Here you will learn addition principle of counting and multiplication principle in permutation and combination with example.

Let’s begin –

If an event A can occur in ‘m’ different ways and another event B can occur in ‘n’ different ways, then the total number of different ways of-

## Multiplication Principle of Counting

Simultaneous occurrences of both events in a definite order is \(m\times n\). This can be extended to any number of events.

Example : There are 15 IITs in India and let each IIT has 10 branches, then the IITJEE topper can select the IIT and branch in \(15\times 10\) = 150 number of ways

## Addition Principle of Counting

Happening exactly one of the events is m + n.

Example : There are 15 IITs & 20 NITs in India, then a student who cleared both IITJEE & AIEEE exams can select an institute in (15 + 20) = 35 number of ways.

## Factorial Notations

(i) A useful notation: n! (factorial n) = n.(n – 1).(n – 2)……….3.2.1; n! = n.(n – 1)! where n \(\in\) N

(ii) 0! = 1! = 1

(iii) Factorial of negative integers are not defined

(iv) (2n)! = \(2^n\).n![1.3.5.7……….(2n -1)]

## Formation of groups

(a) (i) The number of ways in which (m + n) different things can be divided into two groups such that one of them contains m things and other has n things, is \((m+n)!\over {m! n!}\) (\(m\ne n\)).

(ii) If m = n, it means the groups are equal & in this case the number of divisions is \((2n)!\over {n! n! 2!}\)

As in any one ways it is possible to interchange the two groups without obtaining a new distribution.

(iii) If 2n things are to be divided equally between two persons then the number of ways: \((2n)!\over {n! n! 2!}\)\(\times 2!\)

(b) (i) Number of ways in which (m + n + p) different things can be divided into three groups containing m, n & p things respectively is : \((m+n+p)!\over {m! n! p!}\)(\(m\ne n\ne p\)).

(ii) If m = n = p then the number of groups = \((3n)!\over n! n! n! 3!\).

(iii) If 3n things are to be divided equally among three people then the number of ways in which it can be done is \((3n)!\over {(n!)^3}\).

(c) In general, the number of ways of dividing n distinct objects into x groups containing p objects each and m groups containing q objects each is equal to \(n!(x+m)!\over {(p!)^x (q!)^m x! m!}\).

Example : In how many ways can 15 student be divided into 3 groups of 5 students each such that 2 particular students are always together? Also find the number of ways if these groups are to be sent to three different colleges.

Solution : First pen can be put in 6 ways.

Here first we separate those two particular students and make 3 groups of 5, 5 and 3 of the remaining 13 so that these two particular students always go with the group of 3 students.

\(\therefore\) Number of ways = \(13!\over 5!5!3!\).\(1\over 2!\)

Now these groups are to be sent to three different colleges, total number of ways =
\(13!\over 5!5!3!\)\(1\over 2!\).3!