CorrectionKey=NL-B;CA-B 19 . 2 DO NOT EDIT--Changes … · Use permutations and combinations to compute probabilities of compound events and solve problems. Mathematical Practices

© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

Name Class Date

Resource Locker

Resource Locker

Explore Finding the Number of PermutationsA 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.

ABC ACB BAC BCA CAB CBA

You can find the number of permutations with the Fundamental Counting Principle.

Fundamental Counting Principle

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.

There are 7 members in a club. Each year the club elects a president, a vice president, and a treasurer.

A What is the number of permutations of all 7 members of the club?

There are different ways to make the first selection.

Once the first person has been chosen, there are different ways to make the second selection.

Once the first two people have been chosen, there are different ways to make the third selection.

Continuing this pattern, there are permutations of all the members of the club.

B 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 different ways the position of president can be filled.

Once the president has been chosen, there are different ways the position of vice president can be filled. Once the president and vice president have been chosen, there are different ways the position of treasurer can be filled.

So, there are different ways that the positions can be filled.

7

7

7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

7 × 6 × 5 = 210

6

6

5

5

Module 19 961 Lesson 2

19.2 Permutations and ProbabilityEssential Question: When are permutations useful in calculating probability?

DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-B;CA-B

A2_MNLESE385900_U8M19L2.indd 961 8/26/14 2:13 PM

Common Core Math StandardsThe student is expected to:

S-CP.9(+)

Use permutations and combinations to compute probabilities of compound events and solve problems.

Mathematical Practices

MP.7 Using Structure

Language ObjectiveGive an example of a permutation to a partner and explain how you know it is a permutation.

HARDCOVER PAGES 699706

Turn to these pages to find this lesson in the hardcover student edition.

Permutations and Probability

ENGAGE Essential Question: When are permutations useful in calculating probability?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.

PREVIEW: LESSON PERFORMANCE TASKView the Engage section online. Discuss the photograph, asking students what it might have to do with probability. Then preview the Lesson Performance Task.

961

HARDCOVER

Turn to these pages to find this lesson in the hardcover student edition.

S-CP.9(+) Use permutations and combinations to compute probabilities of compound events

and solve problems.

© H

ough

ton

Mif

flin

Har

cour

t Pub

lishi

ng C

omp

any

Name

Class Date

Resource

LockerResource

Locker

Explore 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.

ABCACB

BACBCA

CABCBA

You can find the number of permutations with the Fundamental Counting Principle.

Fundamental Counting Principle

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.

There are 7 members in a club. Each year the club elects a president, a vice president, and a treasurer.

What is the number of permutations of all 7 members of the club?

There are different ways to make the first selection.

Once the first person has been chosen, there are different ways to make the second

selection.

Once the first two people have been chosen, there are different ways to make the third

selection.

Continuing this pattern, there are permutations

of all the members of the club.

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 different ways the position of president can be filled.

Once the president has been chosen, there are different ways the position of vice

president can be filled. Once the president and vice president have been chosen, there are

different ways the position of treasurer can be filled.

So, there are different ways that the positions can be filled.

7

7

7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

7 × 6 × 5 = 210

6

6

5

5

Module 19

961

Lesson 2

19 . 2 Permutations and Probability

Essential Question: When are permutations useful in calculating probability?

DO NOT EDIT--Changes must be made through "File info"

CorrectionKey=NL-B;CA-B

A2_MNLESE385900_U8M19L2.indd 961

8/26/14 2:20 PM

961 Lesson 19 . 2

L E S S O N 19 . 2

DO NOT EDIT--Changes must be made through “File info”CorrectionKey=NL-C;CA-C

© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

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 members remaining. So there are different ways to make the first selection.

Once the first person has been chosen, there are different ways to make the second selection.

Continuing this pattern, there are permutations of the unelected members of the club.

D Divide the number of permutations of all the members by the number of permutations of the unelected members.

There are permutations of all the members of the club.

There are permutations of the unelected members of the club.

The quotient of these two values is .

Reflect

1. How does the answer to Step D compare to the answer to Step B?

2. Discussion Explain the effect of dividing the total number of permutations by the number of permutations of items not selected.

