1. Counting principle and permutations
Chapter 10
Counting principle and permutations

2. Ways to Count- Tree Diagram
A sporting goods store offers 3 types of snowboards (all-mountain, freestyle, and carving) and 2 types of boots (soft and hybrid). How many choices does the store offer for snowboarding equipment?

3. A sporting goods store offers 3 types of snowboards (all-mountain, freestyle, and carving) and 2 types of boots (soft and hybrid). How many choices does the store offer for snowboarding equipment?
All-mountain
Soft
All-mountain board, soft boots
Hybrid
all-mountain board, hybrid boots
Freestyle
Freestyle board, soft boots
Freestyle board, hybrid boots
Carving
Carving board, soft boots
Hard
Carving board, hard boots

4. Ways to count- Tree Diagram
The tree has 6 branches. So, there are 6 possible choices!

5. Use a Tree Diagram
The snowboard shop also offers 3 different types of bicycles (mountain, racing, and BMX) and 3 different wheel sizes (20”, 22”, and 24”). How many bicycle choices does the store offer?

6. ANSWER
9 Bicycles

7. Fundamental Counting Principle
Snowboarding example-
There are 3 options for boards and 2 options for boots.
So the number of possibilities is 32=6
This is called the “Fundamental Counting Principle”

8. Fundamental Counting Principle
Two events-
If one event can occur in m ways and another event can occur in n ways, then the number of ways that both events can occur is mn.
Three or more events-
The fundamental counting principle extends to 3 or more events. The product of possibilities for each event will give the possibility of ALL the events occurring.

9. Use the Fundamental Counting Principle
You are framing a picture. The frames are available in 12 different styles. Each style is available in 55 different colors. You also want a blue mat board, which is available in 11 different shades of blue. How many different ways can you frame the picture?

10. Use the Fundamental Counting Principle
Multiply
# of frame styles (12)
# of frame colors (55)
# of mat boards (11)
12 x 55 x 11 = 7260
There are 7260 different ways to frame the picture!

11. Use the Counting Principle with or without Repetition
The standard configuration for a Texas license plate is 1 letter followed by 2 digits followed by 3 letters

12. Use the Counting Principle with Repetition
How many different license plates are possible if letters and digits can be repeated?

13. Use the Counting Principle with Repetition
How many different license plates are possible if letters and digits can be repeated?
Multiply
# of letters possible (26)
# of digits possible (10)

14. Use the Counting Principle with Repetition
26 x 10 x 10 x 26 x 26 x 26 = 45,697,600

15. Use the Counting Principle without Repetition
How many different license plates are possible if letters and digits cannot be repeated?

16. Use the Counting Principle without Repetition
How many different license plates are possible if letters and digits cannot be repeated?
Multiply
# of letters possible (26)
# of digits possible (10)
# of digits possible w/o the previous digit used (9)
# of letters possible w/o the first letter used (25)
# of letters possible w/o the first 2 letters used (24)
# of letters possible w/o the first 3 letters used (23)

17. Use the Counting Principle without Repetition
26 x 10 x 9 x 25 x 24 x 23 = 32,292,000

18. More Counting Principles with and without Repetition
How would the answers change for the standard configuration of a New York license plate, which is 3 letters followed by 4 numbers?
How many possibilities with repetition?
How many possibilities without repetition?

19. Answer With Repetition- 175,760,000 possibilities
Without Repetition- 78,624,000 possibilities

20. Permutations Question- How many ways can you rearrange the word COW
COW OCW WOC
CWO OWC WCO

21. Permutations The answer is 6 times.
The ordering of a certain number of n objects is called a PERMUTATION.

22. Permutations
Back to the question… “How many ways can you rearrange the word COW?”
You can use the fundamental counting principle to find the number of permutations of C, O, and W.
First letter- 3 choices
Second letter- 2 choices
Third letter- 1 choice

23. Permutations
Therefore, we find the number of permutations is 3 x 2 x 1 = 6

24. Permutations
How many permutations for the word:
SEA
ROCK
MATH

25. Factorials! 3 x 2 x 1 can also be written as 3!
! is the symbol used for factorial
3! is read “three factorial”

26. Factorials!
In general, n! is defined where n is a positive integer as follows:
n! = n  (n-1)  (n-2)  …  3  2  1

27. Factorials!
5! = 5  4  3  2  1 = 120
12! = 12  11  10  9  8  7  6  5  4  3  2  1 = 479,001,600

28. Permutations using Factorials
Ten teams are competing in the final round of the Olympic four-person bobsledding competition. In how many different ways can the bobsledding teams finish the competition? (Assume no ties)

29. Permutations with Factorials
There are 10! ways that teams can finish.
10! = 10  9  8  7  6  5  4  3  2  1 = 3,628,800

30. Permutations with Factorials
How many different ways can 3 of the bobsledding teams finish first, second, and third to win the gold, silver, and bronze medals?

