## Fundamental Counting Principle (Theorem) of Addition |

## 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 1^{st} alternative-event can be accomplished) + (No. of ways in which the 2 ^{nd} alternative-event can be accomplished) + (No. of ways in which the 3 ^{rd} alternative-event can be accomplished) + .... + .... |

## 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.

Event E can be accomplished in three alternative ways. #### » Event E_{A1}

#### » Event E_{A2}

#### » Event E_{A3}

*E*: Choosing the Committee with 2 Women and 2 Men_{A1}*E*: Choosing the Committee with 3 Women and 1 Man_{A2}*E*: Choosing the Committee with 4 Women and 0 Men_{A3}

Events E_{A1}, E_{A2}, E_{A3} 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.

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

- E
_{A1}: Choosing the 2 women and 2 men [Total Event] - E
_{A1a}: Choosing the 2 women from the available 6 [1^{st}sub-event] - E
_{A1b}: Choosing the 2 men from the available 5 [2^{nd}sub-event]

Working Table

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

Available | 6 | 5 | 11 | |

To Choose | 2 | 2 | 4 | |

Choices | ^{6}C_{2} | × | ^{5}C_{2} |

By the fundamental counting theorem of multiplication,

n_{EA1} | = | n_{EA1a} × n_{EA1b} | |||||||

= | ^{6}C_{2} × ^{5}C_{2} | ||||||||

= |
| ||||||||

= | 15 × 10 | ||||||||

= | 150 |

No. of ways in which the committee with 2 women and 2 men can be chosen | = | 150 |

⇒ n(E_{A1}) | = | 150 |
---|

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

- E
_{A2}: Choosing the 3 women and 1 man [Total Event] - E
_{A2a}: Choosing the 3 women from the available 6 [1^{st}sub-event] - E
_{A2b}: Choosing the 1 man from the available 5 [2^{nd}sub-event]

Working Table

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

Available | 6 | 5 | 11 | |

To Choose | 3 | 1 | 4 | |

Choices | ^{6}C_{3} | × | ^{5}C_{1} |

By the fundamental counting theorem of multiplication,

n_{EA2} | = | n_{EA2a} × n_{EA2b} | |||||||

= | ^{6}C_{3} × ^{5}C_{1} | ||||||||

= |
| ||||||||

= | 20 × 5 | ||||||||

= | 100 |

No. of ways in which the committee with 3 women and 1 man can be chosen | = | 100 |

⇒ n(E_{A2}) | = | 100 |
---|

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

- E
_{A3}: Choosing the 4 women and 0 men (Total Event) - E
_{A3a}: Choosing the 4 women from the available 6 (1^{st}sub-event) - E
_{A3b}: Choosing the 0 men from the available 5 (2^{nd}sub-event)

Working Table

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

Available | 6 | 5 | |

To Choose | 4 | 0 | |

Choices | ^{6}C_{4} | × | ^{5}C_{0} |

By the fundamental counting theorem of multiplication,

n_{EA3} | = | n_{EA3a} × n_{EA3b} | |||||

= | ^{6}C_{4} × ^{5}C_{0} | ||||||

= |
| ||||||

= | 15 × 1 | ||||||

= | 15 |

No. of ways in which the committee with 4 women and 0 men can be chosen | = | 15 |

⇒ n(E_{A3}) | = | 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 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 1^{st} alternative event can be accomplished)

+ (No. of ways in which the 2^{nd} alternative event can be accomplished)

+ (No. of ways in which the 3^{rd} alternative event can be accomplished)

+ (No. of ways in which the 2

+ (No. of ways in which the 3

⇒ n(E) | = | n(E_{A1}) + n(E_{A2}) + n(E_{A3}) |
---|---|---|

= | 150 + 100 + 15 | |

= | 265 |

... 91011 ... |

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