Fundamental Counting Principle (Theorem) of Multiplication

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

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)
  × ....
  × ....

⇒ nE = nE1 × nE2 × nE3 × ....

Illustration » 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".

In how many ways can a person travel from "New Delhi" to "New York" via "London"?

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

[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, 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
    nE1 = 4
  • From "London" to "New York" can be taken up = 5
    nE2 = 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" (1st sub-event) can be accomplished)
        × (No. of ways in which the the task of traveling from "London" to "New York" (2nd sub-event) can be accomplished)
nE = nE1 × nE2
= 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

Each choice of the first can be combined with every choice of the second.

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Illustration » 2  
 
Consider the experiment of drawing 9 balls from a bag.

There are 6 blue, 4 red and 7 white balls in the bag.

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

  • Total number of balls = 17 [6 blue + 4 red + 7 white]
  • Number of balls drawn = 9 [3 blue + 2 red + 4 white]

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

To find this

Let us divide the total event of drawing the 9 balls into three independent sub-events.
  • Total Event (E) = Drawing 9 balls from the total 17
  • 1st sub-event (E1) = Drawing 3 blue balls from the available 6
  • 2nd sub-event (E2) = Drawing 2 red balls from the available 4
  • 3rd 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

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

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

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