Download Permutations and Probability

LESSON
19.2
Name
Permutations and
Probability
Class
Date
19.2 Permutations and Probability
Essential Question: When are permutations useful in calculating probability?
Resource
Locker
Common Core Math Standards
The student is expected to:
Explore
S-CP.9(+)
Use permutations and combinations to compute probabilities of
compound events and solve problems.
ABC
Mathematical Practices
CAB
CBA
There are 7 members in a club. Each year the club elects a president, a vice president, and a treasurer.
A
ENGAGE
View the Engage section online. Discuss the
photograph, asking students what it might have to do
with probability. Then preview the Lesson
Performance Task.
BCA
Fundamental Counting Principle
Give an example of a permutation to a partner and explain how you know
it is a permutation.
What is the number of permutations of all 7 members of the club?
There are 7 different ways to make the first selection.
Once the first person has been chosen, there are 6 different ways to make the second
selection.
Once the first two people have been chosen, there are 5 different ways to make the third
selection.
© Houghton Mifflin Harcourt Publishing Company
PREVIEW: LESSON
PERFORMANCE TASK
BAC
If there are n items and a 1 ways to choose the first item, a 2 ways to select the second item after the first item has
been chosen, and so on, there are a 1 × a 2 ×…×a n ways to choose n items.
Language Objective
Permutations are selections of groups of objects in
which order is important. So, permutations are used
to find the probability of a group of objects being
selected in a particular order.
ACB
You can find the number of permutations with the Fundamental Counting Principle.
MP.7 Using Structure
Essential Question: When are
permutations useful in calculating
probability?
Finding the Number of Permutations
A permutation is a selection of objects from a group in which order is important. For example, there are 6
permutations of the letters A, B, and C.
Continuing this pattern, there are
of all the members of the club.
B
7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
permutations
The club is holding elections for a president, a vice president, and a treasurer. How many
different ways can these positions be filled?
There are 7 different ways the position of president can be filled.
Once the president has been chosen, there are 6 different ways the position of vice
president can be filled. Once the president and vice president have been chosen, there are 5
different ways the position of treasurer can be filled.
So, there are
Module 19
7 × 6 × 5 = 210
different ways that the positions can be filled.
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© Houghto
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
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C
What is the number of permutations of the members of the club who were not elected as
officers?
After the officers have been elected, there are
different ways to make the first selection.
EXPLORE
4 members remaining. So there are 4
Finding the Number of Permutations
Once the first person has been chosen, there are 3 different ways to make the second
selection.
INTEGRATE TECHNOLOGY
Continuing this pattern, there are 4 × 3 × 2 × 1 = 24 permutations of the unelected
members of the club.
D
Students have the option of doing the Explore activity
either in the book or online.
Divide the number of permutations of all the members by the number of permutations of
the unelected members.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Discuss why developing a counting rule for a
There are 5040 permutations of all the members of the club.
24 permutations of the unelected members of the club.
There are
5040
= 210
The quotient of these two values is 24
.
_
Reflect
1.
How does the answer to Step D compare to the answer to Step B?
The values are the same.
2.
Discussion Explain the effect of dividing the total number of permutations by the number of
permutations of items not selected.
This method can be used to calculate the number of permutations of a group of objects
group of objects is useful. Check that students
recognize how quickly some visualizations become
unwieldy, such as making a list or drawing a tree
diagram, as the number of objects grows.
selected from a larger set.
Explain 1
QUESTIONING STRATEGIES
When finding probabilities of groups,
what is the effect of specifying that order
matters? Order distinguishes between groups with
the same set of elements. For example, AB is not the
same as BA if order matters, but it is the same if
order does not matter. This affects numbers of
outcomes, which affect probabilities of events.
Finding a Probability Using Permutations
© Houghton Mifflin Harcourt Publishing Company
The results of the Explore can be generalized to give a formula for permutations. To do so,
it is helpful to use factorials. For a positive integer n, n factorial, written n!, is defined as follows.
n! = n × (n - 1) × (n - 2) ×…× 3 × 2 × 1
That is, n! is the product of n and all the positive integers less than n. Note that 0! is defined to be 1.