31. Permutations with Factorials
Any of the 10 teams can finish first, then any of the remaining 9 teams can finish second, and finally any of the 8 remaining teams can finish third.
So number of ways teams can win medals is
10  9  8 = 720

32. Permutations with Factorials
How would the answers change if there were 12 bobsledding teams competing in the final round of he competition?
A- How many was can the teams finish?
B- How many ways to win medals?

33. Answer
A- 12! = 479,001,600
B- 12  11  10 = 1320

34. Permutations of n Objects Taken r at a Time
The answer to the second part of the bobsledding questions is called the number of permutations of 10 objects taken 3 at a time.
Denoted as 10P3
10P3 = 10  9  8 =
= =

35. Permutations with Repetition
The general equation for the number of permutations of n objects taken r at a time is:

36. Use the Permutation Equation
You a burning a demo CD for your band. Your band has 12 songs stored on your computer. However, you want to put only 4 songs on the demo CD. In how many orders can you burn 4 of the 12 songs onto the CD?

37. Use the Permutation Equation
Try:
5P3
4P1
8P5

38. Practice Answers
5P3 = 60
4P1 = 4
8P5 = 6720

39. Permutations with Repetition
How many ways can you write the word
MOO
Note: Each “O” is a distinct letter.

40. Permutations with Repetition
MOO OMO OOM There are 6 permutations of the letters M, O, and O. However, if the two occurrences of O are considered interchangeable, then there are only three distinguishable permutations: MOO OMO OOM

41. Permutations with Repetition
Each of these permutations corresponds to two of the original six permutations because there are 2! , or 2, permutations of O and O.
So the number of permutations of M, O, and O can be written as

42. Permutations with Repetition
To generalize the situation we just explored we can say that the number of distinguishable permutations of n objects where one object is repeated s1 times, another is repeated s2 times, and so on, is:

43. Permutations with Repetition
How many ways can you rearrange the letters of the following words:
MATHEMATICS
ALGEBRA
MISSISSIPPI
TALLAHASSEE

44. Permutations with Repetition
MATHEMATICS
ALGEBRA

45. Permutations with Repetition
MISSISSIPPI
TALLAHASSEE

46. Combinations

47. Permutations vs Combinations
With permutations, we learned that order is important for some counting problems.
For other counting problems, order is not important.
For example, if you purchase a package of trading cards, the order of the cards inside the package is not important.

48. Permutations vs. Combinations
COMBINATION- a selection of r objects from a group of n objects where order is NOT important.
Think of a combination pizza; you have various kinds of toppings (pepperoni, olives, green peppers, etc.) and you can put them on in several different orders, but in the end, order doesn’t matter. Its just one delicious pizza!

49. Combination Equation
The number of combinations of r objects taken from a group of n distinct objects is denoted by nCr and is given by this formula:
nCr is read “n choose r”

50. Find Combinations
A standard deck of 52 playing cards has 4 suits with 13 different cards in each suit.
SUIT- hearts, clubs, spades, and diamonds

51. Find Combinations Each suit has the following cards called “kinds”:
Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King
There is one of each of these kinds in each suit
Face Cards- Cards that have faces on them (Jack, Queen, King)

52. Find Combinations
If the order in which the cards are dealt is not important, how many different 5-card hands are possible?
Remember…
In this example, n= 52 cards and r = 5 cards

53. Find Combinations
Therefore,

54. Find Combinations
In how many 5-card hands are all 5 cards of the same color?

55. Find Combinations
For all 5 cards to be the same color, you need to choose 1 of the 2 colors and then 5 of the 26 cards in that color. So, the number of possible hands is:

56. Multiple Events
When finding the number of ways both an event A and an event B can occur, you need to multiply.
When finding the number of ways that event A or event B can occur, you add instead.

57. Add or Multiply Combinations
William Shakespeare wrote 38 plays that can be divided into 3 genres. Of the 38 plays, 18 are comedies, 10 are histories, and 10 are tragedies.
How many different sets of exactly 2 comedies and 1 tragedy can you read?
How many different sets of at most 3 plays can you read?

