Name: ____________________________________ Date: ________________
Period: _______
Integrated Advanced Algebra
Notes: Fundamental Counting Principle
Textbook: Lessons 6.1, Pages 338 – 341
Homework: Set A, Page 340: #2 – 10 even and #11
Set B, Page 341: #2 – 8a even (omit 8b)
Essential Question: How can you use the tree diagrams and
the counting principle to count the number of outcomes for
an event?
There are many real life situations or problems where you want to count the
number of possibilities for certain events.
For instance, suppose you and
two friends decide to get an ice cream cone.
There are 3 kinds of ice
cream available (chocolate, strawberry and vanilla). How many different
combinations of favors are possible?
One way to answer this question is to use a tree diagram. A tree diagram is a
graphic organizer used to list all possibilities of a sequence of events in a systematic way. Tree
diagrams are one method for calculating the total number of outcomes in a sample space (the
set of all possible outcomes).
V – vanilla; C – chocolate; SB – strawberry
From counting down the right hand side of the
tree diagram, you can see that there are 27
different ways for the flavors to be chosen.
Another way to count the number of possible
favor combinations is to use the fundamental
counting principle.
The Fundamental Counting Principle:
The Multiplication Counting Principle
 If you have 2 events, 1 event can occur m ways and another event can
occur n ways, then the number of ways that both can occur is m * n.
This principle can be extended to three or more events.
Example 1 (using the example with the ice cream favors above):
 Event 1 = Your cone choice (3 options)
 Event 2 = Friend #1 cone choice (3 options)
 Event 3 = Friend #2 cone choice (3 options)
How many different ice cream choice combinations are possible?
3 * 3 * 3 = 27 different orders
Example 2:
At a restaurant at Cedar Point, you have the choice of 8 different
entrees, 2 different salads, 12 different drinks, and 6 different
Name: ____________________________________ Date: ________________
Period: _______
desserts.
choose?
How many different dinners (one choice of each) can you
8 entrees * 2 salads * 12 drinks * 6 desserts = 1152
different dinners
Example 3:
You are packing for a trip. You have 6 pairs of pants, 4 shirts, and 2
pairs of shoes. How many different outfits can you put together for the
trip? When you change the shoes, shirt, or pants, the whole outfit
changes.
6 * 4 * 2 = 48 different outfits
The Addition Counting Principle

If the possibilities being counted can be divided into groups with no
possibilities in common, then the total number of possibilities is
the sum of the numbers of possibilities in each group.
Example 4:
When ordering food at a take out restaurant the customer is given a
randomly generated order code. The code has 3 symbols (letters or
digits). How many codes can be generated if at least one number is used
in each?
 1 – digit: There are 10 choices for each digit and
26 choices for each letter. So, there are 10 * 26 *
26 = 6760 digit-letter-letter possibilities. The
digit can be in any of the three positions, so there
are 3 * 6760 = 20,280 total 1-digit possibilities.
 2 – digits: There are 10 * 10 * 26 = 2600 digitdigit-letter possibilities. The letter can be in
any of the three positions, so there are 3 * 2600 =
7800 total 2-digit possibilities.
 3 – digits: There are 10 * 10 * 10 = 1000 digitdigit-digit possibilities.
So, there are 20,280 + 7800 + 1000 = 29,080 possible
order codes.
Example 5:
Suppose in example 4 the only digits allowed were odd numbers.
many codes would be generated?
 1 – digit: 5 * 26 * 26 = 3380; 3380 * 3 = 10140
possibilities
 2 – digits: 5 * 5 * 26 = 650; 650 * 3 = 1950
possibilities
 3 – digits: 5 * 5 * 5 = 125
How
Name: ____________________________________ Date: ________________
Period: _______
So, there are 10140 + 1950 + 125 = 12215 possible
codes
Example 6:
High school faculty members are to be issued special coded
identification cards that consist of four symbols (letters or digits).
The codes must contain at least two letters. How many different ID
cards can be issued if the letters and digits cannot be repeated on a
given card?
 26 * 25
 26 * 25
 26 * 25
So, there
possible IDs
* 10 * 9 = 58500; 58500 * 6 = 351000
* 24 * 10 = 156000; 156000 * 4 = 624000
* 24 * 23 = 358800
are 351000 + 624000 + 358800 = 1,333,800