# Fundamental Principal of Counting

**Rule of Product**

If one experiment has n possible outcomes and another experiment has m possible outcomes, then there are m Ã— n possible outcomes when both of these experiments are performed.

In other words if a job has n parts and the job will be completed only when each part is completed and the first part can be completed in a_{1} ways, the second part can be completed in a_{2} ways and so on ...... then nth part can be completed in an ways, then the total number of ways of doing the job is a_{1}a_{2}a_{3} ... a_{n}. This is known as the rule of product.

A college offers 7 courses in the morning and 5 in the evening. Find the possible number of choices with the student who wants to study one course in the morning and one in the evening.

The student has seven choices from the morning courses out of which he can select one course in 7 ways.

For the evening course, he has 5 choices out of which he can select one in 5 ways.

Hence the total number of ways in which he can make the choice of one course in the morning and one in the evening = 7 Ã— 5 = 35.

A man has five friends. In how many ways can be invite one or more of them to a tea party.

**Methods I**

Case I: The man can invite one friend in ^{5}C_{1} ways

Case II: The man can invite two friend in ^{5}C_{2} ways

Case III: The man can invite three friend in ^{5}C_{3} ways

Case IV: The man can invite four friend in ^{5}C_{4} ways

Case V: The man can invite five friend in ^{5}C_{5} ways

All these five cases are mutually exclusive i.e. the man can either invite one friend or two friends or ............., but cannot simultaneously invite one friend and two friends.

**Method II.**

This problem can also be solved as follows:

Each of his friends may be dealt in two ways - either invited or not invited.

Therefore The total number of ways of inviting all friends is

= 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 = 32 ways.

But this includes the case, in which all the friends are not invited. Hence, the total number of ways in which the man invites one or more friends is 25 - 1 = 31 ways.

Note: In the above illustration, we have found the total number of combinations of n dissimilar things taking any number of them at a time, which is 2

^{n}- 1.

# Rule of Sum

If one experiment has n possible outcomes and another has m possible outcomes, then there are (m + n) possible outcomes when exactly one of these experiments is performed.

In other words if a job can be done by n methods and by using the first method can be done in a1 ways or by second method in a2 ways and so on ... by the nth method in an ways, then the number of ways to get the job done is (a_{1} + a_{2} + ...... + a_{n}).

How many straight lines can be formed from six points, no three of which are collinear?

To form a straight line, we need to select two points out of six points. This can be done in ^{6}C_{2} ways = 6.5/2.1 = 15 ways.

In how many ways can a committee of five be formed from amongst four boys and six girls so as to include exactly two girls?

We have to select two girls from six girls and three boys from four boys.

Number of ways of selecting girls = ^{6}C_{2} = 6.5/2.1 = 15

Number of ways of selecting boys = ^{4}C_{3} = 4

Number of ways of forming the committee = 15 Ã— 4 = 60.

**Note:**Here we have used multiplication rule.

A college offers 7 courses in the morning and 5 in the evening. Find the number of ways a student can select exactly one course, either in the morning or in the evening.

The student has seven choices from the morning courses out of which he can select one course in 7 ways.

For the evening course, he has 5 choices out of which he can select one in 5 ways.

Hence he has total number of 7 + 5 = 12 choices.

** ** ** **

How many (i) 5 - digit, (ii) 3 - digit numbers can be formed by using 1, 2, 3, 4, 5 without repetition of digits.

(i)Making a 5-digit number amounts to filling 5 places.

** Places: **

** Number of Choices: 5 4 3 2 1**

The first place can be filled in 5 ways using any of the given digits.

The second place can be filled in 4 ways using any of the remaining 4 digits.

Similarly, we can fill the 3^{rd}, 4^{th} and 5^{th} place.

No. of ways to fill all the five places.

= 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 = 120

=> 120 5-digit numbers can be formed.

(ii)Making a 3 - digit number amounts to filling 3 places.

** Places: **

**Number of choices: 5 4 3**

Number of ways to fill all the three places = 5 Ã— 4 Ã— 3 = 60.

Hence the total possible 3 - digit numbers = 60.