10-9 Permutations

Course 3

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Warm UpFind the number of possible outcomes.

1. bagels: plain, egg, wheat, onion meat: turkey, ham, roast beef, tuna

16

Course 3

10-9Permutations

Warm UpFind the number of possible outcomes.

2. eggs: scrambled, over easy, hard boiled meat: sausage patty, sausage link, bacon, ham

12

Course 3

10-9Permutations

Warm UpFind the number of possible outcomes.

3. How many different 4–digit phone extensions are possible?

10,000

Course 3

10-9Permutations

Problem of the Day

What is the probability that a 2-digit whole number will contain exactly one 1?

Course 3

10-9Permutations

1790

Learn to find permutations.

Course 3

10-9Permutations

Vocabulary

factorialpermutation

Insert Lesson Title Here

Course 3

10-9Permutations

Course 3

10-9Permutations and Combinations

The factorial of a number is the product of all the whole numbers from the number down to 1. The factorial of 0 is defined to be 1.

5!5! = 55 • 44 • 33 • 22 • 11

Read 5! as “five factorial.”

Reading Math

Evaluate each expression.

Example 1: Evaluating Expressions Containing Factorials

Course 3

10-9Permutations

A. 9!

9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 362,880

8!6!

8 •7 • 6 • 5 • 4 • 3 • 2 • 1 6 • 5 • 4 • 3 • 2 • 1

Write out each factorial and simplify.

8 • 7 = 56

B.

Multiply remaining factors.

Example 1: Evaluating Expressions Containing Factorials

Course 3

10-9Permutations and Combinations

Subtract within parentheses.

10 • 9 • 8 = 720

10!7!

10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 7 6 5 4 3 2 1

C. 10!

(9 – 2)!

Evaluate each expression.

Check It Out: Example 1

Course 3

10-9Permutations

A. 10!

10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 3,628,800

7!5!

7 • 6 • 5 • 4 • 3 • 2 • 1 5 • 4 • 3 • 2 • 1

Write out each factorial and simplify.

7 • 6 = 42

B.

Multiply remaining factors.

Check It Out: Example 1

Course 3

10-9Permutations

Subtract within parentheses.

9 • 8 • 7 = 504

9!6!

9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 6 5 4 3 2 1

C. 9!

(8 – 2)!

Course 3

10-9Permutations

A permutation is an arrangement of things in a certain order.

If no letter can be used more than once, there are 6 permutations of the first 3 letters of the alphabet: ABC, ACB, BAC, BCA, CAB, and CBA.

first letter

?

second letter

?

third letter

?

3 choices 2 choices 1 choice

The product can be written as a factorial.

• •

3 • 2 • 1 = 3! = 6

Course 3

10-9Permutations

If no letter can be used more than once, there are 60 permutations (orders) of the first 5 letters of the alphabet, when taken 3 at a time: ABE, ACD, ACE, ADB, ADC, ADE, and so on.

first letter

?

second letter

?

third letter

?

5 choices 4 choices 3 choices

5 • 4 • 3 = 60 permutations

Course 3

10-9Permutations

ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE

ACB ADB AEB ADC AEC AED BDC BEC BED CED

BAC BAD BAE CAD CAE DAE CBD CBE DBE DCE

BCA BDA BEA CDA CEA DEA DBC CEB DEB DEC

CAB DAB EAB DAC EAC EAD DCB EBC EBD ECD

CBA DBA EBA DCA ECA EDA DBC ECB EDB EDC

These 6 permutations are all the same combination.

In the list of 60 permutations, each combination is repeated 6 times. The number of combinations is = 10.

60 6

Jim has 6 different books.

Example 2A: Finding Permutations

Course 3

10-9Permutations

Find the number of orders in which the 6 books can be arranged on a shelf.

7206 • 5 • 4 • 3 • 2 • 1 =

There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

Course 3

10-9Permutations

= 5040 7! = 7 • 6 • 5 • 4 • 3 • 2 • 1

Use 7!

There are 5040 orders in which to arrange 7 soup cans.

Check It Out: Example 2A

Find the number of orders in which all 7 soup cans can be arranged on a shelf.

There are 7 soup cans in the pantry.

Example 2B: Finding Permutations

Course 3

10-9Permutations

If the shelf has room for only 3 of the books, find the number of ways 3 of the 6 books can be arranged.

There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways.

6 • 5 • 4 =6P3

The number of books is 6.

The books are arranged 3 at a time. = 120

Course 3

10-9Permutations

There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways.

= 7 • 6 • 5 • 47P4

The number of cans is 7.

The cans are arranged 4 at a time. = 840

There are 7 soup cans in the pantry.Check It Out: Example 2B

If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged.

Evaluate each expression.

1. 9!

2.

3. There are 8 hot air balloons in a race. In how many possible orders can all 8 hot air balloons finish the race?

Lesson Quiz

3024

362,880

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40,320

Course 3

10-9Permutations

9!5!