Here you will learn addition principle of counting and multiplication principle in permutation and combination with example.<\/p>\n
Let’s begin –<\/p>\n
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-<\/p>\n
Simultaneous occurrences of both events in a definite order is \\(m\\times n\\). This can be extended to any number of events.<\/p>\n\n\n
Example : <\/span> 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<\/p>\n\n\n Happening exactly one of the events is m + n.<\/p>\n\n\n Example : <\/span>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.<\/p>\n\n\n (i)\u00a0 A useful notation: n! (factorial n) = n.(n – 1).(n – 2)\u2026\u2026\u2026.3.2.1; n! = n.(n – 1)! where n \\(\\in\\) N<\/p>\n (ii) 0! = 1! = 1<\/p>\n (iii) Factorial of negative integers are not defined<\/p>\n (iv) (2n)! = \\(2^n\\).n![1.3.5.7…\u2026\u2026.(2n -1)]<\/p>\n (a)\u00a0 (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\\)).<\/p>\n (ii) If m = n, it means the groups are equal & in this case the number of divisions is \\((2n)!\\over {n! n! 2!}\\)<\/p>\n As in any one ways it is possible to interchange the two groups without obtaining a new distribution.<\/p>\n (iii) If 2n things are to be divided equally between two persons then the number of ways: \\((2n)!\\over {n! n! 2!}\\)\\(\\times 2!\\)<\/p>\n (b)\u00a0 (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\\)).<\/p>\n (ii)\u00a0 If m = n = p then the number of groups = \\((3n)!\\over n! n! n! 3!\\).<\/p>\n (iii)\u00a0 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}\\).<\/p>\n (c)\u00a0 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!}\\).<\/p>\n\n\n Example : <\/span>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.<\/p>\n Solution : <\/span>First pen can be put in 6 ways. Addition Principle of Counting<\/h2>\n
Factorial Notations<\/h2>\n
Formation of groups<\/h2>\n
\n 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.
\n \t\t \\(\\therefore\\) Number of ways = \\(13!\\over 5!5!3!\\).\\(1\\over 2!\\)
\n \t\t Now these groups are to be sent to three different colleges, total number of ways = \n \t\t \\(13!\\over 5!5!3!\\)\\(1\\over 2!\\).3!\n\t\t <\/p>\n\n\n\n