# 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.

Number of ways in which the total event can be accomplished

= (Number of ways in which the 1st event alternative can be accomplished) + (Number of ways in which the 2nd event alternative can be accomplished) + (Number of ways in which the 3rd event alternative can be accomplished) + ...

⇒ nE = nEA1 + nEA2 + nEA3 + ....

# Illustration 1 - Problem

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?

# Illustration 1 - Solution

Total number of members

= 11 [6 men + 5 women]

Number of members to be selected

= 4

Experiment : Choosing 4 members

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
• 1st event-alternative (EA1)

Choosing the Committee with 2 Women and 2 Men

• 2nd event-alternative (EA2)

Choosing the Committee with 3 Women and 1 Man

• 3rd event-alternative (EA3)

Choosing the Committee with 4 Women and 0 Men

EA1, EA2 and EA3 are Mutually Exclusive Events.

Occurrence of one of these events 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.

## EA1

The total event of choosing the committee of 4 members from the total 11 can be sub divided into two independent sub events as
• 1st sub-event (SEA11)

Choosing 2 men from the total 5

• 2nd sub-event (SEA12)

Choosing 2 women from the total 6

 Men Women Total Available 5 6 11 To Choose 2 2 4 Choices 5C2 6C2 11C4

By the fundamental counting theorem of multiplication

Number of ways in which the committee can be chosen with 2 women and 2 men

= (Number of ways in which the 1st sub event of choosing the 2 men from a total 5 can be accomplished) × (Number of ways in which the 2nd sub event of choosing the 2 women from a total 6 can be accomplished)

⇒ nEA1 = nSEA11 × nSEA12
= 5C2 × 6C2
=
 5 × 4 2 × 1
×
 6 × 5 2 × 1
= 10 × 15
= 150

## EA2

The total event of choosing the committee of 4 members from the total 11 can be sub divided into two independent sub events as
• 1st sub-event (SEA21)

Choosing 1 man from the total 5

• 2nd sub-event (SEA22)

Choosing 3 women from the total 6

 Men Women Total Available 5 6 11 To Choose 1 3 4 Choices 5C1 6C3 11C4

By the fundamental counting theorem of multiplication

Number of ways in which the committee can be chosen with 3 women and 1 man

= (Number of ways in which the 1st sub event of choosing the 1 man from a total 5 can be accomplished) × (Number of ways in which the 2nd sub event of choosing the 3 women from a total 6 can be accomplished)

⇒ nEA2 = nSEA21 × nSEA22
= 5C1 × 6C3
=
 5 1
×
 6 × 5 × 4 3 × 2 × 1
= 5 × 20
= 100

## EA3

The total event of choosing the committee of 4 members from the total 11 can be sub divided into two independent sub events as
• 1st sub-event (SEA31)

Choosing 0 men from the total 5

• 2nd sub-event (SEA32)

Choosing 4 women from the total 6

 Men Women Total Available 5 6 11 To Choose 0 4 4 Choices 5C0 6C4 11C4

By the fundamental counting theorem of multiplication

Number of ways in which the committee can be chosen with 4 women and 0 men

= (Number of ways in which the 1st sub event of choosing 0 men from a total 5 can be accomplished) × (Number of ways in which the 2nd sub event of choosing the 4 women from a total 6 can be accomplished)

⇒ nEA3 = nSEA31 × nSEA32
= 5C0 × 6C4
= 1 ×
 6 × 5 × 4 × 3 4 × 3 × 2 × 1
= 1 × 15
= 15

By the fundamental counting theorem of addition,

The number of ways in which the committee of 4 members be chosen such that it consists of at least 2 women

= (Number of ways in which the 1st alternative event of choosing the 2 men and 2 women can be accomplished) + (Number of ways in which the 2nd alternative event of choosing the 1 man and 3 women can be accomplished) + (Number of ways in which the 3rd alternative event of choosing the 0 men and 4 women can be accomplished)

 ⇒ nE = nEA1 + nEA2 + nEA3 = 150 + 100 + 15 = 265