Fundamental Principles of Counting | Mathematics (Maths) Class 11 - Commerce PDF Download

Fundamental Principles of Counting

I. The Addition Rule

Let us have two events, namely A and B. The number of ways in which event A can occur/ the number of possible outcomes of event A is n(A) and similarly, for the event B, it is n(B). Also, the events A and B are mutually exclusive events i.e. they have no outcome common to each other.
Let E be an event describing the situation in which either event A occurs, OR event B occurs. Then, the number of ways in which the event E can occur or the number of possible outcomes of the event E is given by:

n(E) = n(A) + n(B)

This is known as the Addition Rule of Counting. Let’s clarify our concepts with a suitable example.

Question: Jacob goes to a shop to buy some ping pong balls. He wishes to choose one ball from the amateur section, which had a total of five balls; or one ball from the professional section, which had a total of three balls. How many ways are possible in which he can buy a ball i.e. he can buy one ball from the amateur section OR one ball from the professional section?
Solution: n(Jacob buying a ball) = n(Jacob buys one ball from the amateur section) + n(Jacob buys one ball from the professional section)

n(Jacob buying a ball) = ⁵C₁ + ³C₁
= 5 + 3 = 8
Thus there are 8 possible ways in which Jacob can buy a ball from the store, according to his specific wishes.

II. The Product Rule

In similarity to the events defined as in the Addition Rule, let us have two events namely A and B; such that both are mutually independent of each other i.e. one event’s outcome does not affect the other event’s outcome. (We’ll show this physically through our solved example)
Let E be an event describing the situation in which either event A occurs, AND event B occur i.e. both event A and event B must occur (note the difference from the previously mentioned case). Then, the number of ways in which the event E can occur or the number of possible outcomes of the event E is given by:

n(E) = n(A)×n(B)

This is The Multiplication Rule of Counting or The Fundamental Counting Principle. Let’s try and understand it with an example.

Question: Jacob goes to a sports shop to buy a ping pong ball and a tennis ball. There is a total of five ping pong balls and 3 tennis balls available in the shop. In how many ways can Jacob buy a ping pong ball and a tennis ball?
Solution: Clearly; the phenomenon of Jacob buying a ping pong ball is independent of the phenomenon of Jacob buying a tennis ball. Both are completely separate events!

n(Jacob buying one tennis ball and a ping-pong ball) = n(Jacob buys a ping pong ball) × n(Jacob buys a tennis ball)
n(Jacob buys both one tennis ball and a ping-pong ball) = ⁵C₁ × ³C₁ =   5 × 3 = 15
Thus there are 15 different ways in which Jacob can buy a ping pong ball and a tennis ball from the sports shop.

Generalisation of the Addition and the Product Rule
In general, if there are several mutually exclusive events P₁, P₂, P₃, P₄……P_n …etc. with the respective number of ways given as n (P₁), n(P₂), n(P₃), n(P₄)….n(Pn),  then the number of ways in which either P₁ and P₂ and ….. … P_n can occur is given by,

n(E) = n(P₁) + n(P₂) …….. + n(P_n) 

Similarly, if there are several mutually independent events P₁, P₂, P₃, P₄……P_n…etc.
with the respective number of ways given as n (P₁), n(P₂), n(P₃), n(P₄)….n(P_n) , then the number of ways in which  P₁ and P₂ and ….. … Pn  can occur is given by,

n(E) = n(P₁) × n(P₂) …….. × n(P_n) 

We must note that all the possible number of ways derived thus, all of them will represent the unique and distinct ways in which the event E will take place.

Event and Its Trials
Suppose we have an event E with ‘m’ possible outcomes. If we conduct this event n number of times, then the number of outcomes of ‘n trials of the event’ will be, mn
Clearly, this is only valid when all the outcomes of the experiment/event E are independent of each other. This would ensure that everytime the event takes place, any of its outcomes are possible.

Solved Example

Question: A coin is tossed 50 times. What is the number of possible outcomes of this experiment?
Solution: A coin toss has two possible outcomes: ‘A heads’ or ‘A tails’. Every toss of the coin is independent of every other toss of the coin, since whenever you toss the coin; there would be two possible outcomes with equal probability. Thus, the no. of possible outcomes of the experiment = 2⁵⁰