1. Chapter 5 Section 5
Counting
Techniques

2. Chapter 5 – Section 5 Learning objectives
Solving counting problems using the Multiplication Rule
Solving counting problems using permutations
Solving counting problems using combinations
Solving counting problems involving permutations with nondistinct items
Compute probabilities involving permutations and combinations 1 2 3 5 4

3. Chapter 5 – Section 5 Learning objectives
Solving counting problems using the Multiplication Rule
Solving counting problems using permutations
Solving counting problems using combinations
Solving counting problems involving permutations with nondistinct items
Compute probabilities involving permutations and combinations 1 2 3 5 4

4. Chapter 5 – Section 5
The classical method, when all outcomes are equally likely, involves counting the number of ways something can occur
This section includes techniques for counting the number of results in a series of choices, under several different scenarios

5. Chapter 5 – Section 5 Example
If there are 3 different colors of paint (red, blue, green) that can be used to paint 2 different types of toy cars (race car, police car), then how many different toys can there be?
3 colors … 2 cars … 3 • 2 = 6 different toys
This can be shown in a table or in a tree diagram
Example
If there are 3 different colors of paint (red, blue, green) that can be used to paint 2 different types of toy cars (race car, police car), then how many different toys can there be?

6. Chapter 5 – Section 5 A table of the different possibilities
This is a rectangle with 2 rows and 3 columns … 2 • 3 = 6 entries
Green Police Car
Blue Police Car
Red Police Car
Police Car
Green Race Car
Blue Race Car
Red Race Car
Race Car
Green
Blue
Red

7. Chapter 5 – Section 5 A tree diagram of the different possibilities
This also shows that there are 6 possibilities
Paint
Car
Race
Police
Blue Race Car
Blue Police Car
Green Race Car
Green Police Car
Red Race Car
Red Police Car
Red
Blue
Green

8. Chapter 5 – Section 5
The Multiplication Rule of Counting applies to this type of situation
If a task consists of a sequence of choices
With p selections for the first choice
With q selections for the second choice
With r selections for the third choice …
Then the number of different tasks is
p • q • r • …

9. Chapter 5 – Section 5 Example Part A
A child is coloring a picture of a shirt and pants
There are 5 different colors of markers
How many ways can this be colored?
By the multiplication rule
5 • 5 = 25
Example Part A
A child is coloring a picture of a shirt and pants
There are 5 different colors of markers
How many ways can this be colored?
Example Part A
A child is coloring a picture of a shirt and pants
There are 5 different colors of markers
Example Part A
A child is coloring a picture of a shirt and pants

10. Chapter 5 – Section 5 Example Part B
A child is coloring a picture of a shirt and pants
There are 5 different colors of markers
The child wants to use 2 different colors
How many ways can this be colored?
By the multiplication rule
5 • 4 = 20
Example Part B
A child is coloring a picture of a shirt and pants
There are 5 different colors of markers
The child wants to use 2 different colors
How many ways can this be colored?
Example Part B
A child is coloring a picture of a shirt and pants
There are 5 different colors of markers
Example Part B
A child is coloring a picture of a shirt and pants
Example Part B
A child is coloring a picture of a shirt and pants
There are 5 different colors of markers
The child wants to use 2 different colors

11. Chapter 5 – Section 5 Example continued
Allowing the same marker to be used twice
5 • 5 = 25
Requiring that there be two different markers
5 • 4 = 20
There are 5 selections for the first choice for both Part A and Part B of this example
But they differ for the second choice … there are only 4 selections for Part B
Example continued
Allowing the same marker to be used twice
5 • 5 = 25
Requiring that there be two different markers
5 • 4 = 20

12. Chapter 5 – Section 5 Example continued
Part A, allowing the same marker to be used twice, is called counting with repetition and has formulas such as
5 • 5 • 5 • …
Part B, requiring that there be two different markers, is called counting without repetition and has formulas such as
5 • 4 • 3 • …
Example continued
Part A, allowing the same marker to be used twice, is called counting with repetition and has formulas such as
5 • 5 • 5 • …

