is the probability that you are assigned a password consisting of only letters? Round you answer to the nearest tenth of a percent.
7.
AVOID COMMON ERRORS Students often confuse permutations and combinations. They may not recognize which should be applied or they may apply the wrong formula. Have students begin by deciding whether the order is important or not. Then have them look up the formula. Note that this process may have to be repeated several times to solve a probability problem.
Let S be the sample space, which consists of combinations of 6 characters chosen from 36 characters, and let A be the event that you are assigned a password consisting of only letters.
_
36! 36! = 1,947,792 n(S) = 36C 6 = __ = 6!30! 6!(36 - 6)! n(A) = 26C 6 = P(A) =
26! 26! __ = _ = 230,230 6!(26 - 6)!
6!20!
n(A) _ 230,230 _ = ≈ 11.8%
1,947,792 n(S) A bouquet of 6 flowers is made up by randomly choosing between roses and carnations. What is the probability the bouquet will have at most 2 roses?
8.
© Houghton Mifflin Harcourt Publishing Company
Let S be the sample space, which consists of combinations of 6 flowers (each of which is a rose or a carnation), and let A be the event that the bouquet will have at most 2 roses. n(S) = 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 = 2 6 = 64
n(A) = 6C 2 + 6C 1 + 6C 0 = 15 + 6 + 1 = 22
Because a flower is equally likely to be a rose or a carnation, P(A) =
n(A) _ 11 22 _ _ . = = n(S)
64
32
A bag of fruit contains 10 pieces of fruit, chosen randomly from bins of apples and oranges. What is the probability the bag contains at least 6 oranges?
9.
Let S be the sample space, which consists of combinations of 3 tiles chosen from 26 tiles, and let A be the event that the bag contains at least 6 oranges. n(S) = 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 = 2 10 = 1024
n(A) = 10C 6 + 10C 7 + 10C 8 + 10C 9 + 10C 10 = 210 + 120 + 45 + 10 + 1 = 386 Because a piece of fruit is equally likely to be an apple or an orange, n(A) 193 386 . P(A) = = = 512 1024 n(S)
_ _ _
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Mathematical Practices
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3 Strategic Thinking
MP.3 Logic
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3 Strategic Thinking
MP.3 Logic
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3 Strategic Thinking
MP.3 Logic
10. You flip a coin 10 times. What is the probability that you get at most 3 heads?
PEERTOPEER
Let S be the sample space, which consists of the results of 10 coin flips (each of which is either heads or tails), and let A be the event that you get at most 3 heads.
Have students work with a partner to write a probability problem about numbers, letters, or both in which a combination is needed to count the objects. Have students solve their problems. Then ask them to exchange problems with another pair to solve. Have the pairs review each other’s solution methods.
n(S) = 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 = 2 10 = 1024
n(A) = 10C 3 + 10C 2 + 10C 1 + 10C 0 = 120 + 45 + 10 + 1 = 176 Because a flip is equally likely to result in heads or tails, P(A) =
n(A) _ 176 11 _ = = _. n(S)
1024
64
11. You flip a coin 8 times. What is the probability you will get at least 5 heads?
Let S be the sample space, which consists of the results of 8 coin flips (each of which is either heads or tails), and let A be the event that you get at least 5 heads. n(S) = 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 = 2 8 = 256
n(A) = 8C 5 + 8C 6 + 8C 7 + 8C 8 = 56 + 28 + 8 + 1 = 93 Because a flip is equally likely to result in heads or tails, P(A) =
AVOID COMMON ERRORS
n(A) _ 93 _ . =
256 n(S) 12. You flip a coin 5 times. What is the probability that every result will be tails? Let S be the sample space, which consists of the results of 5 coin flips (each of which is either heads or tails), and let A be the event that every result will be tails.
