9-6 Permutations and Combinations

Course 3

Warm Up

Problem of the Day

Lesson Presentation

Warm UpFind the number of possible outcomes.

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

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

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

16

10,000

12

Course 3

9-6 Permutations and Combinations

Problem of the Day

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

Course 3

9-6 Permutations and Combinations

1790

Learn to find permutations and combinations.

Course 3

9-6 Permutations and Combinations

Vocabulary

factorialpermutationcombination

Insert Lesson Title Here

Course 3

9-6 Permutations and Combinations

Course 3

9-6 Permutations 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 • 4 • 3 • 2 • 1

Read 5! as “five factorial.”

Reading Math

Evaluate each expression.

Additional Example 1A & 1B: Evaluating Expressions Containing Factorials

Course 3

9-6 Permutations and Combinations

A. 8!

8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 40,320

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.

Additional Example 1C: Evaluating Expressions Containing Factorials

Course 3

9-6 Permutations 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.

Try This: Example 1A & 1B

Course 3

9-6 Permutations and Combinations

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.

Try This: Example 1C

Course 3

9-6 Permutations and Combinations

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

9-6 Permutations and Combinations

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

9-6 Permutations and Combinations

If no letter can be used more than once, there are 60 permutations 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

Notice that the product can be written as a quotient of factorials.

60 = 5 • 4 • 3 =

= 60 permutations

5 • 4 • 3 • 2 • 12 • 1

=5! 2!

Course 3

9-6 Permutations and Combinations

Jim has 6 different books.

Additional Example 2A: Finding Permutations

Course 3

9-6 Permutations and Combinations

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

720 6!(6 – 6)!

= 6!0! = 6 • 5 • 4 • 3 • 2 • 1

1=6P6 =

The number of books is 6.

The books are arranged 6 at a time.

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

Additional Example 2B: Finding Permutations

Course 3

9-6 Permutations and Combinations

B. 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 6!(6 – 3)!

= 6!3! = 6 • 5 • 4 • 3 • 2 • 1

3 • 2 • 1=6P3 =

The number of books is 6.

The books are arranged 3 at a time. = 120

Course 3

9-6 Permutations and Combinations

= 5040 7!

(7 – 7)!= 7!

0! = 7 • 6 • 5 • 4 • 3 • 2 • 1 1

7P7 =

The number of cans is 7.

The cans are arranged 7 at a time.

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

Try This: Example 2A

A. 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.

Course 3

9-6 Permutations and Combinations

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

= 7 • 6 • 5 • 4

7!(7 – 4)!

= 7!3! = 7 • 6 • 5 • 4 • 3 • 2 • 1

3 • 2 • 17P4 =

The number of cans is 7.

The cans are arranged 4 at a time. = 840

There are 7 soup cans in the pantry.Try This: Example 2B

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

Course 3

9-6 Permutations and Combinations

A combination is a selection of things in any order.

Course 3

9-6 Permutations and Combinations

If no letter is used more than once, there is only 1 combination of the first 3 letters of the alphabet. ABC, ACB, BAC, BCA, CAB, and CBA are considered to be the same combination of A, B, and C because the order does not matter.

If no letter is used more than once, there are 10 combinations of the first 5 letters of the alphabet, when taken 3 at a time. To see this, look at the list of permutations on the next slide.

Course 3

9-6 Permutations and Combinations

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

Course 3

9-6 Permutations and Combinations

Additional Example 3A: Finding Combinations

Course 3

9-6 Permutations and Combinations

Mary wants to join a book club that offers a choice of 10 new books each month.

A. If Mary wants to buy 2 books, find the number of different pairs she can buy.10 possible books

2 books chosen at a time

10!2!(10 – 2)!

= 10!2!8!

= 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1(2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1)

10C2 =

= 45

There are 45 combinations. This means that Mary can buy 45 different pairs of books.

Additional Example 3B: Finding Combinations

Course 3

9-6 Permutations and Combinations

B. If Mary wants to buy 7 books, find the number of different sets of 7 books she can buy.

10 possible books

7 books chosen at a time

10!7!(10 – 7)!

= 10!7!3!10C7 =

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

= = 120

There are 120 combinations. This means that Mary can buy 120 different sets of 7 books.

Try This: Example 3A

Course 3

9-6 Permutations and Combinations

Harry wants to join a DVD club that offers a choice of 12 new DVDs each month.

A. If Harry wants to buy 4 DVDs, find the number of different sets he can buy.

12 possible DVDs

4 DVDs chosen at a time

12!4!(12 – 4)!

= 12!4!8!

= 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1(4 • 3 • 2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1)

12C4 =

= 495

Try This: Example 3A Continued

Course 3

9-6 Permutations and Combinations

There are 495 combinations. This means that Harry can buy 495 different sets of 4 DVDs.

Try This: Example 3B

Course 3

9-6 Permutations and Combinations

B. If Harry wants to buy 11 DVDs, find the number of different sets of 11 DVDs he can buy.

12 possible DVDs

11 DVDs chosen at a time

12!11!(12 – 11)! =

12!11!1!

= 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1(11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1)(1)

12C11 =

= 12

Try This: Example 3B Continued

Course 3

9-6 Permutations and Combinations

There are 12 combinations. This means that Harry can buy 12 different sets of 11 DVDs.

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?

4. A group of 12 people are forming a committee. How many different 4-person committees can be formed?

Lesson Quiz

3024

362,880

Insert Lesson Title Here

40,320

495

Course 3

9-6 Permutations and Combinations

9!5!