58. Add or Multiply Combination
How many different sets of exactly 2 comedies and 1 tragedy can you read?
You can choose 2 of the 18 comedies and 1 of the 10 tragedies. So, the number of possible sets of plays is:

59. Add or Multiply Combinations
How many different sets of at most 3 plays can you read?
You could read 0, 1, 2, or 3 plays. Because there are 38 plays that can be chosen, the number of possible sets of plays is:

60. Practice
How many different sets of exactly 3 tragedies and 2 histories can you read?
8C3
10C6
7C2

61. Practice Answers
5400 sets
8C3 = 56
10C6 = 210
7C2 = 21

62. Subtracting Possibilities
Counting problems that involve phrases like “at most” or “at least” are sometimes easier to solve by subtracting possibilities you do not want from the total number of possibilities.
The portion NOT included in the number of possibilities is also called the COMPLIMENT.

63. Subtracting Possibilities
During the school year, the girl’s basketball team is scheduled to play 12 home games. You want to attend at least 3 of the games. How many different combinations of games can you attend?

64. Subtracting Possibilities
Of the 12 home games, you want to attend 3 games, or 4 games, or 5 games, and so on. So, the number of combinations of games you can attend is:

65. Subtracting Possibilities
Instead of adding these combinations, use the following reasoning. For each of the 12 games, you can choose to attend or not attend the game, so there are 212 total combinations. If you attend at least 3 games, you do not attend only a total of 0, 1, or 2 games. So, the number of ways you can attend at least 3 games is:

66. Pascal’s Triangle
If you arrange the values of nCr in a triangular pattern in which each row corresponds to a value of n, you get what is called Pascal’s triangle, named after the French mathematician Blaise Pascal.

67. Pascal’s Triangle
Pascal’s triangle is show below with its entries represented by combinations and with its entries represented by numbers The first and last numbers in each row are 1. Every number other than 1 is the sum of the closest two numbers in the row directly above it.
Pascal’s Triangle as numbers
Pascal’s Triangle as combinations

68. Use Pascal’s Triangle
The 6 members of a Model UN club must choose 2 representatives to attend a state convention. Use Pascal’s triangle to find the number of combinations of 2 members that can be chosen as representatives.

69. Use Pascal’s Triangle
Because you need to find 6C2, write the 6th row of Pascal’s triangle by adding numbers from the previous row.
n = 5 (5th row)
n = 6 (6th row)
The value of 6C2 is the 3rd number in the 6th row of Pascal’s triangle, as show above. Therefore, 6C2 = 15. There are 15 combinations of representatives for the convention.

70. Binomial Expansions
There is an important relationship between powers of binomials and combinations. The numbers in Pascal’s triangle can be used to find coefficients in binomial expansions.

71. Binomial Expansions
For example, the coefficients in the expansion (a + b)4 are the numbers of combinations in the row of Pascal’s triangle for n = 4:
(a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4
4C C C C C4
This result is generalized in the binomial theorem.

72. Binomial Theorem
For any positive integer n, the binomial expansion of (a + b)n is:
(a + b)n = nC0anb0 + nC1an-1b1 + nC2an-2b2 +…+ nCna0bn
Notice that each term in the expansion of (a + b)n has the form nCran-rbr where r is an integer from 0 to n.

73. Expand a power of a binomial sum
Use the binomial theorem to write the binomial expansion for:
(x2 + y)3

74. Expand a power of a binomial sum

75. Expand a power of a binomial difference
Use the binomial theorem to write the binomial expansion of:
(a-2b)4

76. Expand a power of a binomial difference

77. Powers of Binomial Differences
To expand a power of a binomial difference, you can rewrite the binomial as a sum. The resulting expansion will have terms whose signs alternate between + and -.

78. Use the binomial theorem to write the binomial expansion
(a + 2b)4
(2p – q)4
(5 – 2y)3

79. (x + 3)5
(a + 2b)4

80. Find a coefficient in an expansion
Find the coefficient of x4 in the expansion of (3x + 2)10

81. Find a coefficient in an expansion
From the binomial theorem, you know the following:
Each term in the expansion has the form
The term containing x4 occurs when r = 6:

82. Find the coefficient in an expansion
So the coefficient of x4 is 1,088,640
Find the coefficient of:
x5 in the expansion of (x – 3)7
x3 in the expansion of (2x + 5)8

83. Practice Answers x5 in the expansion of (x – 3)7 189
1,400,000