Students sometimes attempt to simplify permutations or combinations by canceling factors. Remind students of the meaning of factorials, and suggest that they write out the multiplication to determine which factors actually cancel.
n(S) = 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 = 2 5 = 32
n(A) = 5C 5 = 1
Because a flip is equally likely to result in heads or tails, P(A) =
n(S)
32
© Houghton Mifflin Harcourt Publishing Company • Image Credits: (t) ©JGI/ Jamie Grill/Blend Images/Alamy; (b) ©Borge Svingen/Flickr/Getty Images
13. There are 12 balloons in a bag: 3 each of blue, green, red, and yellow. Three balloons are chosen at random. Find the probability that all 3 balloons are green. Let S be the sample space, which consists of combinations of the 3 balloons chosen from the 12 balloons, and let A be the event that all 3 balloons are green. 12! = 220 n(S) = 12C 3 = 3!(12 - 3)! n(A) = 3C 3 = 1
n(A) _ 1 _ = .
__
P(A) =
n(A) _ 1 _ =
220 n(S) 14. There are 6 female and 3 male kittens at an adoption center. Four kittens are chosen at random. What is the probability that all 4 kittens are female? Let S be the sample space, which consists of the combinations of 4 kittens chosen from the 9 kittens, and let A be the event that all 4 kittens are female. 9! 9! = 126 = n(S) = 9C 4 = 4!5! 4!(9 - 4)!
_ _
n(A) = 6C 4 =
6! 6! _ = _ = 15 4!(6 - 4)!
n(A) 5 15 P(A) = _ = _ = _ Module 19
n(S)
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126
4!2!
42
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Combinations and Probability 980
There are 21 students in your class. The teacher wants to send 4 students to the library each day. The teacher will choose the students to go to the library at random each day for the first four days from the list of students who have not already gone. Answer each question. 15. What is the probability you will be chosen to go on the first day?
Let S be the sample space, which consists of the combinations of 4 students chosen from the 21 students, and let A be the event that you will be chosen to go on the first day. 21! = 21! = 5985 n(S) = 21C 4 = 4!17! 4!(21-4)!
_ _
Since you are one group member, the rest of the group members can be made up of any of the 20 remaining students in your class. n(A) = 20C 3 =
20! 20! __ = _ = 1140 3!(20 - 3)!
n(A) 1140 4 P(A) = _ = _ = _
3!17!
21 5985 n(S) 16. If you have not yet been chosen to go on days 1–3, what is the probability you will be chosen to go on the fourth day? 12 students have already gone, leaving 9 students to go to the library. Let S be the sample space, which consists of the combinations of 4 students chosen from the 9 students, and let A be the event that you will be chosen to go on the fourth day. n(S) = 9C 4 =
9! 9! _ = _ = 126 4!(9 - 4)!
4!5!
Since you are one group member, the rest of the group members can be made up of any of the 8 remaining who have not yet gone.
© Houghton Mifflin Harcourt Publishing Company
n(A) = 8C 3 =
3!5!
Let S be the sample space, which consists of the combinations of 2 students chosen from the 30 students, and let A be the event that you and your best friend are chosen. n(S) = 30C 2 = 435
n(A) = 2C 2 = 1
Module 19
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Lesson 19.3
3!(8 - 3)!
17. Your teacher chooses 2 students at random to represent your homeroom. The homeroom has a total of 30 students, including your best friend. What is the probability that you and your best friend are chosen?
P(A) =
981
8! 8! _ = _ = 56
n(A) _ 1 _ = n(S)
435
981
Lesson 3
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There are 12 peaches and 8 bananas in a fruit basket. You get a snack for yourself and three of your friends by choosing four of the pieces of fruit at random. Answer each question. 18. What is the probability that all 4 are peaches? The sample space S consists of the combinations of 4 pieces of fruit from the 20 pieces of fruit, and event A consists of combinations of 4 peaches.
n(S) = 20C 4 = n(A) = 12C 4 P(A) =
20! 20! __ = _ = 4845 4!(20 - 4)!
4!16!
4!(12 - 4)!
4!8!