In the Explore, the number of permutations of the 7 objects taken 3 at a time is
This can be generalized as follows.
7×6×5×4×3×2×1
7! = _
7!
7 × 6 × 5 = ___ = _
4×3×2×1
4!
(7 - 3)!
Permutations
n! .
The number of permutations of n objects taken r at a time is given by n P r = _
(n - r)!
EXPLAIN 1
Finding a Probability Using
Permutations
AVOID COMMON ERRORS
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PROFESSIONAL DEVELOPMENT
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Learning Progressions
Combinatorics is the branch of mathematics that deals with arranging and
counting finite collections of items. This lesson introduces students to
permutations, which are groupings of items in which the order of the items
matters. In the next lesson, students learn to count combinations, which are
groupings of items in which the order does not matter. It is important for students
to be able to distinguish between the two counting methods. Both provide useful
formulas for counting large collections of items that can be applied to computing
the probabilities of events with many equally likely outcomes.
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Students may assume that the probability is the ratio
of 1 to the permutation. Review the definition of the
probability ratio and why a different permutation is
needed to count the elements in the numerator and
the elements in the denominator.
Permutations and Probability
962
Example 1
INTEGRATE TECHNOLOGY

You may want to have students use their
calculators to check their work. Graphing
calculators have a built-in function that calculates
permutations. For example, to find 10P 4, first enter 10.
Then press MATH and use the arrow keys to choose
the PRB menu. Select 2:nPr and press ENTER.
Now enter 4 and then press ENTER to see that
10P 4 = 5040.
Use permutations to find the probabilities.
A research laboratory requires a four-digit security code to gain access to the facility.
A security code can contain any of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, but no digit is
repeated. What is the probability that a scientist is randomly assigned a code with the digits
1, 2, 3, and 4 in any order?
The sample space S consists is the number of permutations of 4 digits selected from 10 digits.
10! = 5040
10!
=_
n(S) = 10 P 4 = _
6!
(10 - 4)!
Event A consists of permutations of a security code with the digits 1, 2, 3, and 4.
4! = 24
4!
n(A) = 4 P 4 = _
=_
0!
(4 - 4)!
The probability of getting a security code with the digits 1, 2, 3, and 4 is
n(A) _
1 .
P(A) =
= 24 = _
5040
210
n(S)
_

A certain motorcycle license plate consists of 5 digits that are randomly selected. No digit is
repeated. What is the probability of getting a license plate consisting of all even digits?
The sample space S consists of permutations of
n(S) =
QUESTIONING STRATEGIES
Why is it important to be able to recognize a
permutation to count an arrangement of
objects? Why can’t you just make a list or use a tree
diagram? As the number of objects increases, the
number of arrangements in a permutation gets
large very quickly, and it becomes impractical and
difficult to make a list or use a tree diagram.
P
5
10!
= _ = 30, 240
5!
Event A consists of permutations of a license plate with
n(A) =
5
P
all even digits .
5!
5
_ = 120
0!
The probability of getting a license plate with
© Houghton Mifflin Harcourt Publishing Company
The formula for a permutation is a ratio. Can
a permutation ever be a fraction less than 1?
Explain. No, although the formula is a ratio, it
always simplifies to a whole number because it is a
method for counting arrangements.
10
5 digits selected from 10 digits .
all even digits is
120
1
n(A)
P(A) = _ = _ = _ .
n(S)
30, 240
252
Your Turn
There are 8 finalists in the 100-meter dash at the Olympic Games. Suppose 3 of the
finalists are from the United States, and that all finalists are equally likely to win.
3.
What is the probability that the United States will win all 3 medals in this event?
8!
3!
n(S) = 8P 3 = _ = 336; n(A) = 3P 3 = _ = 6
5!
0!
n(A)
6
1
.
=
=
The probability of the United States winning all 3 medals is P(A) =
56
336
n(S)
_ _ _
4.
What is the probability that the United States will win no medals in this event?
8!
5!
n(S) = 8P 3 = _ = 336; n(A) = 5P 3 = _ = 60
5!
2!
n(A)
60
5
=
=
.
