# Fundamental Counting Principle (Theorem) of Addition

# Fundamental Counting Principle of Addition

Number of ways in which the total event can be accomplished

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

⇒ n_{E} = n_{EA1} + n_{EA2} + n_{EA3} + ....

# Illustration 1 - Problem

# 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

**1**^{st}event-alternative (EA_{1})Choosing the Committee with 2 Women and 2 Men

**2**^{nd}event-alternative (EA_{2})Choosing the Committee with 3 Women and 1 Man

**3**^{rd}event-alternative (EA_{3})Choosing the Committee with 4 Women and 0 Men

EA_{1}, EA_{2} and EA_{3} 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.

## EA_{1}

**1**^{st}sub-event (SEA_{11})Choosing 2 men from the total 5

**2**^{nd}sub-event (SEA_{12})Choosing 2 women from the total 6

Men | Women | Total | |
---|---|---|---|

Available | 5 | 6 | 11 |

To Choose | 2 | 2 | 4 |

Choices | ^{5}C_{2} | ^{6}C_{2} | ^{11}C_{4} |

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

⇒ n_{EA1} | = | n_{SEA11} × n_{SEA12} | ||||

= | ^{5}C_{2} × ^{6}C_{2} | |||||

= |
| |||||

= | 10 × 15 | |||||

= | 150 |

## EA_{2}

**1**^{st}sub-event (SEA_{21})Choosing 1 man from the total 5

**2**^{nd}sub-event (SEA_{22})Choosing 3 women from the total 6

Men | Women | Total | |
---|---|---|---|

Available | 5 | 6 | 11 |

To Choose | 1 | 3 | 4 |

Choices | ^{5}C_{1} | ^{6}C_{3} | ^{11}C_{4} |

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

⇒ n_{EA2} | = | n_{SEA21} × n_{SEA22} | ||||

= | ^{5}C_{1} × ^{6}C_{3} | |||||

= |
| |||||

= | 5 × 20 | |||||

= | 100 |

## EA_{3}

**1**^{st}sub-event (SEA_{31})Choosing 0 men from the total 5

**2**^{nd}sub-event (SEA_{32})Choosing 4 women from the total 6

Men | Women | Total | |
---|---|---|---|

Available | 5 | 6 | 11 |

To Choose | 0 | 4 | 4 |

Choices | ^{5}C_{0} | ^{6}C_{4} | ^{11}C_{4} |

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

⇒ n_{EA3} | = | n_{SEA31} × n_{SEA32} | ||

= | ^{5}C_{0} × ^{6}C_{4} | |||

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

⇒ n_{E} | = | n_{EA1} + n_{EA2} + n_{EA3} |

= | 150 + 100 + 15 | |

= | 265 |