Probability : Fundamental Counting Principle (Theorem) of Multiplication

Counting Numbers

The numbers that we use to count things or objects. Natural Numbers are called counting numbers.

N = {1, 2, 3, ... ∞ }

Fundamental 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)
× ....
× ....

⇒ n_E = n_E1 × n_E2 × n_E3 × ....

Fundamental Counting Principle/Theorem of Multilication : Illustration » 1

In how many ways can a person travel from "New Delhi" to "New York" via "London", when there are four routes from New Delhi to London and five routes from London to New York.

Considering 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 this,

Let us divide the total event of traveling from "New Delhi" to "New York" into two independent sub-events.

Total Event (E) : Traveling from "New Delhi" to "New York"
1^st sub-event (E1) : Traveling from "New Delhi" to "London"
2^nd sub-event (E2) : Traveling from "London" to "New York"

What route is taken in one part of the journey is not influenced by what route has been (or has to be) taken in the other part of the journey. Thus, we can say that the sub-events are independent.

"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
⇒ n_E1 = 4
From "London" to "New York" can be taken up = 5
⇒ n_E2 = 5

Therefore,

The number of ways in which the total task of travelling from "New Delhi" to "New York" can be accomplished
= (No. of ways in which the task of traveling from "New Delhi" to "London" (1^st sub-event) can be accomplished)
× (No. of ways in which the the task of traveling from "London" to "New York" (2^nd sub-event) can be accomplished)

⇒ n_E	=	n_E1 × n_E2
	=	4 × 5
	=	20

» Rationale

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

A1 A2 A3 A4 A5
B1 B2 B3 B4 B5
C1 C2 C3 C4 C5
D1 D2 D3 D4 D5

Each choice of the first can be combined with each of the four choices (every choice) of the second.

Fundamental Counting Principle/Theorem of Multilication : Illustration » 2

In how many ways can 3 blue, 2 red and 4 white balls be drawn from a bag containing 6 blue, 4 red and 7 white balls.
(Or)
What are the total number of choices for the event of drawing 3 blue, 2 red and 4 white balls from a bag containing 6 blue, 4 red and 7 white balls.

Considering the experiment of drawing 9 balls from the 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]

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

Total Event (E) : Drawing 9 balls from the total 17
1^st sub-event (E1) : Drawing 3 blue balls from the available 6
2^nd sub-event (E2) : Drawing 2 red balls from the available 4
3^rd sub-event (E3) : Drawing 4 white balls from the available 4

Drawing 9 balls

E1
Drawing 3 blue balls
From the 6 Blue balls E2
Drawing 2 red balls
From the 4 Red balls E3
Drawing 4 white balls
From the 7 White balls

Drawing 9 balls
E1 Drawing 3 blue balls From the 6 Blue balls	E2 Drawing 2 red balls From the 4 Red balls	E3 Drawing 4 white balls From the 7 White balls

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 3 blue balls (1^st sub-event) can be drawn from the total 6 blue balls)
× (No. of ways in which 2 red balls (2^nd sub-event) can be drawn from the total 4 red balls)
× (No. of ways in which 4 white balls (3^rd sub-event) can be drawn from the total 7 white balls)

⇒ n_E = n_E1 × n_E2 × n_E3

= 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

Working Table

The above idea can be represented in a working table as

Blue × Red × White

Available 6 4 7

To Choose 3 2 4

Choices ⁶C₃ ⁴C₂ ⁷C₄

	Blue	×	Red	×	White
Available	6	4	7
To Choose	3	2	4
Choices	⁶C₃	⁴C₂	⁷C₄

Disclaimer ♣ Privacy Policy ♣ Terms of Service ♣ Who We Are