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Basic Principle of Counting

BASIC PRINCIPLE OF COUNTING

Chapter 6 of the sources is titled Basic Principle of Counting. It introduces the fundamental mathematical rules used to determine the number of ways events can occur , which serves as a prerequisite for understanding probability.


1. Addition Rule of Counting

πŸ’‘

Addition Rule (n1+n2n_1 + n_2)

This rule is used when you must choose between different actions that cannot happen at the same time.

If an action AA can occur in n1n_1 ways and action BB can occur in n2n_2 ways, the total number of ways of occurrence for either AA or BB is n1+n2n_1 + n_2.

Example

You have a gift card to buy one item: either a shirt or a pair of trousers.

  • Shirt choices: 4
  • Trouser choices: 3
  • Total choices: 4+3=74 + 3 = \mathbf{7} ways.

2. Multiplication Rule of Counting

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Multiplication Rule (n1Γ—n2n_1 \times n_2)

This rule is used when multiple actions occur together.

If action AA can occur in n1n_1 ways and action BB can occur in n2n_2 ways, the total number of ways for both AA and BB to occur together is n1Γ—n2n_1 \times n_2.

Example

A gift card allows you to buy one shirt and one pair of trousers.

  • Shirt choices: 4
  • Trouser choices: 3
  • Total choices: 4Γ—3=124 \times 3 = \mathbf{12} ways.

3. Generalisation of the Multiplication Rule

If rr actions are performed in a specific order, and there are n1n_1 possibilities for the first, n2n_2 for the second, and so on…

πŸ’‘

General Formula

Total possibilities = n1Γ—n2Γ—β‹―Γ—nrn_1 \times n_2 \times \dots \times n_r

Example: Buying one shirt (4 choices), one pair of trousers (3 choices), and one pair of shoes (2 choices).

  • Solution: 4Γ—3Γ—2=244 \times 3 \times 2 = \mathbf{24} ways.

Solved Practice Examples

QQ1

Password Creation (Repetition Allowed)

Create a six-digit alphanumeric password: the first two characters must be upper-case alphabets and the remaining four must be numbers. Repetition is allowed.

View Detailed Solution β–Ό

There are 26 alphabets and 10 digits (0–9).

  • Ways: 26Γ—26Γ—10Γ—10Γ—10Γ—10=6,760,00026 \times 26 \times 10 \times 10 \times 10 \times 10 = \mathbf{6,760,000}.
QQ2

Password Creation (No Repetition)

Same as above, but characters and numbers cannot be repeated.

View Detailed Solution β–Ό

The pool of available choices decreases after each selection.

  • Ways: 26Γ—25Γ—10Γ—9Γ—8Γ—7=3,276,00026 \times 25 \times 10 \times 9 \times 8 \times 7 = \mathbf{3,276,000}.
QQ3

Race Finishers

Eight athletes compete in a 100m race. In how many ways can they finish (no ties)?

View Detailed Solution β–Ό

Any of the 8 can come first, then 7 for second, 6 for third, and so on.

  • Ways: 8Γ—7Γ—6Γ—5Γ—4Γ—3Γ—2Γ—1=40,3208 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \mathbf{40,320}.

Unsolved Problems for Practice

QA

The Sports Shop (Either/Or)

Narendra is buying sports items. He has a choice of 4 cricket bats, 4 cricket balls, 8 brands of stumps, 5 brands of jerseys, and 10 brands of shoes. If he can only purchase one item from any of these categories, how many ways can he choose?

View Detailed Solution β–Ό

This uses the Addition Rule.

  • 4+4+8+5+10=314 + 4 + 8 + 5 + 10 = \mathbf{31} ways.
QB

The Sports Shop (One of Each)

Using the same choices as above, how many ways can Narendra purchase one item from each category?

View Detailed Solution β–Ό

This uses the Multiplication Rule.

  • 4Γ—4Γ—8Γ—5Γ—10=64004 \times 4 \times 8 \times 5 \times 10 = \mathbf{6400} ways.
QC

Class Elections

A class has 60 students. A cricket captain and a class representative must be elected. If a student can only hold one position at a time, what are the total ways?

View Detailed Solution β–Ό

Multiplication rule without repetition. 60 choices for the first role, 59 for the second.

  • 60Γ—59=354060 \times 59 = \mathbf{3540} ways.
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