Explain 1 Finding a Probability Using PermutationsThe 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

7 × 6 × 5 = 7 × 6 × 5 × 4 × 3 × 2 × 1 ___ 4 × 3 × 2 × 1 = 7! _ 4! = 7! _ (7 - 3) !

This can be generalized as follows.

Permutations

The number of permutations of n objects taken r at a time is given by n P r = n! _ (n - r) ! .

4

3

24

5040

5040 _ 24

= 210

4

4 × 3 × 2 × 1 = 24

The values are the same.

This method can be used to calculate the number of permutations of a group of objects

selected from a larger set.





EXPLORE Finding the Number of Permutations

INTEGRATE TECHNOLOGYStudents have the option of doing the Explore activity either in the book or online.

INTEGRATE MATHEMATICAL PRACTICESFocus on ReasoningMP.2 Discuss why developing a counting rule for a 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.

QUESTIONING STRATEGIESWhen 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.

EXPLAIN 1 Finding a Probability Using Permutations

AVOID COMMON ERRORSStudents 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.

PROFESSIONAL DEVELOPMENT

Learning ProgressionsCombinatorics 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.

Permutations and Probability 962


© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

Example 1 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?

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 selected from .

n (S) = P = _ =

Event A consists of permutations of a license plate with .

n (A) = P _ =

The probability of getting a license plate with is

P (A) = n (A)

_ n (S)

= _ = _ .

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?

4. What is the probability that the United States will win no medals in this event?

The sample space S consists is the number of permutations of 4 digits selected from 10 digits.

n (S) = 10 P 4 = 10! _ (10 - 4)!

= 10! _ 6! = 5040

Event A consists of permutations of a security code with the digits 1, 2, 3, and 4.

n (A) = 4 P 4 = 4! _ (4 - 4)!

= 4! _ 0! = 24

The probability of getting a security code with the digits 1, 2, 3, and 4 is

P (A) = n (A)

_ n (S)

= 24 _ 5040 = 1 _ 210 .

120

10!

5!

5!

0!

30, 240

5 digits 10 digits

all even digits

all even digits

30, 240

120

252

1

n (S) = 8 P 3 = 8! _ 5! = 336; n (A) = 3 P 3 = 3! _

0! = 6

The probability of the United States winning all 3 medals is P (A) = n (A)

_ n (S)

= 6 _ 336

= 1 _ 56

.

n (S) = 8 P 3 = 8! _ 5! = 336; n (A) = 5 P 3 = 5! _

2! = 60

The probability of the United States winning no medals is P (A) = n (A)

_ n (S)

= 60 _ 336

= 5 _ 28

.

10

5

5

5


DO NOT EDIT--Changes must be made through "File info"CorrectionKey=NL-B;CA-B

A2_MNLESE385900_U8M19L2 963 8/27/14 10:07 PM

COLLABORATIVE LEARNING

Peer-to-Peer ActivityHave 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.

INTEGRATE TECHNOLOGYYou 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 10 P 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 10 P 4 = 5040.

QUESTIONING STRATEGIESThe 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.

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.

963 Lesson 19 . 2


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

Explain 2 Finding the Number of Permutations with RepetitionUp 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.

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 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.

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 Find the number of permutations.

How many different permutations are there of the letters in the word ARKANSAS?

One of the zip codes for Anchorage, Alaska, is 99522. How many permutations are there of the numbers in this zip code?

There are digits in the zip code, and there are , and in the zip code, so the number of permutations of the zip code is

_ = .

Your Turn

5. How many different permutations can be formed using all the letters in MISSISSIPPI?

6. One of the standard telephone numbers for directory assistance is 555–1212. How many different permutations of this telephone number are possible?

There are 8 letters in the word, and there are 3 A’s and 2 S’s, so the number of permutations

of the letters in ARKANSAS is 8! _ 3!2! = 3360.

11! _ 4!4!2! = 34, 650

7! _ 3!2!2! = 210

2!2!30

5!

5 2 nines 2 twos





DIFFERENTIATE INSTRUCTION

Multiple RepresentationsHave 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.