The probability of the United States winning no medals is P(A) =
336
28
n(S)
_ _ _
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COLLABORATIVE LEARNING
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Peer-to-Peer Activity
Have students work in pairs to confirm that the permutation formulas work.
Suggest that they make a list of the number of ways the letters in the word CLUE
can be arranged and then find the arrangements using the permutations formula.
Next, have them examine what happens to the number of permutations as the
number of letters in a word increases by starting a list for the word CLUES. Repeat
for the words COOL and TOOT. Have students comment on the usefulness of the
permutation formulas when counting items.
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Explain 2
Finding the Number of Permutations with Repetition
EXPLAIN 2
Up to this point, the problems have focused on finding the permutations of distinct objects. If some
of the objects are repeated, this will reduce the number of permutations that are distinguishable.
Finding the Number of Permutations
with Repetition
For example, here are the permutations of the letters A, B, and C.
ABC
ACB
BAC
BCA
CAB
CBA
Next, here are the permutations of the letters M, O, and M. Bold type is used to show the different positions of the
repeated letter.
MOM
MOM
MMO
MMO
OMM
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Discuss the importance of identifying the
OMM
Shown without the bold type, here are the permutations of the letters M, O, and M.
MOM
MOM
MMO
MMO
OMM
OMM
Notice that since the letter M is repeated, there are only 3 distinguishable permutations of the letters.
This can be generalized with a formula for permutations with repetition.
permutation needed to count the sample space
separately from the permutation needed to calculate
the event when using permutations with probability.
Permutations with Repetition
The number of different permutations of n objects where one object repeats a times, a second
object repeats b times, and so on is
n!
_
a!× b!×...
Example 2

QUESTIONING STRATEGIES
How is the formula for the number of
permutations where one object repeats
different from the formula for the number of
permutations with no repeats? How is it similar? In
both formulas the numerator is n! The denominator
of the formula for no repeats is (n - r)!, while the
denominator for repeats is a! • b! and so on to count
the repeating objects.
Find the number of permutations.
How many different permutations are there of the letters in the word ARKANSAS?
There are 8 letters in the word, and there are 3 A’s and 2 S’s, so the number of permutations
8! = 3360.
of the letters in ARKANSAS is _
3!2!

© Houghton Mifflin Harcourt Publishing Company
One of the zip codes for Anchorage, Alaska, is 99522. How many permutations are there of
the numbers in this zip code?
There are 5 digits in the zip code, and there are 2 nines , and 2 twos in the zip
code, so the number of permutations of the zip code is
5!
_=
2!2!
30 .
Your Turn
5.
How many different permutations can be formed using all the letters in MISSISSIPPI?
11!
= 34, 650
4!4!2!
6.
One of the standard telephone numbers for directory assistance is 555–1212. How many
different permutations of this telephone number are possible?
7!
= 210
3!2!2!
_
_
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DIFFERENTIATE INSTRUCTION
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Multiple Representations
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Have students calculate the number of permutations in various situations “by
hand”; that is, by reasoning about the situation and using multiplication, by
making a list, or by drawing tree diagrams. Then ask students what patterns they
notice. A common element of every solution should be the product of a string of
descending, consecutive integers, letters, or other items. Discuss how students can
use this process to help them recall how to find the number of permutations.
Permutations and Probability
964
Explain 3
EXPLAIN 3
Finding a Probability Using Permutations with
Repetition
Permutations with repetition can be used to find probablilities.
Finding a Probability Using
Permutations with Repetition
Example 3

The school jazz band has 4 boys and 4 girls, and they are randomly lined
up for a yearbook photo.
Find the probability of getting an alternating boy-girl arrangement.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Discuss how students are able to identify n, r,
8! = 70
n(S) _
4!4!
Event A consists of permutations that alternate boy-girl or girl-boy. The possible permutations
are BGBGBGBG and GBGBGBGB.
a, b, and so on to use permutations with repetition to
find probabilities.
n(A)
1.
2 =_
The probability of getting an alternating boy-girl arrangement is P(A) = _ = _
70
35
n(S)
The sample space S consists of permutations of 8 objects, with 4 boys and 4 girls.
n(A) = 2

Find the probability of getting all of the boys grouped together.