13. Chapter 5 – Section 5
One way to help write these products is using the factorial symbol n!
n! = n • (n-1) • (n-2) • … • 2 • 1
We start off by saying that
0! = 1 and 1! = 1
For example
5! = 5 • 4 • 3 • 2 • 1 = 120
Notice how 5! looks like the 5 • 4 • 3 from the previous example
One way to help write these products is using the factorial symbol n!
n! = n • (n-1) • (n-2) • … • 2 • 1
One way to help write these products is using the factorial symbol n!
n! = n • (n-1) • (n-2) • … • 2 • 1
We start off by saying that
0! = 1 and 1! = 1
For example
5! = 5 • 4 • 3 • 2 • 1 = 120

14. Chapter 5 – Section 5 Learning objectives
Solving counting problems using the Multiplication Rule
Solving counting problems using permutations
Solving counting problems using combinations
Solving counting problems involving permutations with nondistinct items
Compute probabilities involving permutations and combinations 1 2 3 5 4

15. Chapter 5 – Section 5
The problem of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is an example of a permutation
A permutation is an
ordered arrangement, in which
r different objects are chosen out of
n different objects with
repetition not allowed
The number of ways is written nPr
The problem of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is an example of a permutation
The problem of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is an example of a permutation
A permutation is an
ordered arrangement, in which
r different objects are chosen out of
n different objects with
repetition not allowed

16. Chapter 5 – Section 5
The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20
In general, since there are n choices for the first selection, n-1 choices for the second, etc.
nPr = n • (n-1) • (n-2) • … • (n-(r-1))‏
The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20
In general, since there are n choices for the first selection, n-1 choices for the second, etc.
nPr = n • (n-1) • (n-2) • … • (n-(r-1))‏
is the formula for permutations
The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20
In general, since there are n choices for the first selection, n-1 choices for the second
nPr = n • (n-1) • (n-2) • … • (n-(r-1))‏
The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20
The number of ways of choosing one marker out of 5 and then choosing a second marker out of the 4 remaining is 5P4 = 20
In general, since there are n choices for the first selection
nPr = n • (n-1) • (n-2) • … • (n-(r-1))‏
Third choice = (n – 2)‏
rth choice = (n – (r-1))‏
Second choice = (n – 1)‏
First choice = (n – 0)‏

17. Chapter 5 – Section 5
A mathematical way to write the formula for the number of permutations is
This is a very convenient mathematical way to write a formula for nPr, but it is not a particularly efficient way to actually compute it
In particular, n! gets rapidly gets very large

18. Chapter 5 – Section 5 Learning objectives
Solving counting problems using the Multiplication Rule
Solving counting problems using permutations
Solving counting problems using combinations
Solving counting problems involving permutations with nondistinct items
Compute probabilities involving permutations and combinations 1 2 3 5 4

19. Chapter 5 – Section 5
For some problems, the order of choice does not matter
Order matters example
Choosing one person to be the president of a club and another to be the vice-president
Two different roles
Order does not matter example
Choosing two people to go to a meeting
The same role
For some problems, the order of choice does not matter
For some problems, the order of choice does not matter
Order matters example
Choosing one person to be the president of a club and another to be the vice-president
Two different roles

20. Chapter 5 – Section 5
When order does not matter, this is called a combination
A combination is an
unordered arrangement, in which
r different objects are chosen out of
n different objects with
repetition not allowed
The number of ways is written nCr
When order does not matter, this is called a combination
When order does not matter, this is called a combination
A combination is an
unordered arrangement, in which
r different objects are chosen out of
n different objects with
repetition not allowed

21. Chapter 5 – Section 5
Comparing the description of a permutation with the description of a combination
The only difference is whether order matters
No repetition
Out of n objects
Choose r objects
Order does not matter
Order matters
Combination
Permutation

