Here you will learn addition principle of counting and multiplication principle in permutation and combination with example.
Letโs begin โ
If an event A can occur in โmโ different ways and another event B can occur in โnโ 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)ย A useful notation: n! (factorial n) = n.(n โ 1).(n โ 2)โฆโฆโฆ.3.2.1; n! = n.(n โ 1)! where n \(\in\) N
(ii) 0! = 1! = 1
(iii) Factorial of negative integers are not defined
(iv) (2n)! = \(2^n\).n![1.3.5.7โฆโฆโฆ.(2n -1)]
Formation of groups
(a)ย (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)ย (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)ย If m = n = p then the number of groups = \((3n)!\over n! n! n! 3!\).
(iii)ย 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)ย 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.
\(\therefore\)ย ย 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!