4 boys and 4 girls .
The sample space S consists of permutations of 8 students, with
QUESTIONING STRATEGIES
8!
n(S) = _ = 70
4!4!
How can you use what you know about
probability to check that the ratio makes sense
when finding probability using permutations with or
without repetition? The probability must be
between 0 and 1.
Event A consists of permutations with all 4 boys in a row . The possible permutations
are BBBBGGGG, GBBBBGGG, GGBBBBGG, GGGBBBBG, and GGGGBBBB .
n(A) = 5
© Houghton Mifflin Harcourt Publishing Company
5
1
n(A)
The probability of getting all the boys grouped together is P(A) = _ = _ = _.
n(S)
70
14
Your Turn
7.
There are 2 mystery books, 2 romance books, and 2 poetry books to be randomly placed
on a shelf. What is the probability that the mystery books are next to each other, the
romance books are next to each other, and the poetry books are next to each other?
n(S) =
6!
_
= 90
2!2!2!
Event A consists of permutations with books from each category next to each other.
The possible permutations are MMRRPP, MMPPRR, RRMMPP, RRPPMM, PPMMRR, and
PPRRMM.
n(A) = 6
The probability of getting all the books from each category next to each other is
n(A)
6
1
.
=
=
P(A) =
15
90
n(S)
_ _ _
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Lesson 2
LANGUAGE SUPPORT
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Connect Vocabulary
Have students look up the meaning of the verb permute (it is to change the
order of ). Connect the definition to the mathematical definition of permutation,
which is an arrangement of objects in a definite order.
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8.
What is the probability that a random arrangement of the letters in the word
APPLE will have the two P’s next to each other?
ELABORATE
The sample space S consists of permutations of the letters in APPLE, and there are 2 P’s,
5!
so n(S) = __
= 60. Consider the two P’s as a single block, so there are 4 positions for the
2!
AVOID COMMON ERRORS
letters to occupy.
n(A) = 4! = 24
The probability of getting the two P’s next to each other is P(A) =
n(A)
Students may be able to identify the permutation
needed for one part of the probability ratio but not
the other. Remind students to first decide what the
sample space is and then use a permutation to count
the elements in the sample space. Next, they should
use a permutation as needed to count the elements in
the event. Review why n(S) is always in the
denominator.
2
24 _
_=_
= .
n(S)
60
5
Elaborate
9.
If nP a = nP b , what is the relationship between a and b? Explain your answer.
n!
n!
=
. This occurs only when a = b.
The equation is true if
(n - a)!
(n - b)!
_ _
10. It was observed that there are 6 permutations of the letters A, B, and C. They are ABC,
ACB, BAC, BCA, CAB, and CBA. If the conditions are changed so that the order of
selection does not matter, what happens to these 6 different groups?
They would become a single group of the letters, ABC. The other five groups are duplicates
SUMMARIZE THE LESSON
of this result.
What are permutations and how can you use
them to calculate probabilities? A
permutation is a group of objects in which order is
important. You can use permutations to find the
number of outcomes in a sample space or in an
event.
11. Essential Question Check-In How do you determine whether choosing a group
of objects involves permutations?
Permutations are used when the order of selection matters.
Evaluate: Homework and Practice
An MP3 player has a playlist with 12 songs. You select the shuffle option, which
plays each song in a random order without repetition, for the playlist. In how
many different orders can the songs be played?
• Online Homework
• Hints and Help
• Extra Practice
12! = 479,001,600
The songs can be played in 479,001,600 different orders
2.
There are 10 runners in a race. Medals are awarded for 1st, 2nd, and 3rd place. In
how many different ways can the medals be awarded?
10 × 9 × 8 = 720
There are 720 possibilities for awarding medals.
3.
© Houghton Mifflin Harcourt Publishing Company
1.
EVALUATE
ASSIGNMENT GUIDE
There are 9 players on a baseball team. In how many different ways can the coach
choose players for first base, second base, third base, and shortstop?
9 × 8 × 7 × 6 = 3024
There are 3024 ways to arrange players.
Module 19
Exercise
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Depth of Knowledge (D.O.K.)
