{"id":4024,"date":"2021-08-14T20:02:34","date_gmt":"2021-08-14T20:02:34","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4024"},"modified":"2022-01-16T16:55:25","modified_gmt":"2022-01-16T11:25:25","slug":"total-number-of-combinations","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/total-number-of-combinations\/","title":{"rendered":"Total Number of Combinations | Permutation and Combination"},"content":{"rendered":"
Here you will learn how to find total number of combinations and distribution of distinct objects and alike objects.<\/p>\n
Let’s begin –<\/p>\n
(a) Given n different objects, the number of ways of selecting at least one of them is,<\/p>\n
\n\\(^{n}C_1\\) + \\(^{n}C_2\\) + \\(^{n}C_3\\) +…….+ \\(^{n}C_n\\)<\/p>\n<\/blockquote>\n
This can also be stated as the total number of combinations of n distinct things.<\/p>\n
(b) (i) Total number of ways in which it is possible to make a selection by taking some or all out of p + q + r +…….things, where p are alike of one kind, q alike of a second kind, r alike of third kind and so on is given by : <\/p>\n
\n(p + 1)(q + 1)(r + 1)……….-1<\/p>\n<\/blockquote>\n
(ii) The total number of ways of selecting one or more things from p identical things of one kind, q identical things of second kind, r identical things of third kind and n different things is given by:<\/p>\n
\n(p + 1)(q + 1)(r + 1)\\(2^n\\) – 1<\/p>\n<\/blockquote>\n\n\n
Example : <\/span> There are 3 books of mathematics, 4 of science and 5 of english. How many different collections can be made such that each collection consists of-
\n (i) one book of each subject?
\n \t\t (ii) at least one book of each subject?
\n \t\t (iii) at least one book of english?<\/p>\nSolution : <\/span>(i) \\(^{3}C_1\\times ^{4}C_1\\times ^{5}C_1\\) = 60
\n \t\t (ii) \\((2^3 – 1)(2^4 – 1)(2^5 -1)\\) = \\(7\\times 15\\times 31\\) = 3255
\n \t\t (iii) \\((2^5 – 1)(2^3)(2^4)\\) = \\(31\\times 128\\) = 3968
<\/p>\n\n\nDistribution of Distinct Objects and Alike Objects<\/h2>\n
(a) Distribution of distinct objects<\/strong><\/p>\n
Number of ways in which n distinct things can be distributed to p persons if there is no restriction to the number of things recieved by them is given by :<\/p>\n
\n\\(p^n\\)<\/p>\n<\/blockquote>\n
(b) Distribution of alike objects<\/strong><\/p>\n
Number of ways to distribute n alike things among p persons so that each may get none, one or more thing(s) is given by<\/p>\n
\n\\(^{n+p-1}C_{p-1}\\)<\/p>\n<\/blockquote>\n\n\n
Example : <\/span>In how many ways can 5 different mangoes, 4 different oranges & 3 different apples be distributed among 3 children such that each gets at least one mango ?<\/p>\n
Solution : <\/span>5 different mangoes can be distributed by following ways among 3 children such that each gets at least 1 :
\n Total number of ways : (\\(5!\\over 3! 1! 1! 2!\\) + \\(5!\\over 2! 2! 2!\\))\\(\\times 3!\\)
\n\t\t Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children =\n\t\t \\(3^7\\) (as each fruit has three options).\n\t\t
\n\t\t \\(\\therefore\\) Total no. of ways = (\\(5!\\over 3! 2!\\) + \\(5!\\over {(2!)^3}\\))\\(\\times 3!\\times 3^7\\)
<\/p>\n\n\n\n