1. 10-9 Permutations Warm Up Problem of the Day Lesson Presentation
Course 3
Warm Up
Problem of the Day
Lesson Presentation

2. 10-9 Permutations Warm Up Find the number of possible outcomes.
Course 3
10-9
Permutations
Warm Up
Find the number of possible outcomes.
1. bagels: plain, egg, wheat, onion
meat: turkey, ham, roast beef, tuna 16

3. 10-9 Permutations Warm Up Find the number of possible outcomes.
Course 3
10-9
Permutations
Warm Up
Find the number of possible outcomes.
2. eggs: scrambled, over easy, hard boiled
meat: sausage patty, sausage link, bacon, ham 12

4. 10-9 Permutations Warm Up Find the number of possible outcomes.
Course 3
10-9
Permutations
Warm Up
Find the number of possible outcomes.
3. How many different 4–digit phone extensions are possible?
10,000

5. 10-9 Permutations Problem of the Day
Course 3
10-9
Permutations
Problem of the Day
What is the probability that a 2-digit whole number will contain exactly one 1? 17 90

6. Course 3
10-9
Permutations
Learn to find permutations.

7. Insert Lesson Title Here
Course 3
10-9
Permutations
Insert Lesson Title Here
Vocabulary
factorial
permutation

8. 5! = 5 • 4 • 3 • 2 • 1 10-9 Permutations and Combinations
Course 3
10-9
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

9. Example 1: Evaluating Expressions Containing Factorials
Course 3
10-9
Permutations
Example 1: Evaluating Expressions Containing Factorials
Evaluate each expression.
A. 9!
9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 362,880 8! B. 6!
Write out each factorial and simplify.
8 •7 • 6 • 5 • 4 • 3 • 2 • 1
6 • 5 • 4 • 3 • 2 • 1
Multiply remaining factors.
8 • 7 = 56

10. Example 1: Evaluating Expressions Containing Factorials
Course 3
10-9
Permutations and Combinations
Example 1: Evaluating Expressions Containing Factorials
10!
(9 – 2)! C.
10! 7!
Subtract within parentheses.
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
7  6  5  4  3  2  1
10 • 9 • 8 = 720

11. 10-9 Permutations Check It Out: Example 1 Evaluate each expression.
Course 3
10-9
Permutations
Check It Out: Example 1
Evaluate each expression.
A. 10!
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 3,628,800 7! B. 5!
Write out each factorial and simplify.
7 • 6 • 5 • 4 • 3 • 2 • 1
5 • 4 • 3 • 2 • 1
Multiply remaining factors.
7 • 6 = 42

12. 10-9 Permutations Check It Out: Example 1 9! C. (8 – 2)! 9!
Course 3
10-9
Permutations
Check It Out: Example 1 9!
(8 – 2)! C. 9! 6!
Subtract within parentheses.
9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
6  5  4  3  2  1
9 • 8 • 7 = 504

13. Course 3
10-9
Permutations
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

14. Course 3
10-9
Permutations
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  
= 60 permutations
5 • 4 • 3

15. 10-9 Permutations ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
Course 3
10-9
Permutations
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

16. Example 2A: Finding Permutations
Course 3
10-9
Permutations
Example 2A: Finding Permutations
Jim has 6 different books.
Find the number of orders in which the 6 books can be arranged on a shelf.
6 • 5 • 4 • 3 • 2 • 1 =
720
There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

17. 10-9 Permutations Check It Out: Example 2A
Course 3
10-9
Permutations
Check It Out: Example 2A
There are 7 soup cans in the pantry.
Find the number of orders in which all 7 soup cans can be arranged on a shelf.
Use 7!
7! =
7 • 6 • 5 • 4 • 3 • 2 • 1
= 5040
There are 5040 orders in which to arrange 7 soup cans.

18. Example 2B: Finding Permutations
Course 3
10-9
Permutations
Example 2B: Finding Permutations
If the shelf has room for only 3 of the books, find the number of ways 3 of the 6 books can be arranged.
The number of books is 6. =
6P3
6 • 5 • 4
The books are arranged 3 at a time.
= 120
There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways.

19. 10-9 Permutations Check It Out: Example 2B
Course 3
10-9
Permutations
Check It Out: Example 2B
There are 7 soup cans in the pantry.
If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged.
The number of cans is 7. =
7 • 6 • 5 • 4
7P4
The cans are arranged 4 at a time.
= 840
There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways.

20. Insert Lesson Title Here
Course 3
10-9
Permutations
Insert Lesson Title Here
Lesson Quiz
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?
362,880 9! 5!
3024
40,320