Mathematical Practices
1-11
1 Recall of Information
MP.2 Reasoning
12–15
1 Recall of Information
MP.1 Problem Solving
16–23
2 Skills/Concepts
MP.4 Modeling
24
3 Strategic Thinking
MP.3 Logic
25
3 Strategic Thinking
MP.2 Reasoning
26
3 Strategic Thinking
MP.6 Precision
27
3 Strategic Thinking
MP.3 Logic
04/09/14 6:43 PM
Concepts and Skills
Practice
Explore
Finding the Number of Permutations
Exercises 1–3
Example 1
Finding a Probability Using
Permutations
Exercises 4–7,
18–19, 22–23,
26–27
Example 2
Finding the Number of Permutations
With Repetition
Exercises 8–24
Example 3
Finding a Probability Using
Permutations With Repetition
Exercises 12–17,
20–21, 25
Permutations and Probability
966
4.
GRAPHIC ORGANIZERS
Have students make a graphic organizer to
summarize what they know about permutations. Ask
them to include the formulas, with descriptions and
examples.
A bag contains 9 tiles, each with a different number from 1 to 9. You choose a tile
without looking, put it aside, choose a second tile without looking, put it aside, then
choose a third tile without looking. What is the probability that you choose tiles with
the numbers 1, 2, and 3 in that order?
Let S be the sample space and let A be the event that you choose tiles
with the numbers 1, 2, and 3 in that order.
9!
9!
= 504
=
n(S) = 9P 3 =
(9 - 3)! 6!
n(A) = 1
n(A)
1
=
P(A) =
504
n(S)
_ _
_ _
5.
There are 11 students on a committee. To decide which 3 of these students will attend
a conference, 3 names are chosen at random by pulling names one at a time from a
hat. What is the probability that Sarah, Jamal, and Mai are chosen in any order?
Let S be the sample space and let A be the event that Sarah,
Jamal, and Mai are chosen in any order.
11!
11!
n(S) = 11P 3 =
= 990
=
8!
(11 - 3)!
3!
3!
=6
=
n(A) = 3P 3 =
(3 - 3)! 0!
n(A)
6
1
=
=
P(A) =
165
990
n(S)
_ _
_ _
_ _ _
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Chad
McDermott/Shutterstock
6.
A clerk has 4 different letters that need to go in 4 different envelopes. The clerk places
one letter in each envelope at random. What is the probability that all 4 letters are
placed in the correct envelopes?
Let S be the sample space and let A be the event that all 4 letters are
placed in the correct envelopes.
n(S) = 4P 4 =
4!
4!
_
= _ = 24
(4 - 4)! 0!
n(A) = 1
n(A)
1
=
P(A) =
24
n(S)
A swim coach randomly selects 3 swimmers from a team of
8 to swim in a heat. What is the probability that she will
choose the three strongest swimmers?
_ _
7.
Let S be the sample space and let A be the event
that the coach chooses the three strongest
swimmers.
n(S) = 8P 3 =
n(A) = 3P 3 =
8!
8!
_
= _ = 336
(8 - 3)!
5!
(3 - 3)!
0!
3!
3!
_
=_=6
n(A)
6
1
P(A) = _ = _ = _
n(S)
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8.
How many different sequences of letters can be formed using all the letters in
ENVELOPE?
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Discuss how to recognize when a probability
The three letter E’s are not distinguishable.
8!
8P 8
=
= 6720
3!
3!
6720 sequences
_ _
9.
Yolanda has 3 each of red, blue, and green marbles. How many possible ways can the
9 marbles be arranged in a row?
9!
9P 9
=
= 1680
3!3!3!
3!3!3!
1680 ways
evaluated using a permutation will be equal to the
1
.
ratio ___
_ _
n(S)
10. Jane has 16 cards. Ten of the cards look exactly the same and have the number 1 on
them. The other 6 cards look exactly the same and have the number 2 on them. Jane
is going to make a row containing all 16 cards. How many different ways can she
order the row?
16!
16P 16
=
= 8008
10!6!
10!6!
8008 row arrangements
AVOID COMMON ERRORS
Students may make errors when simplifying with
factorials. Encourage students to write out the
product a factorial represents, at least until they are
sure which numbers to cancel.
