Counting

1) tree diagram2) sample space3) event4) Fundamental Counting Principle5) Factorial

Counting OutcomesCounting Outcomes

Count outcomes using a tree diagram.

Count outcomes using the Fundamental Counting Principle.


One method used for counting the number of possible outcomes is to draw atree diagram.

The last column of a tree diagram shows all of the possible outcomes.

The list of all possible outcomes is called the sample space, while any collection of one or more outcomes in the sample space is called an event.


A football team uses red jerseys for road games, white jerseys for home games,and gray jerseys for practice games. The team uses gray or black pants, and black and white shoes. Use a tree diagram to determine the number of possible uniforms.



Jersey Pants Shoes Outcomes

Red

White

Gray




Red

White

Gray

Gray

Black

Gray

Black

Gray

Black




Red

White

Gray

Gray

Black

Gray

Black

Gray

Black

BlackWhiteBlackWhite






Red

White

Gray

Gray

Black

Gray

Black

Gray

Black


RGBRGWRBBRBW

WGBWGWWBBWBW

GGBGGWGBBGBW






Red

White

Gray

Gray

Black

Gray

Black

Gray

Black


RGBRGWRBBRBW

WGBWGWWBBWBW

GGBGGWGBBGBW



The tree diagram shows that there are 12 possible outcomes.


In the previous example, the number of possible outcomes could also be found bymultiplying the number of choices for each item.



The team could choose from:

3different

colored jerseys

2different

colored pants

2different

colored shoes




3different

colored jerseys

2different

colored pants

2different

colored shoes

There are 3 X 2 X 2 or 12 possible uniforms.




3different

colored jerseys

2different

colored pants

2different

colored shoes


This example illustrates the Fundamental Counting Principle.




3different

colored jerseys

2different

colored pants

2different

colored shoes


This example illustrates the Fundamental Counting Principle.

If an event M can occur in m ways, and is followed by an event N that can occurin n ways, then the event M followed by event N can occur m X n ways.


A deli offers a lunch special in which you can choose a sandwich, a side dish, an abeverage.

If there are 10 different sandwiches, 12 different side dishes, and 7 different beverages,from which to choose, how many different lunch specials can be ordered?




Multiply to find the number of lunch specials.

sandwichchoices

side dishchoices

beveragechoices

number ofspecialsX X =




Multiply to find the number of lunch specials.

sandwichchoices

side dishchoices

beveragechoices

number ofspecialsX X =

10 X 12 X 7 = 840

The number of different lunch specials is 840.


A.J. is setting up a display of the ten most popular video games from the previousweek. If he places the games side-by-side on a shelf, in how many different wayscan he arrange them?



The number of ways to arrange the games can be found by multiplying the numberof choices for each position.




A.J. has 10 games from which to choose for the first position.





After choosing a game for the first position, there are nine games left from which to choose for the second position.






There are now eight choices for the third position.







This process continues until there is only one choice left for the last position.








Let n represent the number of arrangements.

n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 or 3,628,800








Let n represent the number of arrangements.

n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 or 3,628,800

There are 3,628,800 different ways to arrange the video games.


The expression n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 used in the previous examplecan be written as 10! using a factorial.

The expression n!, read n factorial, where n is greater than zero, is the

product of all positive integers beginning with n and counting backward to 1.

n! = n(n – 1)* (n – 2) * . . . 3 * 2 * 1

Example: 5! = 5 * 4 * 3 * 2 * 1 or 120


0! is defined as being equal to 1. Let’s see why.



4

!4!3

Writing this out using the definition of factorials.

4

1*2*3*4

4

1*2*3*4!3



4

!4!3


4

1*2*3*4

4

1*2*3*4!3

3

!3!2



4

!4!3


4

1*2*3*4

4

1*2*3*4!3

3

!3!2

2

!2!1



4

!4!3


4

1*2*3*4

4

1*2*3*4!3

3

!3!2

2

!2!1

1

!1!0 so, the next logical conclusion is that

Counting

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