There are some basic counting techniques which will be useful in determining the number through different ways of arranging or selecting objects. The basic counting principles are given below.

**Principles
of Counting**

There are
some basic counting techniques which will be useful in determining the number through
different ways of arranging or selecting objects. The basic counting principles
are given below.

__1. Addition
principle__

If there
are two selections such that they can be done **independently** in **m** ways
and **n**
ways respectively, then either of the two selections can be done in **(m** **+** **n)** ways.

Let us learn
about this addition principle of counting as given below :

*Situation:*

In class VIII, there are 16 boys and 9 girls. The teacher wants to select either a boy or a girl as the class leader. Let us see, in how many ways can the teacher select the class leader.

The teacher
can select the class leader in any one of the following ways.

(i) In the
first choice, the teacher can select **a boy
among 16 boys in 16 ways** (who ever may be of the 16 boys).

(ii) In the
second choice, the teacher can select **a girl
among 9 girls in 9 ways **(who ever may be of the 9 girls).

Hence, the
teacher can select the class leader who is **a
boy** or **a girl** in **25 different**
**ways (16 boys
+ 9 girls).**

Thus, we
come to know, if a selection **A** can occur
in ** m** ways and another selection

**Example 7.1**

If you are
going to a hotel to have food and the hotel offers different food items as shown
in Fig 7.2. Find how many ways are possible to have either tiffin or meals?

*Solution:*

From the
above Fig. 7.2, we come to know

(i) For tiffin,
we can choose one among **7 items** in 7
ways

(ii) For
meals, we can choose one among **9 items**
in 9 ways.

Therefore,
there are **16**
**(7**
**tiffin items +** **9** **meals items)** **different ways** by which we can choose any one food.

** **

__2. Multiplication
principle__

If a selection
can be performed in ** m** ways, following which another
selection can be performed in

Now, we shall
learn about multiplication principle of counting from the following situation.

*Situation:*

There are
3 places in a city namely A,B and C. There are 3 routes **a1, a2 and a3** from A to B. There are 2 different routes, **b1 and b2** from B to C as shown in the Fig
7.3.

Suppose a person wants to travel from A to C via B. Lets us see the number of ways he can go from place A to C via B.

(i) In the
first way, he can go from **A** **to B in 3 routes **and

(ii) In the
second way, he can go from **B to C in 2 different
routes.**

Therefore,
the total number of ways in which he can travel is **6** **(3 × 2)** routes as shown in Fig 7.4.

Thus, we
come to know, if a selection **A** can occur
in ** m** ways and another selection

**Example 7.2**

Praveen bought
**3 shirts, 2 jeans and 3 pairs of shoes**
for his birthday. In Fig. 7.5 shows the different ways of wearing the dress. In
how many different ways can Praveen wear a dress on his birthday?

*Solution:*

Here, Praveen
has 3 shirts , 2 jeans and 3 pairs
of shoes.

He can wear
a dress either this way or he can have the choices as shown in the
Fig. 7.6

Therefore,
Praveen can wear his dress in **18** **(3×2×3)**
**different possible
ways** on his birthday.

**Example 7.3**

In class
VIII, a math club has four members **M,A,T**
and **H.** Find the number of different ways,
the club can elect

(i) a leader,

(ii) a leader
and an assistant leader.

*Solution:*

**(i) To elect a leader**

In class
VIII, a math club has four members namely **M, A, T** and **H.**

Therefore,
there are **4**
*(***4** **×** **1)** **different ways**
by which they can be elected a leader.

**(ii) To elect a leader
and an assistant leader**

In the
Fig. 7.7, the **red**
shaded boxes show that same member comes twice. As, one person cannot have two
leadership, therefore, the red shaded boxes cannot be counted. So there are
only **12**
((4 × 1 × 4) – 4) **different ways** (shown in yellow boxes and
green boxes) to choose a leader and an assistant leader for the math club.

**Activity**

**1. Determine the
number of two digit numbers that can be formed using the digits 1, 3 and 5 with
repetition of digits allowed.**

**The activity consists
of two parts**

**(i) Choose a one's
digit.**

**(ii) Choose a ten's
digit.**

**Complete the table
given beside**

**Solution:**

**2. Find the three
digit numbers that can be formed using the digits 1, 3 and 5 without repetition
of digits.**

**Complete the tree
diagram given below to the numbers**

**Solution:**

A **password** using 6
characters is created where the first 2 characters are any of the **alphabets**, the third character is any
one special character **like @, #, $, %,
&, _,+,~, * or -** and the last 3 characters are any of the numbers from** 0 to 9**. For that, there are **26 × 26 × 10 × 10 × 10 × 10 = 67,60,000** number of different
ways possible to create that password.

**Example 7.4**

In how many
ways, can the students answer 3 true or false type questions in a slip test?

*Solution:*

(i) Assuming
that the question **Q _{1}** is answered True, questions

(ii) Assuming
that the question **Q _{1}** is answered False,

Thus, as
each question has only two options (True or False) in 2 ways, the number of ways
of answering these 3 questions in a slip test is **8 (2 x 2 x
2) possible ways **as shown in Fig. 7.8.

**Example 7.5**

Madhan wants
to a buy a new car. The following choices are available for him.

• There are
**2 types** of cars as shown in the Fig.
7.9

• There are
**5 colours** available in each type as shown
in Fig. 7.9.

• There are
**3 models** available in each colour

**(i) GL **(standard model)

(ii) **SS** (sports model)

(iii) **SL** (luxury model)

(i) In how
many different ways can Madhan buy any one of the new car?

(ii) If the
**white colour** is not available in **Type 2**, then in how many ways can Madan
buy a new car among the given option?

*Solution:*

(i) To buy
any one of the new car from the given choices

Here, we
have **2 types** of car with **5 different colours** and **3 models** in each colour.

Therefore,
there are **30**
**[2 (1 × 5 × 3)]** **different ways** to buy a new car by
Madhan.

(ii) If the
**white colour** is not available in **Type 2,** then...

(i) For **Type 1,** we have 5 colours and 3 models and
hence there are **1 × 5 × 3 = 15 choices.**

(ii) For
**Type 2,** we have only 4 colours and 3
models and hence there are **1 × 4 × 3 = 12
choices.** Therefore, there are **27** **(15 +
12)** **different
ways** to buy a new car by Madhan

The above
example illustrates both the addition and multiplication principles.

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