_ _
11. Ramon has 10 cards, each with one number on it. The numbers are 1, 2, 3, 4, 4, 6, 6,
6, 6, 6. Ramon is going to make a row containing all 10 cards. How many different
ways can he order the row?
10!
10P 10
=
= 15,120
2!5!
2!5!
15,120 row arrangements
_ _
12. A grocer has 5 apples and 5 oranges for a window display. The grocer makes a row
of the 10 pieces of fruit by choosing one piece of fruit at random, making it the first
piece in the row, choosing a second piece of fruit at random, making it the second
piece in the row, and so on. What is the probability that the grocer arranges the fruits
in alternating order? (Assume that the apples are not distinguishable and that the
oranges are not distinguishable.)
Let S be the sample space and let A be the event that the grocer arranges
the fruits in alternating order.
P
10!
_
= _ = 252
10
© Houghton Mifflin Harcourt Publishing Company
n(S) =
10
5!5!
5!5!
n(A) = 2, AOAOAOAOAO or OAOAOAOAOA
P(A) =
n(A) _
2
1
_
=
=_
n(S)
252
126
13. The letters G, E, O, M, E, T, R, Y are on 8 tiles in a bag, one letter on each tile. If you
select tiles randomly from the bag and place them in a row from left to right, what is
the probability the tiles will spell out GEOMETRY?
Let S be the sample space and let A be the event that the tiles spell out
GEOMETRY again. The two letter E’s are not distinguishable.
n(S) =
P
8!
_
= _ = 20,160
8
8
2!
2!
n(A) = 1, GEOMETRY
P(A) =
Module 19
n(A) _
1
_
=
n(S)
A2_MNLESE385900_U8M19L2.indd 968
20,160
968
Lesson 2
8/26/14 2:35 PM
Permutations and Probability
968
14. There are 11 boys and 10 girls in a classroom. A teacher chooses a student at random
and puts that student at the head of a line, chooses a second student at random and
makes that student second in the line, and so on, until all 21 students are in the line.
What is the probability that the teacher puts them in a line alternating boys and girls?
where no two of the same gender stand together?
PEERTOPEER
Have students work with a partner to write a
probability problem about numbers, letters, or both
in which a permutation is needed to count the
objects. Have students solve their problems. Then ask
them to exchange problems with another pair and
solve those. Have the pairs review each other’s
solution methods.
n(S) =
P
21!
_
= _ = 352,716
21
21
11!10!
11!10!
Because there are more boys than girls, there is only one way to alternate
them: BGBGBGBGBGBGBGBGBGBGB. If a girl started first, the end would
have too many boys. So n(A) = 1.
P(A) =
n(A) _
1
_
=
n(S)
352, 716
15. There are 4 female and 4 male kittens are sleeping together in a row. Assuming that
the arrangement is a random arrangement, what is the probability that all the female
kittens are together, and all the male kittens are together?
P
8!
n(S) = 8 8 =
= 70, n(A) = 2, FFFFMMMM or MMMMFFFF
4!4!
4!4!
n(A)
2
1
=
=
P(A) =
70
35
n(S)
VISUAL CUES
_ _
_ _ _
Discuss the locations of and relationship between n
and r in the rule for permutations, and how they can
be used to help students remember the formula
for nP r. Make sure students do not confuse the P in
the permutation formula for the P used to denote
probability.
16. If a ski club with 12 members votes to choose 3 group leaders, what is the probability
that Marsha, Kevin, and Nicola will be chosen in any order for President, Treasurer,
and Secretary?
3!
12!
12!
3!
n(S) = 12P 3 =
= 1320, n(A) = 3P 3 =
=
= __
=6
9!
(12 - 3)!
(3 - 3)! 0!
n(A)
6
1
=
=
P(A) =
1320
220
n(S)
_ _
_
_ _ _
© Houghton Mifflin Harcourt Publishing Company
17. There are 7 books numbered 1–7 on the summer reading list. Peter randomly chooses
2 books. What is the probability that Peter chooses books numbered 1 and 2, in either
order?
n(S) = 7P 2 =
n(S)
42
5!
P2 =
2
2!
2!