EXPLAIN 2 Finding the Number of Permutations with Repetition

INTEGRATE MATHEMATICAL PRACTICESFocus on ModelingMP.4 Discuss the importance of identifying the permutation needed to count the sample space separately from the permutation needed to calculate the event when using permutations with probability.

QUESTIONING STRATEGIESHow 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.



© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

Explain 3 Finding a Probability Using Permutations with Repetition

Permutations with repetition can be used to find probablilities.

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.

Find the probability of getting all of the boys grouped together.

The sample space S consists of permutations of , with .

n (S) = _ =

Event A consists of permutations with . The possible permutations

are BBBBGGGG, GBBBBGGG, .

n (A) =

The probability of getting all the boys grouped together is P (A) = n (A)

_ n (S)

= _ = _ .

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?

The sample space S consists of permutations of 8 objects, with 4 boys and 4 girls.

n (S) 8! _ 4!4! = 70

Event A consists of permutations that alternate boy-girl or girl-boy. The possible permutations are BGBGBGBG and GBGBGBGB.

n (A) = 2

The probability of getting an alternating boy-girl arrangement is P (A) = n (A)

_ n (S)

= 2 _ 70 = 1 _ 35 .

4!4!

1470

708!

155

GGBBBBGG, GGGBBBBG, and GGGGBBBB

all 4 boys in a row

4 boys and 4 girls8 students

n (S) = 6! _ 2!2!2!

= 90

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

P (A) = n (A)

_ n (S)

= 6 _ 90

= 1 _ 15

.



A2_MNLESE385900_U8M19L2 965 8/27/14 10:07 PM

LANGUAGE SUPPORT

Connect VocabularyHave 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.

EXPLAIN 3 Finding a Probability Using Permutations with Repetition

INTEGRATE MATHEMATICAL PRACTICESFocus on CommunicationMP.3 Discuss how students are able to identify n, r, a, b, and so on to use permutations with repetition to find probabilities.

QUESTIONING STRATEGIESHow 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.

965 Lesson 19 . 2


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

• Online Homework• Hints and Help• Extra Practice

Evaluate: Homework and Practice

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

9. If n P a = n P b , what is the relationship between a and b? Explain your answer.

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?

11. Essential Question Check-In How do you determine whether choosing a group of objects involves permutations?

1. 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?

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?

3. 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?

The sample space S consists of permutations of the letters in APPLE, and there are 2 P’s,

so n (S) = 5! __ 2! = 60. Consider the two P’s as a single block, so there are 4 positions for the

letters to occupy.

n (A) = 4! = 24

The probability of getting the two P’s next to each other is P (A) = n (A)

_ n (S)

= 24 _ 60

= 2 _ 5

.

The equation is true if n!_ (n - a) ! = n!_

(n - b) ! . This occurs only when a = b.

They would become a single group of the letters, ABC. The other five groups are duplicates

of this result.

Permutations are used when the order of selection matters.

12! = 479,001,600

The songs can be played in 479,001,600 different orders

10 × 9 × 8 = 720

There are 720 possibilities for awarding medals.

9 × 8 × 7 × 6 = 3024

There are 3024 ways to arrange players.




A2_MNLESE385900_U8M19L2 966 04/09/14 6:43 PMExercise 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

ELABORATE AVOID COMMON ERRORSStudents 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.

SUMMARIZE THE LESSONWhat 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.

EVALUATE

ASSIGNMENT GUIDE

Concepts and Skills Practice

ExploreFinding the Number of Permutations

Exercises 1–3

Example 1Finding a Probability Using Permutations

Exercises 4–7, 18–19, 22–23, 26–27

Example 2Finding the Number of Permutations With Repetition

Exercises 8–24

Example 3Finding a Probability Using Permutations With Repetition

Exercises 12–17, 20–21, 25



© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y • I

mag

e C

red

its:

©C

had

M

cDer

mot

t/Sh

utte

rsto

ck

4. 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?

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?

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?

7. 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?

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.

n (S) = 9 P 3 = 9! _ (9 - 3) !

= 9! _ 6! = 504

n (A) = 1

P (A) = n (A)

_ n (S)

= 1 _ 504

Let S be the sample space and let A be the event that Sarah,

Jamal, and Mai are chosen in any order.

n (S) = 11 P 3 = 11! _ (11 - 3) !

