1. Pre-Algebra HOMEWORK
Page 469 #1-12 ANSWERS

2. Pre-Algebra HOMEWORK
Page 474 #1-16

3. Permutations and Combinations
9-6
Permutations and Combinations
Warm Up
Problem of the Day
Lesson Presentation
Pre-Algebra

4. Permutations and Combinations
Pre-Algebra
9-6
Permutations and Combinations
Warm Up
Find 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 12
10,000

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

6. Today’s Learning Goal Assignment
Learn to find permutations and combinations.

7. Vocabulary
factorial
permutation
combination

8. 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. Additional Example 1A & 1B: Evaluating Expressions Containing Factorials
Evaluate each expression.
A. 8!
8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 40,320 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. Try This: Example 1A & 1B
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

11. Additional Example 1C: 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

12. Try This: Example 1C 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. 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. Notice that the product can be written as a quotient of factorials.
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  
= 60 permutations
Notice that the product can be written as a quotient of factorials.
5 • 4 • 3 • 2 • 1
2 • 1 =
5! 2!
60 = 5 • 4 • 3 =

16. Additional Example 2A: Finding Permutations
Jim has 6 different books.
A. Find the number of orders in which the 6 books can be arranged on a shelf.
The number of books is 6. 6!
(6 – 6)! = 6! 0! =
6 • 5 • 4 • 3 • 2 • 1 1 =
6P6 =
720
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.

17. Try This: Example 2A
There are 7 soup cans in the pantry.
A. Find the number of orders in which all 7 soup cans can be arranged on a shelf.
The number of cans is 7. 7!
(7 – 7)! = 7! 0! =
7 • 6 • 5 • 4 • 3 • 2 • 1 1
7P7 =
= 5040
The cans are arranged 7 at a time.
There are 5040 orders in which to arrange 7 soup cans.

18. Additional Example 2B: Finding Permutations
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.
The number of books is 6. 6!
(6 – 3)! = 6! 3! =
6 • 5 • 4 • 3 • 2 • 1
3 • 2 • 1 =
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. Try This: Example 2B
There are 7 soup cans in the pantry.
B. 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!
(7 – 4)! = 7! 3! =
7 • 6 • 5 • 4 • 3 • 2 • 1
3 • 2 • 1
7P4 =
The cans are arranged 4 at a time.
= 7 • 6 • 5 • 4
= 840
There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways.

20. A combination is a selection of things in any order.

21. 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.

22. 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 10

24. Additional Example 3A: Finding 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
10!
2!(10 – 2)! =
10!
2!8!
10C2 =
2 books chosen at a time
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
(2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) =
= 45
There are 45 combinations. This means that Mary can buy 45 different pairs of books.

25. Try This: Example 3A
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
12!
4!(12 – 4)! =
12!
4!8!
12C4 =
4 DVDs chosen at a time =
12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
(4 • 3 • 2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1)
= 495

26. Try This: Example 3A Continued
There are 495 combinations. This means that Harry can buy 495 different sets of 4 DVDs.

27. Additional Example 3B: Finding Combinations
B. If Mary wants to buy 7 books, find the number of different sets of 7 books she can buy.
10 possible books
10!
7!(10 – 7)! =
10!
7!3!
10C7 =
7 books chosen at a time
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.

28. Try This: Example 3B
B. If Harry wants to buy 11 DVDs, find the number of different sets of 11 DVDs he can buy.
12 possible DVDs
12!
11!(12 – 11)! =
12!
11!1!
12C11 =
11 DVDs chosen at a time =
12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
(11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1)(1)
= 12

29. Try This: Example 3B Continued
There are 12 combinations. This means that Harry can buy 12 different sets of 11 DVDs.

30. Evaluate each expression. 1. 9! 2.
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?
4. A group of 12 people are forming a committee. How many different 4-person committees can be formed?
362,880 9! 5!
3024
40,320
495