# Addition Principle of Counting | Multiplication Principle

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

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!