= 11! _ 8! = 990

n (A) = 3 P 3 = 3! _ (3 - 3) !

= 3! _ 0! = 6

P (A) = n (A)

_ n (S)

= 6 _ 990

= 1 _ 165

Let S be the sample space and let A be the event that all 4 letters are placed in the correct envelopes.

n (S) = 4 P 4 = 4! _ (4 - 4) !

= 4! _ 0! = 24

n (A) = 1

P (A) = n (A)

_ n (S)

= 1 _ 24

Let S be the sample space and let A be the event that the coach chooses the three strongest swimmers.

n (S) = 8 P 3 = 8! _ (8 - 3) !

= 8! _ 5! = 336

n (A) = 3 P 3 = 3! _ (3 - 3) !

= 3! _ 0! = 6

P (A) = n (A)

_ n (S)

= 6 _ 336

= 1 _ 56




GRAPHIC ORGANIZERSHave students make a graphic organizer to summarize what they know about permutations. Ask them to include the formulas, with descriptions and examples.

967 Lesson 19 . 2


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

8. How many different sequences of letters can be formed using all the letters in ENVELOPE?

9. Yolanda has 3 each of red, blue, and green marbles. How many possible ways can the 9 marbles be arranged in a row?

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?

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?

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.)

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?

9 P 9

_ 3!3!3! = 9! _

3!3!3! = 1680

1680 ways

16 P 16

_ 10!6! = 16! _

10!6! = 8008

8008 row arrangements

10 P 10

_ 2!5! = 10! _

2!5! = 15,120

15,120 row arrangements

The three letter E’s are not distinguishable.

8 P 8

_ 3! = 8! _

3! = 6720

6720 sequences

Let S be the sample space and let A be the event that the grocer arranges the fruits in alternating order.

n (S) = 10 P 10

_ 5!5!

= 10! _ 5!5! = 252

n (A) = 2, AOAOAOAOAO or OAOAOAOAOA

P (A) = n (A)

_ n (S)

= 2 _ 252

= 1 _ 126

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) = 8 P 8

_ 2! = 8! _

2! = 20,160

n (A) = 1, GEOMETRY

P (A) = n (A)

_ n (S)

= 1 _ 20,160





INTEGRATE MATHEMATICAL PRACTICESFocus on Math ConnectionsMP.1 Discuss how to recognize when a probability evaluated using a permutation will be equal to the ratio 1 ___

n (S) .

AVOID COMMON ERRORSStudents 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.



© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

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?

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?

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?

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?

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?

n (S) = 21 P 21 _

11!10! = 21! _

11!10! = 352,716

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)

_ n (S)

= 1 _ 352, 716

n (S) = 8 P 8

_ 4!4!

= 8! _ 4!4! = 70, n (A) = 2, FFFFMMMM or MMMMFFFF

P (A) = n (A)

_ n (S)

= 2 _ 70

= 1 _ 35

n (S) = 12 P 3 = 12! _ (12 - 3) !

= 12! _ 9! = 1320, n (A) = 3 P 3 = 3! _

(3 - 3) != 3! __

0! = 6

P (A) = n (A)

_ n (S)

= 6 _ 1320

= 1 _ 220

n (S) = 7 P 2 = 7! _ (7 - 2) !

= 7! _ 5! = 42, n (A) = 2 P 2 = 2! _

(2 - 2) ! = 2! _

0! = 2

P (A) = n (A)

_ n (S)

= 2 _ 42

= 1 _ 21

n (S) = 5 P 5 = 5! _ (5 - 5) !

= 5! _ 0! = 120, n (A) = 1

P (A) = n (A)

_ n (S)

= 1 _ 120




PEERTOPEERHave 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.

VISUAL CUESDiscuss 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 n P r . Make sure students do not confuse the P in the permutation formula for the P used to denote probability.

969 Lesson 19 . 2


© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany • Im

age C

redits: ©

Dan

Kosmayer/Shuttersto

ck

19. A fan makes 6 posters to hold up at a basketball game.Each poster has a letter of the word TIGERS. Six friendssit next to each other in a row. The posters are distributedat random. What is the probability that TIGERS is spelledcorrectly when the friends hold up the posters?