_
=_ =2
(2 - 2)!
0!
21
18. On an exam, students are asked to list 5 historical events in the order in which they
occurred. A student randomly orders the events. What is the probability that the
student chooses the correct order?
5!
5!
_
= _ = 120, n(A) = 1
(5 - 5)! 0!
n(A)
1
P(A) = _ = _
n(S) = 5P 5 =
n(S)
A2_MNLESE385900_U8M19L2.indd 969
Lesson 19.2
(7 - 2)!
n(A)
2
1
P(A) = _ = _ = _
Module 19
969
7!
7!
_
= _ = 42, n(A) =
120
969
Lesson 2
8/26/14 2:37 PM
19. A fan makes 6 posters to hold up at a basketball game.
Each poster has a letter of the word TIGERS. Six friends
sit next to each other in a row. The posters are distributed
at random. What is the probability that TIGERS is spelled
correctly when the friends hold up the posters?
Let S be the sample space and let A be the event
that that TIGERS is spelled correctly when the
friends hold up the posters.
n(S) = 6P 6 =
n(A) = 1
P(A) =
6!
6!
_
= _ = 720
(6 - 6)!
0!
n(A) _
1
_
=
n(S)
720
20. The 10 letter tiles S, A, C, D, E, E, M, I, I, and O are in a bag. What is the probability
that the letters S-A-M-E will be drawn from the bag at random, in that order?
Let S be the sample space and let A be the event that the letters S-A-M-E
will be drawn from the bag at random, in that order.
n(S) =
P
5040
_
= _ = 1260
10
4
2!2!
4
n(A) = 1
P(A) =
n(A) _
1
_
=
n(S)
1260
21. If three cards are drawn at random from a standard deck of 52 cards, what is the
probability that they will all be 7s? (There are four 7s in a standard deck of 52 cards.)
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Dan
Kosmayer/Shutterstock
Let S be the sample space and let A be the event that all four cards are 7s.
52!
52!
n(S) = 52P 3 =
= 132,600
=
(52 - 3)! 49!
_ _
n(A) = 4P 3 =
n(A)
4!
4!
_
= _ = 24
(4 - 3)!
1!
_ _ _
1
24
P(A) =
=
=
132,600
5525
n(S)
22. A shop classroom has ten desks in a row. If there are 6 students in shop class and
they choose their desks at random, what is the probability they will sit in the first
six desks?
Let S be the sample space and let A be the event that the students sit in
the first six desks.
n(S) = 10P 6 =
n(A) = 6P 6
10!
10!
_
= _ = 151,200
(10 - 6)!
4!
6!
6!
= _ = _ = 720
(6 - 6)!
0!
n(A)
720
1
P(A) = _ = _ = _
n(S)
Module 19
A2_MNLESE385900_U8M19L2.indd 970
151,200
210
970
Lesson 2
8/27/14 4:08 PM
Permutations and Probability
970
23. Match each event with its probability. All orders are chosen randomly.
JOURNAL
A. There are 15 floats that will be in a town parade. Event A: The
mascot float is chosen to be first and the football team float is
chosen to be second.
Have students cite real-world examples of
permutations. Have them explain how they know the
examples are permutations.
B. Beth is one of 10 students performing in a school talent show.
Event B: Beth is chosen to be the fifth performer and her best
friend is chosen to be fourth.
C. Sylvester is in a music competition with 14 other musicians.
Event C: Sylvester is chosen to be last, and his two best friends
are chosen to be first and second.
C _
1
1092
A _
1
210
B _
1
90
P(A) =
n(mascot 1st and football 2nd) _
P
1
___
=
=_
P(B) =
n(Beth 5th and friend 4th) _
P
1
___
=
=_
P(C) =
n(Sylvester last, friends 1st and 2nd)
2. P
1
2 =_
____
= _=_
13
n(all floats in parade)
13
210
P 15
15
8
n(all contestants)
8
90
P 10
10
11
n(all musicians)
11
P 14
14
2184
1092
H.O.T. Focus on Higher Order Thinking
24. Explain the Error Describe and correct the error in evaluating the expression.
5
× 4 × 3!
5! = 5_
= 20
P3 = _
3!
