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 1st sub-event can be accomplished)   × (No. of ways in which the 2nd sub-event can be accomplished)   × (No. of ways in which the 3rd sub-event can be accomplished)   × ....   × .... ⇒ nE = nE1 × nE2 × nE3 × ....

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. Considering 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". To find this, Let us divide the total event of traveling from "New Delhi" to "New York" into two independent sub-events. Total Event (E) : Traveling from "New Delhi" to "New York" 1st sub-event (E1) : Traveling from "New Delhi" to "London" 2nd sub-event (E2) : Traveling from "London" to "New York" What route is taken in one part of the journey is not influenced by what route has been (or has to be) taken in the other part of the journey. Thus, we can say that the sub-events are independent. "New Delhi"     →     "New York" E1 "New Delhi"     →     "London" 4 routes E2 "London"     →     "New York" 5 routes There are Four routes from "New Delhi" to "London" and five routes from "London" to "New York" ⇒ The number of ways in which the journey From "New Delhi" to "London" can be taken up = 4 ⇒ nE1 = 4 From "London" to "New York" can be taken up = 5 ⇒ nE2 = 5 Therefore, The number 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" (1st sub-event) can be accomplished)         × (No. of ways in which the the task of traveling from "London" to "New York" (2nd sub-event) can be accomplished) ⇒ nE = nE1 × nE2 = 4 × 5 = 20 » Rationale Let "A", "B", "C" and "D" represent the four routes from "New Delhi" to "London". Let "1", "2", "3", "4" and "5" be the numbers representing the routes from "London" to "New York". The possibilities can be summarised as A1   A2   A3   A4   A5 B1   B2   B3   B4   B5 C1   C2   C3   C4   C5 D1   D2   D3   D4   D5 Each choice of the first can be combined with each of the four choices (every choice) of the second.

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. Considering 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 drawn = 9 [3 blue + 2 red + 4 white] For finding the number of ways in which the 3 blue, 2 red and 4 white balls can be drawn, we divide the total event of drawing the 9 balls into three independent sub-events Total Event (E) : Drawing 9 balls from the total 17 1st sub-event (E1) : Drawing 3 blue balls from the available 6 2nd sub-event (E2) : Drawing 2 red balls from the available 4 3rd sub-event (E3) : Drawing 4 white balls from the available 4 Drawing 9 balls E1 Drawing 3 blue balls From the 6 Blue balls E2   Drawing 2 red balls   From the 4 Red balls E3 Drawing 4 white balls From the 7 White balls Therefore, The number of ways in which the 9 balls can be drawn such that 3 blue, 2 red and 4 white balls are drawn = (No. of ways in which 3 blue balls (1st sub-event) can be drawn from the total 6 blue balls)   × (No. of ways in which 2 red balls (2nd sub-event) can be drawn from the total 4 red balls)   × (No. of ways in which 4 white balls (3rd sub-event) can be drawn from the total 7 white balls) ⇒ nE = nE1 × nE2 × nE3 = 6C3 × 4C2 × 7C4 = 6 × 5 × 4 3 × 2 × 1 × 4 × 3 2 × 1 × 7 × 6 × 5 × 4 4 × 3 × 2 × 1 = 20 × 6 × 35 = 4,200 Working Table The above idea can be represented in a working table as Blue × Red × White Available 6 4 7 To Choose 3 2 4 Choices 6C3 4C2 7C4

Author Credit : The Edifier

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