12! 12! = __ = _ = 495
n(A)
33 495 _=_ =_
323 4845 n(S) 19. What is the probability that all 4 are bananas? The sample space S consists of the combinations of 4 pieces of fruit from the 20 pieces of fruit, and event A consists of combinations of 4 bananas. n(S) = 20C 4 = n(A) = 8C 4 P(A) =
20! 20! __ = _ = 4845 4!(20 - 4)!
4!16!
8! 8! = _ = _ = 70
n(A)
4!(8 - 4)!
4!4!
70 14 _=_ =_
969 4845 n(S) 20. There are 30 students in your class. Your science teacher will choose 5 students at random to create a group to do a project. Find the probability that you and your 2 best friends in the science class will be chosen to be in the group.
_ _ _ _
21. On a television game show, 9 members of the studio audience are randomly selected to be eligible contestants.
a. Six of the 9 eligible contestants are randomly chosen to play a game on the stage. How many combinations of 6 players from the group of eligible contestants are possible? 9! 9! = = 84 9C 6 = 6!3! 6!(9 - 6)! b. You and your two friends are part of the group of 9 eligible contestants. What is the probability that all three of you are chosen to play the game on stage? Explain how you found your answer.
_ _
© Houghton Mifflin Harcourt Publishing Company
Since 3 of the group members are you and your friends, the additional 2 group members can come from any combination of the students left in class. The sample space S consists of the combinations of 5 students from the 30 students, and event A consists of combinations of 2 students from the 27 who are left. n(A) 351 1 27C 2 = = = P(A) = 142,506 406 n(S) 30C 5
The sample space S consists of the combinations of 6 contestants from the 9 who are eligible. n(S) = 9C 6 = 84
After you and your friends are chosen, 3 other contestants from the remaining 6 can be chosen in any combination, so event A consists of combinations of 3 contestants from the 6 who are left. n(A) 5 20 = n(A) = 6C 3 = 20, P(A) = = 21 84 n(S)
_ _ _
Module 19
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Lesson 3
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Combinations and Probability 982
22. Determine whether you should use permutations or combinations to find the number of possibilities in each of the following situations. Select the correct answer for each lettered part.
JOURNAL Have students cite real-world examples of combinations. Have them explain how they know the examples are combinations.
a. Selecting a group of 5 people from a group of 8 people
permutation
combination
b. Finding the number of combinations for a combination lock
permutation
combination
permutation
combination
permutation
combination
c.
Awarding first and second place ribbons in a contest
d. Choosing 3 books to read in any order from a list of 7 books
a. It doesn’t matter in what order the people are selected in. b. Order matters: numbers have to be in a specific order to open the lock. c. Order matters: awarding Sam first place and Elena second is different from awarding Elena first place and Sam second. d. It doesn’t matter in what order the books are chosen.
H.O.T. Focus on Higher Order Thinking
23. Communicate Mathematical Ideas Using the letters A, B, and C, explain the difference between a permutation and a combination.
In permutations, order matters. In combinations, order does not matters. In a permutation of A, B, and C, ABC is different from CBA, so they would be counted as two different permutations. In a combination, ABC is the same as CBA, and would not be counted again. 24. a. Draw Conclusions Calculate 10C 6 and 10C 4.
C6 =
3,628,800 10! 10! __ = _ = _ = 210
C4 =
3,628,800 10! 10! __ = _ = _ = 210
© Houghton Mifflin Harcourt Publishing Company
10
10
6!(10 - 6)!
6!4!
4!(10 - 4)!
4!6!
17,280 17,280
b. What do you notice about these values? Explain why this makes sense.
C 6 = 10C 4 = 210; it makes sense that these values are equal because every combination of 6 objects that are selected has a corresponding combination of 4 objects that are not selected.
10
c.
Use your observations to help you state a generalization about combinations.
In general, nC r = nC n-r.
Module 19
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Lesson 19.3
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Lesson 3
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25. Justify Reasoning Use the formula for combinations to make a generalization about nC n. Explain why this makes sense.
