# Fundamental Counting Principle (Theorem) of Addition

## Fundamental Counting Principle of Addition

If a total event can be accomplished in two or more mutually exclusive alternative ways, then the number of ways in which the total event can be accomplished is given by the sum of the number of ways in which each alternative-event can be accomplished.

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

 = (No. of ways in which the 1st alternative-event can be accomplished) + (No. of ways in which the 2nd alternative-event can be accomplished) + (No. of ways in which the 3rd alternative-event can be accomplished) + .... + ....
⇒ nE = nEA1 + nEA2 + nEA3 + ....

## Fundamental Counting Principle/Theorem of Addition : Illustration ยป 1

In how many ways can a committee 4 members be chosen from a group of 6 men and 5 women such that the committee consists of at least 2 women⇒

In the experiment of choosing a committee of 4 members from a group of 6 men and 5 women such that the committee consists of at least 2 women

• Total number of members = 11 [6 men + 5 women]
• Number of Members to be selected = 4

Let

• E : The event of choosing the committee with at least 2 women in it.

### • For Event E

Event E can be accomplished in three alternative ways.
• EA1 : Choosing the Committee with 2 Women and 2 Men
• EA2 : Choosing the Committee with 3 Women and 1 Man
• EA3 : Choosing the Committee with 4 Women and 0 Men

Events EA1, EA2, EA3 are mutually exclusive events. Occurrence of one of them prevents the occurrence of the others. If the committee is chosen in one of these ways we can say that it was not chosen in the other ways.

#### » Event EA1

The total event of choosing the 2 women and 2 men can be sub-divided into two independent sub-events

• EA1 : Choosing the 2 women and 2 men [Total Event]
• EA1a : Choosing the 2 women from the available 6 [1st sub-event]
• EA1b : Choosing the 2 men from the available 5 [2nd sub-event]

Working Table

 Women Men Total 6 5 11 2 2 4 6C2 × 5C2

By the fundamental counting theorem of multiplication,

nEA1 = nEA1a × nEA1b
= 6C2 × 5C2
=
 6 × 5 2 × 1
×
 5 × 4 2 × 1
= 15 × 10
= 150

 ⇒ n(EA1) No. of ways in which the committee with 2 women and 2 men can be chosen = 150 = 150

#### » Event EA2

The total event of choosing the 3 women and 1 man can be sub-divided into two independent sub-events

• EA2 : Choosing the 3 women and 1 man [Total Event]
• EA2a : Choosing the 3 women from the available 6 [1st sub-event]
• EA2b : Choosing the 1 man from the available 5 [2nd sub-event]

Working Table

 Women Men Total 6 5 11 3 1 4 6C3 × 5C1

By the fundamental counting theorem of multiplication,

nEA2 = nEA2a × nEA2b
= 6C3 × 5C1
=
 6 × 5 × 4 3 × 2 × 1
×
 5 1
= 20 × 5
= 100

 ⇒ n(EA2) No. of ways in which the committee with 3 women and 1 man can be chosen = 100 = 100

#### » Event EA3

The total event of choosing the 4 women and 0 men can be sub-divided into two independent sub-events

• EA3 : Choosing the 4 women and 0 men (Total Event)
• EA3a : Choosing the 4 women from the available 6 (1st sub-event)
• EA3b : Choosing the 0 men from the available 5 (2nd sub-event)

Working Table

 Women Men 6 5 4 0 6C4 × 5C0

By the fundamental counting theorem of multiplication,

nEA3 = nEA3a × nEA3b
= 6C4 × 5C0
=
 6 × 5 × 4 × 3 4 × 3 × 2 × 1
× 1
= 15 × 1
= 15

 ⇒ n(EA3) No. of ways in which the committee with 4 women and 0 men can be chosen = 15 = 15

Therefore,

The number of ways in which the committee can be chosen
= (No. of ways in which the committee with 2 women and 2 men can be chosen)
+ (No. of ways in which the committee with 3 women and 1 man can be chosen)
+ (No. of ways in which the committee with 4 women and 0 men can be chosen)
= (No. of ways in which the 1st alternative event can be accomplished)
+ (No. of ways in which the 2nd alternative event can be accomplished)
+ (No. of ways in which the 3rd alternative event can be accomplished)
 ⇒ n(E) n(EA1) + n(EA2) + n(EA3) = = 150 + 100 + 15 = 265

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