22. Chapter 5 – Section 5 To list a combination of r items
Any of the r items can be written first
Any of the remaining (r-1) items can be written second
Etc.
Each of these r! lists is the same combination but a different permutation
Thus each combination corresponds to r! permutations
To list a combination of r items
Any of the r items can be written first
Any of the remaining (r-1) items can be written second
Etc.
To list a combination of r items
Any of the r items can be written first
Any of the remaining (r-1) items can be written second
Etc.
Each of these r! lists is the same combination but a different permutation

23. Chapter 5 – Section 5 Example
If there are 8 researchers and 3 of them are to be chosen to go to a meeting
A combination since order does not matter
There are 56 different ways that this can be done
Example
If there are 8 researchers and 3 of them are to be chosen to go to a meeting
Example
If there are 8 researchers and 3 of them are to be chosen to go to a meeting
A combination since order does not matter

24. Chapter 5 – Section 5
Because each combination corresponds to r! permutations, the formula nCr for the number of combinations is

25. Chapter 5 – Section 5 Is a problem a permutation or a combination?
One way to tell
Write down one possible solution (i.e. Roger, Rick, Randy)‏
Switch the order of two of the elements (i.e. Rick, Roger, Randy)‏
Is this the same result?
If no – this is a permutation – order matters
If yes – this is a combination – order does not matter
Is a problem a permutation or a combination?
Is a problem a permutation or a combination?
One way to tell
Write down one possible solution (i.e. Roger, Rick, Randy)‏
Switch the order of two of the elements (i.e. Rick, Roger, Randy)‏

26. Chapter 5 – Section 5 Learning objectives
Solving counting problems using the Multiplication Rule
Solving counting problems using permutations
Solving counting problems using combinations
Solving counting problems involving permutations with nondistinct items
Compute probabilities involving permutations and combinations 1 2 3 5 4

27. Chapter 5 – Section 5
Our permutation and combination problems so far assume that all n total items are different
Sometimes we have a permutations but not all of the n items are different
This is a more complicated problem
How many ways are there?

28. Chapter 5 – Section 5 Example
How many ways to put 3 A’s, 2 N’s, and 2 T’s to try to make a seven letter sequence?
____ ____ ____ ____ ____ ____ ____
Each of the blanks can be filled in with either an A or a N or a T
The three A’s are the same … the two N’s are the same … the two T’s are the same
Example
How many ways to put 3 A’s, 2 N’s, and 2 T’s to try to make a seven letter sequence?
____ ____ ____ ____ ____ ____ ____
Example
How many ways to put 3 A’s, 2 N’s, and 2 T’s to try to make a seven letter sequence?
____ ____ ____ ____ ____ ____ ____
Each of the blanks can be filled in with either an A or a N or a T

29. Chapter 5 – Section 5 Example continued Where can the A’s go?
There are 7 possible places
Any 3 of them are possible
Order does not matter
So 7C3 different ways to put in the A’s

30. Chapter 5 – Section 5 Example continued Where can the N’s go?
There are 4 possible places (since 3 of the 7 have been taken by the A’s already)‏
Any 2 of them are possible
Order does not matter
So 4C2 different ways to put in the N’s
And there are 2C2 different ways to put in the T’s

31. Chapter 5 – Section 5 Example continued Altogether there are
7C3 • 4C2 • 2C2
different ways
This is
Notice that the denominator is 3, 2, 2 … the numbers of each letter
Example continued
Altogether there are
7C3 • 4C2 • 2C2
different ways
Example continued
Altogether there are
7C3 • 4C2 • 2C2
different ways
This is

32. Chapter 5 – Section 5
The general formula for the number of permutations of
n total objects where there are
n1 of the first kind
n2 objects of the second kind
… and
nk of the kth kind is

33. Chapter 5 – Section 5 Learning objectives
Solving counting problems using the Multiplication Rule
Solving counting problems using permutations
Solving counting problems using combinations
Solving counting problems involving permutations with nondistinct items
Compute probabilities involving permutations and combinations 1 2 3 5 4

