# Number Blunder

### Counting & Probability

## To start with:

## 1. Factorial Notation

**Factorial notation** is used to write the product of all the positive whole numbers up to a given number.

**Definition of n!**

** n factorial** is the product of all the integers from 1 to

*n*

"*n* factorial" is written with an exclamation mark as follows: n!

*n*! = (*n*)(*n* − 1)(*n* − 2)...

**Example:**

5! = 5 × 4 × 3 × 2 × 1 = 120

## 2. Counting

**Number of Outcomes of an Event**

E = an event

n(E) = number of outcomes of event E

**Example:**

*E* = "hours per day"

n(E)=24

So there are 24 hours per day

**Addition Rule**

Let *E*1 and *E*2 be **mutually exclusive** events.

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)

**Example:**

In how many ways can a number be chosen from 1 to 22 such that it is a multiple of

3 or 8?

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

**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

**Example:**

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?

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

*) =*

*E**n*(

*E*1) ×

*n*(

*E*2) = 2 × 4 × 3 = 24 combinations

## 3. Permutations

An arrangement or ordering of a set of objects is called a **permutation**.

In a permutation, the **order** that we arrange the objects in** is important.**

**Theorem 1 - Arranging n Objects**

In general, *n* distinct objects can be arranged in n! ways.

**Example:**

In how many ways can 4 different resistors be arranged in series?

Since there are 4 objects, the number of ways is

4!=24 ways

**Theorem 2 - Number of Permutations**

The number of permutations of *n* distinct objects taken *r* at a time, denoted by Pnr, where **repetitions are not allowed**, is given by Pnr=n(n−1)(n−2)...(n−r+1)=n! ÷ (n−r)!

**Example:**

In how many ways can a supermarket manager display 5 brands of cereals in 3 spaces on a shelf?

This is asking for the number of permutations, since we don't want repetitions. The number of ways is:

P53=5! ÷ (5−3)! =5! ÷ 2! =60

**Theorem 3 - Permutations of Different Kinds of Objects**

The number of different permutations of *n* objects of which *n*1 are of one kind, *n*2 are of a second kind, ... *nk*are of a *k*-th kind is n! ÷ (n1!×n2!×n3×...×nk!)

**Example:**

In how many ways can the six letters of the word "mammal" be arranged in a row?

Since there are 3 "m"s and 2 "a"s in the word "mammal", we have:

6! ÷ 3!2!=60

There is one "L" in "mammal", but it does not affect the answer, since 1! = 1.

**Theorem 4 - Arranging Objects in a Circle**

There are (n−1)! ways to arrange *n* distinct objects in a circle.

**Example:**

In how many ways can 5 people be arranged in a circle?

(5−1)!=4!=24 ways

## 4. Combinations

**combination**of

*n*objects taken

*r*at a time is a selection which does not take into account the arrangement of the objects. That is, the

**order is not important**.

**Number of Combinations**

The number of ways (or **combinations**) in which *r* objects can be selected from a set of *n* objects, where repetition is **not** allowed, is denoted by:

Cnr=n! ÷ r!(n−r)!

**Example:**

Find the number of ways in which 3 components can be selected from a batch of 20 different components.

C203 =20! ÷ 3!(20−3)! =20! ÷ 3!17! =1140

## 5. Probability

**Definition of a Probability**

Suppose an event *E* can happen in *r* ways out of a total of *n* possible equally likely ways.

Then the probability of occurrence of the event is denoted by P(E)=r ÷ n

The probability of **non-occurrence** of the event is denoted by P(E⎯)=(n−r)÷n=(1−r)÷n

Notice the bar above the *E*, indicating the event does **not **occur.

Thus, P(E⎯)+P(E)=1

In words, this means that the sum of the probabilities in any experiment is 1.

**Definition of Probability using Sample Spaces**

When an experiment is performed, a sample space of all possible outcomes is set up.

In a sample of *N* equally likely outcomes, a chance of 1 ÷ N is assigned to each outcome.

The probability of an event for such a sample is the number of outcomes favorable to *E *divided by the total number of equally likely outcomes in the sample space *S* of the experiment.

P(E)=n(E) ÷ n(S)

where:

n(E) is the number of outcomes favorable to *E*

n(S) is the total number of equally likely outcomes in the sample space *S* of the experiment.

**Properties of Probability**

(a) 0 ≤ *P*(event) ≤ 1

In words, the probability of an event must be a number between 0 and 1.

(b) *P*(impossible event) = 0

In words: The probability of an impossible event is 0.

(c) *P*(certain event) = 1

In words: The probability of an absolutely certain event is 1.

**Example:**

What is the probability of...

(a) Getting an ace if I choose a card at random from a standard pack of 52 playing cards.

There are 4 aces in a normal pack. So the probability of getting an ace is:

P(ace)=4/52=1/13

(b) Getting a 5 if I roll a die.

A die has 6 numbers.

There is only one 5 on a die, so the probability of getting a 5 is given by:

P(5)=1/6

(c) Getting an even number if I roll a die.

Even numbers are 2,4,6. So

P(even)=3/6=1/2

(d) Having one Tuesday in this week?

Each week has a Tuesday, so probability = 1.

## i hope you did not spend all your energy on trying to grasp the theories above as you have reached the fun interactive part where you should put all your effort in order to master the concept of counting and probability

## Testing Your Understanding

Solution: Let E be the event "landing on blue". The experiment of the sample space shows that there one "blue sector" therefore n(E) = 1 and n(S) = 4. Hence the probability of event E occuring is P(E) = 1/4.