Tarryn has five different skirts, four different tops and three pairs of shoes. Assuming that all the colours complement each other, how many different outfits can she put together?
The fundamental counting principle
Previous
Tools and Techniques

Next
Factorial notation

10.4 The fundamental counting principle (EMCJZ)
Mathematics began with counting. Initially, fingers, beans and buttons were used to help with counting, but these are only practical for small numbers. What happens when a large number of items must be counted?
This section focuses on how to use mathematical techniques to count different assortments of items.
Introduction (EMCK2)
An important aspect of probability theory is the ability to determine the total number of possible outcomes when multiple events are considered.
For example, what is the total number of possible outcomes when a die is rolled and then a coin is tossed? The roll of a die has six possible outcomes (\(1;2;3;4;5\) or \(\text{6}\)) and the toss of a coin, \(\text{2}\) outcomes (heads or tails). The sample space (total possible outcomes) can be represented as follows:
\[S = \left\{\begin{array}{cccccc} (1;H); & (2;H); & (3;H); & (4;H); & (5;H); & (6;H); \\ (1;T); & (2;T); & (3;T); & (4;T); & (5;T); & (6;T) \end{array}\right\}\]Therefore there are \(\text{12}\) possible outcomes.
The use of lists, tables and tree diagrams is only feasible for events with a few outcomes. When the number of outcomes grows, it is not practical to list the different possibilities and the fundamental counting principle is used instead.
 The fundamental counting principle

The fundamental counting principle states that if there are \(n(A)\) outcomes in event \(A\) and \(n(B)\) outcomes in event \(B\), then there are \(n(A) \times n(B)\) outcomes in event \(A\) and event \(B\) combined.
If we apply this principle to our previous example, we can easily calculate the number of possible outcomes by multiplying the number of possible die rolls with the number of outcomes of tossing a coin: \(6 \times 2 = 12\) outcomes. This allows us to formulate the following:
If there \(n_{1}\) possible outcomes for event \(A\) and \(n_{2}\) outcomes for event \(B\), then the total possible number of outcomes for both events is \(n_1 \times n_2\)
This can be generalised to \(k\) events, where \(k\) is the number of events. The total number of outcomes for \(k\) events is:
\[{n}_{1}\times {n}_{2}\times {n}_{3}\times \cdots \times {n}_{k}\]
The order in which the experiments are done does not affect the total number of possible outcomes.
Worked example 10: Choices without repetition
A takeaway has a 4piece lunch special which consists of a sandwich, soup, dessert and drink for R \(\text{25,00}\). They offer the following choices for:
Sandwich: chicken mayonnaise, cheese and tomato, tuna mayonnaise, ham and lettuce
Soup: tomato, chicken noodle, vegetable
Dessert: icecream, piece of cake
Drink: tea, coffee, Coke, Fanta, Sprite
How many possible meals are there?
Determine how many parts to the meal there are
There are 4 parts: sandwich, soup, dessert and drink.
Identify how many choices there are for each part
Meal component  Sandwich  Soup  Dessert  Drink 
Number of choices  \(\text{4}\)  \(\text{3}\)  \(\text{2}\)  \(\text{5}\) 
Use the fundamental counting principle to determine how many different meals are possible
\(4\times 3\times 2\times 5=120\)So there are \(\text{120}\) possible meals.
In the previous example, there were a different number of options for each choice. But what happens when the number of choices is unchanged each time you choose?
For example, if a coin is flipped three times, what is the total number of different results? Each time a coin is flipped, there are two possible outcomes, namely heads or tails. The coin is flipped \(\text{3}\) times. We can use a tree diagram to determine the total number of possible outcomes:
From the tree diagram, we can see that there is a total of 8 different possible outcomes.
Drawing a tree diagram is possible to draw for three different coin flips, but as soon as the number of events increases, the total number of possible outcomes increases to the point where drawing a tree diagram is impractical.
For example, think about what a tree diagram would look like if we were to flip a coin six times. In this case, using the fundamental counting principle is a far easier option. We know that each time a coin is flipped that there are two possible outcomes. So if we flip a coin six times, the total number of possible outcomes is equivalent to multiplying 2 by itself six times:
\[2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{6} = 64\]Another example is if you have the letters A, B, C, and D and you wish to discover the number of ways of arranging them in threeletter patterns if repetition is allowed, such as ABA, DCA, BBB etc. You will find that there are \(\text{64}\) ways. This is because for the first letter of the pattern, you can choose any of the four available letters, for the second letter of the pattern, you can choose any of the four letters, and for the final letter of the pattern you can choose any of the four letters. Multiplying the number of available choices for each letter in the pattern gives the total available arrangements of letters:
\[4 \times 4 \times 4 = 4^{3} = 64\]This allows us to formulate the following:
When you have \(n\) objects to choose from and you choose from them \(r\) times, then the total number of possibilities is \[n \times n \times n \ldots \times n \enspace (r \text{ times}) = n^{r}\]
Worked example 11: Choices with repetition
A school plays a series of \(\text{6}\) soccer matches. For each match there are \(\text{3}\) possibilities: a win, a draw or a loss. How many possible results are there for the series?
Determine how many outcomes you have to choose from for each event
There are \(\text{3}\) outcomes for each match: win, draw or lose.
Determine the number of events
There are \(\text{6}\) matches, therefore the number of events is \(\text{6}\).
Determine the total number of possible outcomes
There are \(\text{3}\) possible outcomes for each of the \(\text{6}\) events. Therefore, the total number of possible outcomes for the series of matches is
\[3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 = 729\]Number of possible outcomes if repetition is allowed
Therefore there are \(\text{8 000 000}\) more possible number plates using the new style.
The code for a safe is of the form XXXXYYY where X is any number from 0 to 9 and Y represents the letters of the alphabet. How many codes are possible for each of the following cases:
We exclude the digit \(\text{0}\) and the vowels (A; E; I; O; U), leaving \(\text{9}\) other digits and \(\text{21}\) letters to choose from.
\[9^{4} \times 21^{3} = \text{60 761 421} \text{ possible codes}\]The prime digits are \(\text{2}\), \(\text{3}\), \(\text{5}\) and \(\text{7}\). This gives us 4 possible digits. If we exclude the letters X, Y and Z, we are left with \(\text{23}\) letters to choose from.
\[4^{4} \times 23^{3} = \text{3 114 752} \text{ possible codes}\] A person who eats only a starter has 4 choices
 A person who eats only a main meal has 8 choices
 A person who eats only a dessert has 6 choices
 A person who eats a starter and a main course has \(4 \times 8 = 32\) choices
 A person who eats a starter and a dessert has \(4 \times 6 = 24\) choices
 A person who eats a main meal and a dessert has \(8 \times 6 = 48\) choices
 A person who eats all three courses has \(4 \times 8 \times 6 = 192\) choices.
Previous
Tools and Techniques

Table of Contents 
Next
Factorial notation