AVOID COMMON ERRORS In Question 2, students may incorrectly conclude that the winning candidate received 71.1% of the total of 208,253 votes cast. The problem states, however, that 208,253 votes were 71% of all the votes cast. The part (208,253) and the percent (71.1) are given and the whole is asked for: part 208,253 _______ whole = _______ percent = 0.711 ≈ 292,902
Using the formula for combinations and the fact that n! n! n! = 1; this makes sense because there = 0! = 1, nC n = _ = n!0! n! n!(n - n)! is only 1 combination of n objects taken n at a time.
_ _
26. Explain the Error Describe and correct the error in evaluating 9C 4. 9! = 3024 9! _ =_ 9C 4 = (9 - 4)! 5! 9! 9! The answer given was 9P 4, not 9C 4; 9C 4 = = 126 = 4!5! 4!(9 - 4)!
_ _
Lesson Performance Task 1. In the 2012 elections, there were six candidates for the United States Senate in Vermont. In how many different orders, from first through sixth, could the candidates have finished?
INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 Shown below are the first 6 rows of an array
2. The winner of the Vermont Senatorial election received 208,253 votes, 71.1% of the total votes cast. The candidate coming in second received 24.8% of the vote. How many votes did the second-place candidate receive? Round to the nearest ten.
called Pascal’s Triangle. Each number in the array is found by adding together the two numbers above it.
3. Following the 2012 election there were 53 Democratic, 45 Republican, and 2 Independent senators in Congress. a. How many committees of 5 Democratic senators could be formed?
Row 0 Row 1 Row 2 Row 3 Row 4 Row 5 Row 6
b. How many committees of 48 Democratic senators could be formed? c. Explain how a clever person who knew nothing about combinations could guess the answer to (b) if the person knew the answer to (a). 4. Following the election, a newspaper printed a circle graph showing the make-up of the Senate. How many degrees were allotted to the sector representing Democrats, how many to Republicans, and how many to Independents? © Houghton Mifflin Harcourt Publishing Company
1. 6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720 208,253 2. = 292,901.5471 total votes 0.711
_
292,901.5471 × 0.248 = 72,639.58 ≈ 72,640
__ __
344,362,200 53! 3. a. 53C 5 = = = 2,869,685 120 5! ⋅ (53 - 5)! b.
C 48 =
344,362,200 53! __ = __ = 2,869,685
120 48! ⋅ (53 - 48)! c. The person could reason that for each committee of 5 Democratic Senators in (a) there were 48 who were not on the committee. So, there is a one-to-one correspondence between the 2,869,685 committees of 5 and the 2,869,685 committees of 48. 53
984
1
1 2
3
1 3
1
1 1 1
4 6 4 1 5 10 10 5 1 6 15 20 15 6 1
What is the 14th term in Row 17 of Pascal’s Triangle? 17C 14 = 680
Republicans 360° ⋅ 0.45 = 162°
Module 19
1 1
Find 3C 2, 5C 3, and 6C 4. Then propose a connection between the terms in the triangle and the quantity mC n. (You’ll find the connection easiest to spot by numbering the first term in each row Term 0. So, Term 3 in Row 6 is 20.) mC n equals Term n in Row m.
4. Democrats 360° ⋅ 0.53 = 190.8° Independents 360° ⋅ 0.02 = 7.2°
1
Lesson 3
EXTENSION ACTIVITY A2_MNLESE385900_U8M19L3 984
Each day, Senator Smith leaves his office and walks to the Committee Room along a grid of hallways that forms a 5 by 5 square. He moves only right (R) and down (D). The path shown can be written RRRDDDRDDR. The senator has developed a method to use combinations to find the number of different ways he can complete his walk. What is the method? How many ways can he do it? Find 10! = 252 ways. all possible combinations of five R’s and five D’s; ____ 5!5!
8/26/14 7:58 PM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Combinations and Probability 984