## Fundamental Counting Principle (Theorem) of Multiplication |

## Fundamental Counting Principle (Theorem) of Multiplication |

## Counting Numbers |

The numbers that we use to count things or objects.

Natural Numbers [N = {1, 2, 3, ... ∞ }] are called counting numbers.

We generally start counting with 1. However, at times we may use zero while counting. As such, it is also argued that counting numbers start from zero.

In such cases we may say that whole numbers [W = {0, 1, 2, 3, ... ∞ }] are 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 1^{st} sub-event can be accomplished) × (No. of ways in which the 2 ^{nd} sub-event can be accomplished) × (No. of ways in which the 3 ^{rd} sub-event can be accomplished) × .... × .... |

## Fundamental Counting Principle/Theorem of Multilication : Illustration » 1 |

In how many ways can a person travel from "New Delhi" to "New York" via "London", when there are four routes from New Delhi to London and five routes from London to New York.

In 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".

Dividing the total event into two independent sub-events:

- Total Event (E) : Traveling from "New Delhi" to "New York"
- 1
^{st}sub-event (E_{1}) : Traveling from "New Delhi" to "London" - 2
^{nd}sub-event (E_{2}) : Traveling from "London" to "New York"

The route to be taken up in traveling from "London" to "New York" is not influenced by the route taken in traveling from "New Delhi" to "London". vice versa. As such, we can say that the sub-events are independent.

"New Delhi" ⇒ "New York" | |
---|---|

E_{1}"New Delhi" ⇒ "London" 4 routes | E_{2}"London" ⇒ "New York" 5 routes |

The number of ways in which the journey

- From "New Delhi" to "London" can be taken up = 4 ⇒
**n**_{E1}= 4 - From "London" to "New York" can be taken up = 5 ⇒
**n**_{E2}= 5

Therefore,

No. 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" (1^{st} sub-event) can be accomplished) × (No. of ways in which the the task of traveling from "London" to "New York" (2 ^{nd} sub-event) can be accomplished) |

⇒ n_{E} | = | n_{E1} × n_{E2} |
---|---|---|

= | 4 × 5 | |

= | 20 |

Let

- A, B, C and D represent the four routes from New Delhi to London.
- 1, 2, 3, 4 and 5 represent the routes from London to New York.

The possibilities would be

| Each choice from the first can be combined with each of the five choices from the second. |

## Fundamental Counting Principle/Theorem of Multilication : Illustration » 2 |

In how many ways can 3 blue, 2 red and 4 white balls be drawn from a bag containing 6 blue, 4 red and 7 white balls. (Or)

What are the total number of choices for the event of drawing 3 blue, 2 red and 4 white balls from a bag containing 6 blue, 4 red and 7 white balls.

In the experiment of drawing 9 balls from the bag containing 6 blue, 4 red and 7 white balls.

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

For finding the number of choices in which the 3 blue, 2 red and 4 white balls would be drawn, we divide the total event of drawing the 9 balls into three independent sub-events

- E (Total Event) : Drawing 9 balls from the total 17
- E
_{1}(1^{st}sub-event) : Drawing 3 blue balls from the available 6 - E
_{2}(2^{nd}sub-event) : Drawing 2 red balls from the available 4 - E
_{3}(3^{rd}sub-event) : Drawing 4 white balls from the available 4

Drawing 9 balls | ||
---|---|---|

E_{1}Drawing 3 blue balls From the 6 Blue balls | E_{2}Drawing 2 red balls From the 4 Red balls | E_{3}Drawing 4 white balls From the 7 white balls |

Therefore,

The number of choices in which 3 blue, 2 red and 4 white balls are drawn | = | (No. of ways in which 3 blue balls can be drawn from the total 6 blue balls [1^{st} sub-event]) × (No. of ways in which 2 red balls can be drawn from the total 4 red balls [2 ^{nd} sub-event]) × (No. of ways in which 4 white balls can be drawn from the total 7 white balls [3 ^{rd} sub-event]) | |||||||||||

⇒ n_{E} | = | n_{E1} × n_{E2} × n_{E3} | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

= | ^{6}C_{3} × 4C_{2} × 7C_{4} | ||||||||||||

= |
| ||||||||||||

= | 20 × 6 × 35 | ||||||||||||

= | 4,200 |

For problem solving, the same idea can be represented in the form of a working table as below.

Blue | Red | White | Total | |||
---|---|---|---|---|---|---|

Available | 6 | 4 | 7 | 17 | ||

To Choose | 3 | 2 | 4 | 9 | ||

Choices | ^{6}C_{3} | × | ^{4}C_{2} | × | ^{7}C_{4} |

... 8910 ... |

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