# 2. Basic Principles of Counting

### Later, on this page...

## Counting

An efficient way of **counting** is necessary to handle large masses of statistical data (e.g. the level of inventory at the end of a given month, or the number of production runs on a given machine in a 24 hour period, etc.), and for an understanding of **probability**.

In this section, we shall develop a few counting techniques. Such techniques will enable us to count the following, without having to list all of the items:

- the number of ways,
- the number of samples, or
- the number of outcomes.

Before we learn some of the basic principles of counting, let's see some of the notation we'll need.

## Number of Outcomes of an Event

As an example, we may have an event *E *defined as

E= "day of the week"

We write the "number of outcomes of event *E*" as *n*(*E*).

So in the example,

`n(E) = 7`,

since there are `7` days in the week.

## Addition Rule

Let *E*_{1} and *E*_{2} be **mutually exclusive** events (i.e. there are no common outcomes).

Let event *E* describe the situation where either event *E*_{1} **or** event *E*_{2} will occur.

The number of times event *E* will occur can be given by the expression:

n(E) =n(E_{1}) +n(E_{2})

where

n(E) = Number of outcomes of eventE

n(E_{1}) = Number of outcomes of eventE_{1}

n(E_{2}) = Number of outcomes of eventE_{2}

[We see more on mutually exclusive events later in this chapter.]

### Tip

In counting and probability, **"OR**" usually requires us to **ADD**.

### Example 1

Consider a set of numbers `S = {-4, -2, 1, 3, 5, 6, 7, 8, 9, 10}`

Let the events *E*_{1}, *E*_{2} and *E*_{3} be defined as:

*E *= choosing a negative or an odd number from *S*;

*E*_{1}= choosing a negative number from S;

*E*_{2} = choosing an odd number from S.

Find *n*(*E*).

Answer

*E*_{1} and *E*_{2} are mutually exclusive events (i.e. no common outcomes).

*n*(*E*) = *n*(*E*_{1}) + *n*(*E*_{2})

= 2 + 5

= 7

Get the Daily Math Tweet!

IntMath on Twitter

### Example 2

In how many ways can a number be chosen from `1` to `22` such that

(a) it is a multiple of `3` or `8`?

(b) it is a multiple of `2` or `3`?

Answer

#### Part (a)

Here, *E*_{1} = multiples of `3`:

E_{1}= {3, 6, 9,12, 15, 18, 21}

n(E_{1}) = 7

*E*_{2} = multiples of `8`:

E_{2}= {8, 16}

n(E_{2}) = 2

Events *E*_{1} and *E*_{2} are mutually exclusive.

n(E) =n(E_{1}) +n(E_{2}) = 7 + 2 = 9

#### Part (b)

Here, *E*_{1} = multiples of `2`:

E_{1}= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22}

n(E_{1}) = 11

*E*_{2} = multiples of `3`:

*E*_{2} = {3, 6, 9,12, 15, 18, 21}

*n*(*E*_{2}) = 7

Events *E*_{1} and *E*_{2} are **not** mutually exclusive.

We could proceed as follows:

n(E) =n(E_{1}) +n(E_{2}) −n(E_{1}∩E_{2}) = 11 + 7 − 3 = 15

where *E*_{1} ∩ *E*_{2} means "the intersection of the sets* E*_{1} and *E*_{2}".

## Multiplication Rule

Now consider the case when two events *E*_{1} and *E*_{2} are to be performed and the events *E*_{1} and *E*_{2} are **independent** events i.e. one does not affect the other's outcome.

[We see more on independent events later in this chapter.]

### Example 3

Say the only clean clothes you've got are `2` t-shirts and `4` pairs of jeans. How many different combinations can you choose?

#### Answer

We can think of it as follows:

We have `2` t-shirts and with each t-shirt we could pick `4` pairs of jeans. Altogether there are

`2 × 4 = 8` possible combinations.

We could write

E_{1}= "choose t-shirt" and

E_{2}= "choose jeans"

## Multiplication Rule in General

Suppose that event *E*_{1} can result in any one of *n*(*E*_{1}) possible outcomes; and for each outcome of the event *E*_{1}, there are *n*(*E*_{2}) possible outcomes of event *E*_{2}.

Together there will be *n*(*E*_{1}) × *n*(*E*_{2}) possible outcomes of the two events.

### Tip

In counting and probability, **"AND"** usually requires us to **MULTIPLY**.

That is, if event *E* is the event that both *E*_{1} and *E*_{2} **must** occur, then

n(E) =n(E_{1}) ×n(E_{2})

In our example above,

n(E_{1}) = 2 (since we had 2 t-shirts)

n(E_{2}) = 4 (since there were 4 pairs of jeans)

So total number of possible outcomes is given by:

n(E) =n(E_{1}) ×n(E_{2}) = 2 × 4 = 8

### Example 4

What is the total number of possible outcomes when a pair of coins is tossed?

Answer

The events are described as:

E_{1}= toss first coin (2 outcomes, son(E_{1}) = 2.)

E_{2}= toss second coin (2 outcomes, son(E_{2}) = 2.)

They are independent, since neither toss affects the outcome of the other toss.

So *n*(*E*) = *n*(*E*_{1}) × *n*(*E*_{2}) = 2 × 2 = 4

[We could list the outcomes: HH HT TH TT].

Easy to understand math videos:

MathTutorDVD.com

### Example 5

The life insurance policies of an insurance company are classified by:

- age of the insured:
- under 25 years,
- between 25 years and 50 years,
- over 50 years old;

- sex;
- marital status:
- single or
- married.

What is the total number of classifications?

Answer

The events are described as:

E_{1}= age of the insured: 3 age divisions, son(E_{1}) = 3.

E_{2}= sex: 2 possibilities, son(E_{2}) = 2.

E_{3}= marital status: 2 possibilities, son(E_{3}) = 2.

Each event is independent, so

n(E) =n(E_{1}) ×n(E_{2}) ×n(E_{3}) = 3 × 2 × 2 = 12

### Example 6

For our clothes problem above, say we found `3` caps that we could wear with our `2` t-shirts and `4` pairs of jeans. How many different combinations could we choose from now?

Answer

Our diagram now looks like this:

We have 2 choices in the first row, 4 in the second row and 3 in the third row. Together, we will have

(n) =En(E_{1}) ×n(E_{2}) = 2 × 4 × 3 = 24 combinations

Get the Daily Math Tweet!

IntMath on Twitter

### Example 7

Image source: Monoface

In the excellent Monoface (external site, no longer available), we could change the head, left and right eyes, nose and mouth of some zany guys who work at momo-1.com.

We are told that there are 759,375 possible faces. Where does this number come from?

If they were to let us change **the chin** as well, how many possible combinations would there be?

Answer

There are `15` people in the photos (I got this by counting the number of heads) and there are `5` different options that you can change (head, left eye, right eye, nose and mouth).

So there are

`15 × 15 ×15 ×15 ×15` ` = 15^5` `= 759,375` combinations

If we add another possible event (changing the chin), we have `6` possible options, so the answer is simply:

`15^6= 11,390,625` combinations

### Search IntMath, blog and Forum

### Online Math Solver

This math solver can solve a wide range of math problems.

Go to: Online algebra solver

### Math Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand math lessons on DVD. See samples before you commit.

More info: Math videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!