34. Chapter 5 – Section 5
Probabilities using the classical method involve counting the number of possibilities
Often the number of possibilities is some permutation or some combination
The permutation / combination formulas can be used to calculate probabilities

35. Chapter 5 – Section 5 A permutation example
In a horse racing “Trifecta”, a gambler must pick which horse comes in first, which second, and which third
If there are 8 horses in the race, and every order of finish is equally likely, what is the chance that any ticket is a winning ticket?
Order matters, so this is a permutations problem
A permutation example
In a horse racing “Trifecta”, a gambler must pick which horse comes in first, which second, and which third
If there are 8 horses in the race, and every order of finish is equally likely, what is the chance that any ticket is a winning ticket?

36. Chapter 5 – Section 5 A permutation example continued
There are 8P3 permutations of the order of finish of the horses
The probability that any one ticket is a winning ticket is 1 out of 8P3, or 1 out of 56

37. Chapter 5 – Section 5 A combination example
The Powerball lottery consists of choosing 5 numbers out of 55 and then 1 number out of 42
The grand prize is given out when all 6 numbers are correct
What is the chance of getting the grand prize?
Order does not matter, so this is a combinations problem (for the 5 balls)‏
A combination example
The Powerball lottery consists of choosing 5 numbers out of 55 and then 1 number out of 42
The grand prize is given out when all 6 numbers are correct
What is the chance of getting the grand prize?

38. Chapter 5 – Section 5 A combination example continued
There are 55C5 combinations of the 5 numbers
There are 42 possibilities for the last ball, so the probability of the grand prize is 1 out of
which is pretty small
A combination example continued
There are 55C5 combinations of the 5 numbers

39. Summary: Chapter 5 – Section 5
The Multiplication Rule counts the number of possible sequences of items
Permutations and combinations count the number of ways of arranging items, with permutations when the order matters and combinations when the order does not matter
Permutations and combinations are used to compute probabilities in the classical method

40. Example
Phil Black, a former Navy SEAL instructor, has developed a game called the Push Up Game (PUG). The game consists of a drawing a card from the FitDeck, a pack of 56 cards, and doing the exercise listed on the card (many of these are variations on push-ups.) Black claims the FitDeck cards can be shuffled to produce more than 3 trillion workouts. (Source: Ryckman, L. Scripps Howard News Service, Push Upmanship, Post Register. December 5, 2005.) Assuming all the cards are unique, how many different workouts can be created by shuffling the deck and then doing the exercise on each card in the order they are drawn? Is Black’s claim accurate?

41. (56! ≈ 7.11x1074; yes)

42. Example
The dial on a combination lock shows the integers 0 through 39. Turning the dial to three specific numbers (in order) opens the lock.
a. How many different locks can a manufacturer produce before they have to reuse combinations, if numbers cannot be repeated in a combination (i.e., is not allowed)?
b. How many different locks can a manufacturer produce before they have to reuse the combination if numbers can be repeated in a combination?

43. (40P3 = 59,280)
(403=64,000)

44. Example
A bride is trying to design her wedding ring. Based on her tastes and the groom’s finances, she has limited her choices to: five different bands, three diamonds, and four metals (platinum, gold, white gold, and silver). How many different ring combinations can she choose?
A student places four books in her backpack. How many ways can she order these books?

45. (5·3·4 = 60 ring combinations)
(4! = 24)

46. Example
A statistics professor is preparing an exam from a test bank. If the test bank consists of 30 problems and the professor will include 10 of them, how many different ways can the professor select the problems? (The order of the questions on the exam is not important.)

47. (30C10 = 30,045,015)

48. Example
A statistics professor has written an exam with 10 questions. To discourage cheating, this professor changes the order of the questions and creates several versions of the exam. In how many ways can these questions be arranged?

49. (10! = 3,628,800)