(i) Here, we have 2 types of car with 5 different colors and 3 models in each color.

Total number of different way to buy a new car by Madhan = 2 x 5 x 3 = 30. (You can also find the number of ways from the figure shown above).
(ii) If the white colour is not available in Type 2, then..

For Type 1, we have 5 colors and 3 models and hence there are 5 x 3 = 15 choices.
For Type 2, we have only 4 colors and 3 models and hence there are 4 x 3 = 12 choices.
Therefore, the total number of different ways 15 + 12 = 27 ways.
The above example illustrates both the addition and multiplication principles.

(i) To elect a leader
In class VIII, a math club has four members namely M, A, T, and H,
Therefore, there are 4 different ways by which they can be elected a leader.
(ii) To elect a leader and an assistant leader
In Figure, 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. There are only 12 different ways (either shown in yellow boxes or green boxes) to choose a leader and an assistant leader for a math club.

Assuming that the question Q₁ is answered True, questions Q₂ and Q₃ can be answered as TT, TF, FT, and FF in 4 ways.
Similarly, assuming that the question Q₂ is answered False, Q₂ and Q₃ can also be answered as TT, TF, FT, and FF in 4 ways.
Thus, as each question has only two options (True or False), the number of ways of answering these 3 questions in a slip test is 2 x 2 x 2 = 8 possible ways

There are 6 different outcomes, namely 1, 2, 3, 4, 5 and 6. All the 6 outcomes cannot occur simultaneously. Hence, there are 6 different possible outcomes in rolling a die.

There are 2 different outcomes in tossing a coin namely a Head and a Tail. Both events cannot occur simultaneously.

Number of veg options 2
Number of meat option 5
One can opt for anyone dinner
∴ Total number of ways 2 + 5 = 7 ways

There are 5 stages Question - 1
Question - 2
Question - 3
Question - 4
Question - 5
There are 2 choices for each question (Yes / No)
∴  Total number of possible ways to answer
= 2 x 2 x 2 x 2 x 2 = 32 ways.

There are 5 stages, Number - 1
Number - 2
Number - 3
Letter - 1
Letter - 2
There are 10 possible numbers 0 to 9
There are 26 possible letters A to Z.
∴ We have 10 x 10 x 10 x 26 x 26 = 6,76,000 possible codes.

There are 4 stages
1. Choosing a Sandwich
2. Choosing a side
3. Choosing a dessert
4. Choosing a drink
There are 4 different types of sandwich, 3 different types of side two different type of desserts and five different types of drink.
∴  The number of meal combos possible is = 4 x 3 x 2 x 5 = 120

A letter can be selected from A, B, C, D and E in 5 different ways, :
The last 3 digits from 0 to 9 can be selected in 10 x 10 x 10 different ways.
Therefore the total possible ways to select a roll numbers is 5 x 10 x 10 x 10 different ways
ie., 5000 different ways.
Except Roll numbers are A000, 8000, C000,D000 and E000.
Total number of possible ways = 5000 - 5 :
= 4995 ways

The student has 2 school bags and 3 water bottles.

Therefore he can carry both a school bag and a water bottle in 6 (2 x 3) different possible ways.

The teacher is going to select one student from class VI out of 10 students in 10 ways from class VII out of 15 students in 15 ways and from class VIII out of 20 students in 20 ways.
ஃ Number of ways 3 students can be selected = 10 + 15 + 20 = 45 ways

This is one of the solutions. Try for more

This is one of the solutions. Try for more solutions

This is one of the solutions. Try for more.