3!
5!
5!
The denominator of the first fraction should be (5 - 3)!, not 3!; ______
= __
= 60.
2!
(5 - 3)!
25. Make a Conjecture If you are going to draw four cards from a deck of cards,
does drawing four aces from the deck have the same probability as drawing four 3s?
Explain.
© Houghton Mifflin Harcourt Publishing Company
Yes, they have the same probability. The probability of drawing 4 aces from a deck of cards
4 4
would be P(A) = ___ = ___
because the 4 aces could be drawn in any order, and there are
P
n(A)
n(S)
P
52
4
52 cards in the deck. There are also 4 3s in the deck that could be drawn in any order,
so the probability would be the same.
26. Communicate Mathematical Ideas Nolan has Algebra, Biology, and World
History homework. Assume that he chooses the order that he does his homework at
random. Explain how to find the probability of his doing his Algebra homework first.
If he is doing his Algebra homework first, the only classes that can change homework
order are Biology and World History, so n(A) = 2P 2. The sample space would have all
1
Nolan doing his Algebra homework first is _
.
3
Module 19
A2_MNLESE385900_U8M19L2.indd 971
971
Lesson 19.2
n(A)
n(S)
2
1
2 2
three classes changing, so n(S) = 3P 3. Since P(A) = ___ =___
=_
=_
, the probability of
6
3
P
971
P
3
3
Lesson 2
8/27/14 4:08 PM
27. Explain the Error A student solved the problem shown. The student’s work is also
shown. Explain the error and provide the correct answer.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Question 3 of the Lesson Performance Task
A bag contains 6 tiles with the letters A, B, C, D, E, and F, one letter on each tile. You
choose 4 tiles one at a time without looking and line them up from left to right as you
choose them. What is the probability that your tiles spell BEAD?
Let S be the sample space and let A be the event that the tiles spell BEAD.
6! = 360
6!
n(S) = 6P 4 = _
=_
(6 - 4)! 2!
4!
4! = 24
n(A) = 4P 4 = _
=_
(4 - 4)! 0!
n(A)
1
24 = _
P(A) = _ = _
5
360
n(S)
shows a map containing nine regions that cannot be
colored with three colors; four colors are required.
Challenge students to draw a map containing fewer
than nine regions that requires four colors. What is
the fewest number of regions a map can contain that
requires four colors? 5
n(A) should be 1 since the tiles must appear in the order
1
.
B-E-A-D. The correct probability is
360
_
Lesson Performance Task
How many different ways can a blue card, a red card, and a green card be
arranged? The diagram shows that the answer is six.
B R G
R G B
G B R
B G R
R B G
G R B
AVOID COMMON ERRORS
Some students may draw a map with two regions that
share a vertex point and must be the same color, and
think they have disproved the theorem. Remind
students that regions only require different colors if
they share a boundary, not a corner.
1. Now solve this problem: What is the least number of colors
needed to color the pattern shown here, so that no two
squares with a common boundary have the same color? Draw
a sketch to show your answer.
two
2. Now try this one. Again, find the least number of colors
needed to color the pattern so that no two regions with a
common boundary have the same color. Draw a sketch to
show your answer.
3
three
1
© Houghton Mifflin Harcourt Publishing Company
2
2
3
3. In 1974, Kenneth Appel and Wolfgang Haken solved a
problem that had confounded mathematicians for more than
a century. They proved that no matter how complex a map is,
it can be colored in a maximum of four colors, so that no two
regions with a common boundary have the same color. Sketch
the figure shown here. Can you color it in four colors? Can
you color it in three colors?
Four colors are needed.
Module 19
972
1
2
3
4
1
4
3
2
1
Lesson 2
EXTENSION ACTIVITY
A2_MNLESE385900_U8M19L2 972
Give each student several black-and-white copies of a map of a region containing
a complex array of countries, and crayons or colored pencils in four colors.
Challenge students to color the map in three colors, following the same restriction
stated in the Lesson Performance Task: No two countries sharing a border can
have the same color. Regardless of whether the map can be colored in three colors,
have students show how it can be colored in four colors.
04/09/14 6:46 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.
Permutations and Probability
972