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

)
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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Finding
A2_MNLESE385900_U8M19L3 973
bability
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Name
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HARDCOVER PAGES 707714
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
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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
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This is the
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Find the
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3!
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3P 3
!
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Potato Skins
Beef Chili
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Lesson 3
973
Module 19
973
Lesson 19.3
L3 973
0_U8M19
SE38590
A2_MNLE
8/26/14
7:59 PM
8/26/14 7:58 PM

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)
Module 19
974
EXPLAIN 1
© Houghton Mifflin Harcourt Publishing Company
n
Finding a Probability Using
Combinations
INTEGRATE TECHNOLOGY
Lesson 3
PROFESSIONAL DEVELOPMENT
A2_MNLESE385900_U8M19L3 974
Math Background
8/26/14 7:58 PM
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)
Module 19
975
Lesson 3
COLLABORATIVE LEARNING
A2_MNLESE385900_U8M19L3 975
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.
975
Lesson 19.3
8/26/14 7:58 PM
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) =
5 · 3C 2 +
3!
3!
___
C 3 = 5 · ___
2!1! + 3!0! = 15 + 1 = 16
3
16
2
n(A)
The probability of getting a 6 at least twice in 3 rolls is P(A) = _ = _ = _.
n(S)
27
216
Module 19
976
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
DIFFERENTIATE INSTRUCTION
A2_MNLESE385900_U8M19L3.indd 976
Graphic Organizers
10/06/15 5:37 PM
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
977
Lesson 3
LANGUAGE SUPPORT
A2_MNLESE385900_U8M19L3 977
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
977
Lesson 19.3
8/26/14 7:58 PM
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
Module 19
Exercise
A2_MNLESE385900_U8M19L3 978
Lesson 3
978
Depth of Knowledge (D.O.K.)
Mathematical Practices
1–12
1 Recall of Information
MP.2 Reasoning
13–14
2 Skills/Concepts
MP.4 Modeling
15–20
2 Skills/Concepts
MP.1 Problem Solving
21
2 Skills/Concepts
MP.4 Modeling
22
2 Skills/Concepts
MP.2 Reasoning
23
3 Strategic Thinking
MP.3 Logic
8/26/14 9:18 PM
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)
_ _ _
Module 19
Exercise
A2_MNLESE385900_U8M19L3 979
979
Lesson 19.3
979
Lesson 3
Depth of Knowledge (D.O.K.)
Mathematical Practices
8/27/14 10:18 PM
24
3 Strategic Thinking
MP.3 Logic
25
3 Strategic Thinking
MP.3 Logic
26
3 Strategic Thinking
MP.3 Logic
10. You flip a coin 10 times. What is the probability that you get at most 3 heads?
PEERTOPEER
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)
A2_MNLESE385900_U8M19L3.indd 980
126
4!2!
42
980
Lesson 3
8/27/14 4:20 PM
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
A2_MNLESE385900_U8M19L3 981
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
8/26/14 7:58 PM
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
A2_MNLESE385900_U8M19L3.indd 982
982
Lesson 3
8/27/14 4:20 PM
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
983
Lesson 3
8/26/14 7:58 PM
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