1. Permutations and Combinations
Course 3
10-9
Permutations and Combinations
Warm Up #20
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

2. Homework Solution. Lesson 10.1
10) 1/5 12) 1/6 14) 1/2 16) 1/3 18) 0 26) 17,576,000 27) 6,760,000 28) 17,576,000 29) 6,760,000

3. Homework
Lesson 10.2 _page 641 #45-50 ALL

4. Lesson 10.2 Permutations

5. 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.
** 0! = 1
5! = 5 • 4 • 3 • 2 • 1
Read 5! as “five factorial.”
Reading Math

6. Additional Example 1: Evaluating Expressions Containing Factorials
Course 3
10-9
Permutations and Combinations
Additional 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

7. Additional Example 1: Evaluating Expressions Containing Factorials
Course 3
10-9
Permutations and Combinations
Additional 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

8. Permutations and Combinations
Course 3
10-9
Permutations and Combinations
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

9. Permutations and Combinations
Course 3
10-9
Permutations and Combinations
A permutation is an arrangement of things in a certain order. (THE ORDER DOES MATTER!)
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

10. Permutations and Combinations
Course 3
10-9
Permutations and Combinations
By definition, 0! = 1.
Remember!

11. Permutations and Combinations
Course 3
10-9
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  
= 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 =

12. Additional Example 2A: Finding Permutations
Course 3
10-9
Permutations and Combinations
Additional 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.
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.

13. Additional Example 2B: Finding Permutations
Course 3
10-9
Permutations and Combinations
Additional 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. 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.

14. Permutations and Combinations
Course 3
10-9
Permutations and Combinations
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.
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.

15. Permutations and Combinations
Course 3
10-9
Permutations and Combinations
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!
(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.

16. Permutations with Identical Objects
The number of distinct permutations of n objects with r identical objects is given by
𝒏! 𝒓! , where 1≤𝑟≤𝑛
More identical objects:
𝒏! 𝒓 𝟏 ! 𝒓 𝟐 ! 𝒓 𝟑 !

17. Example 1
Shay is planting 11 colored flowers in a line. In how many ways can she plant 4 red flowers, 5 yellow flowers, and 2 purple flowers? 11! 4!∙5!∙2! = Answer: 6930 ways

18. Example 2
Kelly is planting 10 colored flowers in a line. In how many ways can she plant 4 red flowers and 3 purple flowers?

19. Circular Permutations
If n distinct objects are arranged around a circle, then there are (n-1)! Circular permutations of the n objects.

20. Example 3
In how many different ways can 7 different appetizers be arranged on a circular tray? Answer: 720 distinct ways

21. Example 4
Five different stuffed animals are to be placed on a circular display rack in a department store. In how many ways can this be done?
Answer: 24 ways

22. Review on Permutations

23. Homework Solution lesson 10.2
45) 151,200 46) 34,650 47) 908,107,200 48) 24 49) ) 24

24. Homework
Lesson 10.2_pg 640 #12-22 EVEN

25. Classwork Write down your name, period, date
TITLE: Lesson 10.2 Classwork
Task: your task is to complete each problem. You need to show your work in order to get full classwork points

26. Example
10) Kelly is planting 10 colored flowers in a line. In how many ways can she plant 4 red flowers and 3 purple flowers? 10) 10! 3!4! =25200 ways

27. Warm-Up #21
7P3=
4P3 8P3 =
In how many ways can 3 males (Joe, Jerry John) and 3 females (Jamie, Jenny, Jasmine) be seated in a row if the genders alternate down the row?
210
0.7 72

28. Homework Solution lesson 10.2
12) 2 14) ) ) ) ) About 0.07

29. Homework
Lesson 10.3_pg 647 #24-26 ALL #34-38 ALL

30. Lesson 10.3 Combinations

31. Example
You have 3 letters. A B C. Write all of the combinations. A B C A C A B A C B C A C A B C B A
How many different combinations can I make if the order of the letters does matter?
Permutations: 6 ways
How many different combinations can I make if the order of the letters does not matter?
Combination: 1 way

32. Course 3
10-9
Permutations and Combinations
A combination is a selection of things in any order. The order does not matter.

33. Permutations and Combinations
Course 3
10-9
Permutations and Combinations

34. Additional Example 3A: Finding Combinations
Course 3
10-9
Permutations and Combinations
Additional Example 3A: Finding Combinations
Mary wants to join a book club that offers a choice of 10 new books each month.
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.

35. Additional Example 3B: Finding Combinations
Course 3
10-9
Permutations and Combinations
Additional Example 3B: Finding Combinations
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.

36. Permutations and Combinations
Course 3
10-9
Permutations and Combinations
Check It Out: Example 3A
Harry wants to join a DVD club that offers a choice of 12 new DVDs each month.
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

37. Check It Out: Example 3A Continued
Course 3
10-9
Permutations and Combinations
Check It Out: Example 3A Continued
There are 495 combinations. This means that Harry can buy 495 different sets of 4 DVDs.

38. Permutations and Combinations
Course 3
10-9
Permutations and Combinations
Check It Out: Example 3B
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

39. Check It Out: Example 3B Continued
Course 3
10-9
Permutations and Combinations
Check It Out: Example 3B Continued
There are 12 combinations. This means that Harry can buy 12 different sets of 11 DVDs.

40. Example
How many different ways are there to purchase 2 CDs, 3 cassettes, and 1 videotape if there are 7 CD titles, 5 cassette titles, and 3 videotape titles? Answer: (7C2)(5C3)(3C1)= 630 ways