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?

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.)

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 eventthat that TIGERS is spelled correctly when the friends hold up the posters.

n (S) = 6 P 6 = 6! _ (6 - 6) !

= 6! _ 0! = 720

n (A) = 1

P (A) = n (A)

_ n (S)

= 1 _ 720

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) = 10 P 4

_ 2!2!

= 5040 _ 4

= 1260

n (A) = 1

P (A) = n (A)

_ n (S)

= 1 _ 1260

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 = 10!_ (10 - 6) !

= 10!_4! = 151,200

n (A)= 6P 6 = 6!_ (6 - 6) !

= 6!_0! = 720

P (A)=n (A) _n (S)

= 720_151,200

= 1_210

Let S be the sample space and let A be the event that all four cards are 7s.

n (S) = 52 P 3 = 52! _ (52 - 3) !

= 52! _ 49! = 132,600

n (A) = 4 P 3 = 4! _ (4 - 3) !

= 4! _ 1! = 24

P (A) = n (A)

_ n (S)

= 24 _ 132,600

= 1 _ 5525







© H

oug

hton

Mif

flin

Har

cour

t Pub

lishi

ng

Com

pan

y

23. Match each event with its probability. All orders are chosen randomly.

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.

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.

1 _ 1092

1 _ 210

1 _ 90

H.O.T. Focus on Higher Order Thinking

24. Explain the Error Describe and correct the error in evaluating the expression.

5 P 3 = 5! _ 3! = 5 × 4 × 3! _ 3! = 20

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.

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.

P (A) = n (mascot 1st and football 2nd)

___ n (all floats in parade)

= 13 P 13

_ 15 P 15

= 1 _ 210

P (B) = n (Beth 5th and friend 4th)

___ n (all contestants)

= 8 P 8 _ 10 P 10 = 1 _

90

P (C) = n (Sylvester last, friends 1st and 2nd)

____ n (all musicians)

= 2. 11 P 11 _ 14 P 14

= 2 _ 2184

= 1 _ 1092

C

A

B

The denominator of the first fraction should be (5 - 3) !, not 3!; 5! ______ (5 - 3) !

= 5! __ 2! = 60.

Yes, they have the same probability. The probability of drawing 4 aces from a deck of cards

would be P(A) = n(A) ___

n(S) = 4 P 4

___ 52 P 4 because the 4 aces could be drawn in any order, and there are

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.

If he is doing his Algebra homework first, the only classes that can change homework

order are Biology and World History, so n (A) = 2 P 2 . The sample space would have all

three classes changing, so n (S) = 3 P 3 . Since P (A) = n (A)

___ n (S)

= 2 P 2 ___

3 P 3 = 2 _ 6 = 1 _ 3 , the probability of

Nolan doing his Algebra homework first is 1 _ 3 .




JOURNALHave students cite real-world examples of permutations. Have them explain how they know the examples are permutations.

971 Lesson 19 . 2


B R G R G B G B R

B G R R B G G R B

12 2

3

3

1

1

1

23

3

4 4

2

© H

oug

hton Mifflin H

arcourt Publishin

g Com

pany

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.

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.

n (S) = 6 P 4 = 6! _ (6 - 4) !

= 6! _ 2! = 360

n (A) = 4 P 4 = 4! _ (4 - 4) !

= 4! _ 0! = 24

P (A) = n (A)

_ n (S)

= 24 _ 360 = 1 _ 5

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.

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.

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. 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?

n (A) should be 1 since the tiles must appear in the order

B-E-A-D. The correct probability is 1 _ 360

.

two

three

Four colors are needed.



A2_MNLESE385900_U8M19L2 972 04/09/14 6:46 PM

EXTENSION ACTIVITY

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.

INTEGRATE MATHEMATICAL PRACTICESFocus on ModelingMP.4 Question 3 of the Lesson Performance Task 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

AVOID COMMON ERRORSSome 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.

Scoring Rubric2 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.



CorrectionKey=NL-B;CA-B 19 . 2 DO NOT EDIT--Changes … · Use permutations and combinations to compute probabilities of compound events and solve problems. Mathematical Practices

Documents