Fundamental Principle of Counting: Let’s say you have a number lock. It comprises four wheels, each with ten digits ranging from \(0\) to \(9\), and if four specific digits are arranged in a sequence with no repetition, it can be opened. Let’s say you have forgotten the sequence except for the first digit, \(7\). How many \(3\)-digit sequences will you have to try to open the lock?

To solve this question, you begin by making a list of all conceivable combinations of the \(9\) remaining digits in sets of three. However, this procedure will be tiresome because the number of alternative sequences is likely to be huge. Let us learn some basic counting techniques that will allow us to answer this question without listing all the \(3\)-digit arrangements.

Fundamental Principle of Counting: Overview

Let us first consider the following examples.

Example 1: Sania has two school bags and three water bottles. How many different ways can she take one of each of these goods to school?

• Since there are two types of school bags, a school bag can be chosen in two distinct ways.
• On the other hand, a water bottle can be chosen in three distinct ways.
• There are three options for a water bottle for every option of a school bag. Hence, the number of distinct ways to choose a school bag and water bottle \( = 2 \times 3 = 6\)
• Therefore, Sania can carry these goods to her school in six different ways.
• If we represent the two school bags as \({S_1},{S_2}\) and the three water bottles as \({W_1},{W_2},{W_3}\), the following tree diagram depicts the six distinct combinations:

Thus, \({S_1}{W_1},{S_1}{W_2},{S_1}{W_3},{S_2}{W_1},{S_2}{W_2}\), and \({S_2}{W_3}\) are the possible combinations.

Example 2: The numbers \(1, 2, 3,\) and \(4\) are written on four cards. How many two-digit numbers can be formed when two cards are placed side by side?

• We can choose any card and set it in the tens place to represent the digit.
• Take any of the remaining three cards and set it to the right of the preceding card, which represents a digit at the unit’s place.
• Number of various numbers that can be generated by arranging two cards side by side \( = 4 \times 3 = 12\).

Look at the following tree diagram.

The first set of branches in the diagram above shows the four possible card choices placed at ten’s place, while the second set of branches shows the three possible card choices at unit’s position matching each card choice placed at ten’s place. Two-digit numbers formed are as follows:

\(12\)

\(13\)

\(14\)

\(21\)

\(23\)

\(24\)

\(31\)

\(32\)

\(34\)

\(41\)

\(42\)

\(43\)

Fundamental Principle of Multiplication

If an event can occur in \(a\) different ways, and if it has occurred, a second event can occur in \(b\) different ways, then the total number of different occurrences of the two events is \(a \times b\).

This is known as the principle counting of multiplication.

The needed number of ways to carry a school bag and a water bottle, in example \(1\), was the number of ways for the following events to occur in succession.

1. Selecting a school bag
2. Selecting a water bottle

The counting principle of multiplication can be applied to any finite number of events.

Example: Gopal owns \(8\) shirts, \(5\) pairs of pants, and \(2\) pairs of shoes. How many different ways can he pick his outfits for the day?

Solution:

Number of ways to pick shirts \(=8\)

Number of options for pants \(=5\)

Number of ways to pair shirts and pants \( = 8 \times 5 = 40\) combinations

Number of ways to pick shoes \(=2\)

Number of outfit combinations \( = 40 \times 2 = 80\) distinct ways

The counting principle of multiplication can be generalised as below.

If an event can occur in \(a\) different ways, and if when it has occurred, a second event can occur in \(b\) different ways, following which a third event can occur in \(c\) different ways, and so on, then the total number of different ways of occurrence of all the events is \(a \times b \times c \times \ldots \).

Fundamental Principle of Addition

If two jobs can be performed independently in \(a\) and \(b\) ways respectively, then either of the two jobs can be performed in \((a+b)\) ways.

Example: There are \(11\) boys and \(9\) girls in each class. The teacher wants to choose a student to represent the class at an event. In how many ways can the teacher make this decision?

Solution: The teacher must choose one of the following two tasks:

\((i)\) selecting a boy from a group of \(11\), or \((ii)\) selecting a girl from a group of \(9\).

The first of them can be done in \(11\) different ways, while the second can be done in \(9\) different ways. As a result, either of the two ocan be done in \((11 + 9) = 20\) different ways.

Hence, the teacher has \(20\) options for selecting a class representative.

Solved Examples – Fundamental Principle of Counting

Q.1. How many words can be made from the letters in the word ‘MAGIC’ if all of the letters are used simultaneously (no letters are repeated)?
Ans: We know that \(5\) letters must be arranged in \(5\) places.

• The first spot can be filled in \(5\) different ways, using any of the five letters \(M,A,G,I,C\).
• There are four letters left, and any of them can be used to fill the second spot.
• Hence, the first two spots can be filled in \(5 \times 4\) different ways.
• There are three letters left, and one of them can go in third place.
• There are now two letters left, and one of them can be placed in the fourth position.
• Lastly, one letter remains, that must go into the fifth place.

