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Fundamental Counting Principle (Theorem) of Multiplication

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Counting Numbers

Natural Numbers are called counting numbers.

Counting Principle of Multiplication

If a total event can be sub-divided into two or more independent sub-events, then the number of ways in which the total event can be accomplished is given by the product of the number of ways in which each sub-event can be accomplished.

No. of ways in which the total event can be accomplished

= (No. of ways in which the 1st sub-event can be accomplished)

× (No. of ways in which the 2nd sub-event can be accomplished)

× (No. of ways in which the 3rd sub-event can be accomplished)

× ....

⇒ nE= nE1 × nE2 × nE3 × ....

Eg: 1. Consider the journey from "New Delhi" to "New York" via "London". There are four routes from "New Delhi" to "London" and five routes from "London" to "New York"

To find the number of ways in which a person can travel from New Delhi to New York, we can divide the total event of travelling from "New Delhi" to "New York" into two independent parts.

  • Total Event (E) = Travelling from "New Delhi" to "New York"
  • 1st sub-event (E1) = Travelling from "New Delhi" to "London"
  • 2nd sub-event (E2) = Travelling from "London" to "New York"

[The events are independent because, the route taken in one part journey is not dependent on the route taken on the other i.e. what route is taken is not influenced by what route has been taken in the other part journey]
"New Delhi"     →     "New York"
E1
"New Delhi"     →     "London"
4 routes 		E2
"London"     →     "New York"
5 routes

There are four routes from "New Delhi" to "London" and five routes from "London" to "New York"

⇒ The number of ways in which the journey

  • From "New Delhi" to "London" can be taken up = 4 ⇒ nE1 = 4
  • From "London" to "New York" can be taken up = 5 ⇒ nE2 = 5

Therefore,
The number of ways in which the total journey from "New Delhi" to "New York" can be accomplished

= (No. of ways in which the journey from "New Delhi" to "Lodon" (1st sub-event) can be accomplished)

× (No. of ways in which the journey from "London" to "New York" (2nd sub-event) can be accomplished)

permutations,combinations,quantitative,techniques,methods,operations,research,linear,circular
⇒ nE = nE1 × nE2
= 4 × 5
= 20

Reasoning

Let "A", "B", "C" and "D" represent the four routes from "New Delhi" to "London".

Let "1", "2", "3", "4" and "5" be the numbers representing the routes from "London" to "New York".

The possibilities can be summarised as
A1   A2   A3   A4   A5
B1   B2   B3   B4   B5
C1   C2   C3   C4   C5
D1   D2   D3   D4   D5

2. Drawing 3 blue, 2 red and 4 white balls from a bag containing 6 blue, 4 red and 7 white balls.

Total number of balls = 17 [6 blue + 4 red + 7 white]

Number of balls drawn = 9 [3 blue + 2 red + 4 white]
permutations,combinations,quantitative,techniques,methods,operations,research,linear,circular

To find the number of ways in which the 3 blue, 2 red and 4 white balls are drawn, we can divide the total event of drawing 9 balls into three sub-events

  • Total Event (E) = Drawing 9 balls
  • 1st sub-event (E1) = Drawing 3 blue balls
  • 2nd sub-event (E2) = Drawing 2 red balls
  • 3rd sub-event (E3) = Drawing 4 white balls
Drawing 9 balls
E1
Drawing 3 blue balls
From 6 Blue		 E2
  Drawing 2 red balls  
From 4 Red		 E3
Drawing 4 white balls
From 7 White

Therefore,

The number of ways in which the 9 balls can be drawn such that 3 blue, 2 red and 4 white balls are drawn

= (No. of ways in which the 3 blue balls (1st sub-event) can be drawn from the total 6)

× (No. of ways in which the 2 red balls (2nd sub-event) can be drawn from the total 4)

× (No. of ways in which the 4 white balls (3rd sub-event) can be drawn from the total 7)

⇒ nE = nE1 × nE2 × nE3
= 6C3 × 4C2 × 7C4
=
6 × 5 × 4
3 × 2 × 1
 ×
4 × 3
2 × 1
 ×
7 × 6 × 5 × 4
4 × 3 × 2 × 1
= 20 × 6 × 35
= 4,200

This can be represented as
Blue × Red × White
Available 6 4 7
To Choose 3 2 4
Choices 6C3 4C2 7C4

Author Credit : The Edifier ... Continued Page :: 5

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