As a result, the total number of options to fill five spaces is \( = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

Q.2. How many \(4\)-letter code words are possible using the first \(10\) letters of the English alphabet if
(i) No letter can be repeated
(ii) letters can be repeated
Ans: (i) The first letter of the code can be chosen in \(10\) ways, the second letter can be chosen in \(9\) ways, the third letter can be chosen in \(8\) ways, and the fourth letter can be chosen in \(7\) ways.
Total number of \(4\) letter code words \( = 10 \times 9 \times 8 \times 7 = 5040\)
(ii) The first letter of the code can be chosen in \(10\) different ways. Since the letters can be repeated, there are \(10\) possibilities for each of the other letters in the code.
Total number of \(4\) letter code words \( = 10 \times 10 \times 10 \times 10 = 10000\)

Q.3. How many numbers lying between \(100\) and \(1000\) can be formed with the digits \(0,1,2,3,4,5\) if the repetition of digits is not allowed?
Ans: The numbers lying between \(100\) and \(1000\) consist of \(3\) digits. There are \(6\) different digits with which three places have to be filled. Repetition of digits is not allowed.
As \(3\)-digit numbers are required, the hundred’s place is non-zero. So, it can be filled up in \(5\) ways. The ten’s place can be filled in \(5\) ways with any one of the remaining \(5\) digits as repetition is not allowed. Similarly, the unit’s place can be filled up in \(4\) ways.
By the fundamental principle of counting, the required number of numbers lying between \(100\) and \(1000 = 5 \times 5 \times 4 = 100\).

Q.4. How many ways can \(2\) prizes (in Science and Maths) be awarded to \(15\) students? 
Ans: Number of students \(=15\)
Number of ways of awarding the prize in Science \(=15\)
Number of ways of awarding the prize in Maths \(=15\)
\(\therefore \) Number of ways of awarding \(2\) prizes (in Science and Maths) \( = 15 \times 15 = 225\).

Q.5. Find the number of positive integers between \(6000\) and \(7000\) which are divisible by \(5\), provided that no digit is to be repeated.
Ans: The thousand’s place has to be filled up by the digit \(6\), so there is only one way to fill. As the numbers are divisible by \(5\), the unit’s place must be filled up either by \(5\) or \(0\).  Thus, there are only \(2\) ways to fill the unit’s place. 
Now, the hundred’s place can be filled up in \(8\) ways by any of the remaining \(8\) digits because two digits have already been used at thousand’s place and unit’s place.
Similarly, the ten’s place can be filled up in \(7\) ways.
\(\therefore \) The number of positive integers \( = 1 \times 8 \times 7 \times 2 = 112\)

Summary

There are two fundamental principles of counting: multiplication principle of counting and addition principle of counting. According to the multiplication principle of counting, if there are \(n\) jobs \({J_1},{J_2}, \ldots {J_n}\) such that job \({J_i}\) can be performed independently in \({m_i}\) ways; \(i = 1,2, \ldots ,n\). Then the total number of ways in which all the jobs can be performed is \({m_1} \times {m_2} \times {m_3} \times \ldots \times {m_n}\). And, according to the addition principle of counting, If \({E_1},{E_2}, \ldots {E_k}\) be disjoint events with \({n_1},{n_2}, \ldots {n_k}\) possible outcomes, respectively. Then the total number of outcomes for the event \(“E_1\) or \({E_2}\) or….or \(E_k”\) is \({n_1} + {n_2} + \cdots + {n_k}\).

Frequently Asked Questions (FAQs)

Q.1. How is permutation different from the fundamental principle of counting?
Ans: Multiplying the number of ways to do each activity is the way to count the number of ways to complete a sequence of tasks using the fundamental principle of counting. In contrast, permutation is an arrangement of objects that emphasises the importance of order. Another definition of permutation is the number of different combinations that can be made.

Q.2. What are some of the everyday applications of the fundamental principle of counting?
Ans: Few applications of the Fundamental Principle of Counting are as follows:

• Counting allows you to determine the number of possible events and assists you in making a decision.
• In statistics and data analysis, the fundamental principle of counting is commonly employed.
• More events with more than two options can be thought of, and we can apply the concept of the fundamental principle of counting there.

Q.3. Why is it important to know the fundamental principle of counting?
Ans: By multiplying the number of options for each event, you can find the number of unique ways a combination of events might occur. You can also use exponents if you have the same options in many slots.

Q.4. What is the fundamental principle of counting? Provide an example.
Ans: The fundamental principle of counting states that, “If an event can occur in \(m\) different ways, and if when it has occurred, a second event can occur in \(n\) different ways, then the total number of different ways of occurrence of the two events is \(m \times {n}\)”.
Example: A food stall has \(8\) beverages, and \(6\) desserts. There are \(8 \times 6 = 48\) distinct meals consisting of a beverage and a dessert.

Q.5. What is the fundamental principle of addition?
Ans: The rule of sum, also known as the addition principle, is a fundamental counting principle. It states that, if we have \(A\) number of ways of doing a task and \(B\) number of ways of doing another task, and we cannot do both simultaneously, then there are \(A+B\) ways to choose one of the tasks.

We hope this detailed article on the Fundamental Principle of Counting was helpful. If you have any doubts, let us know about them in the comment section below. Our team will get try to solve your queries at the earliest.