Total Number of Combinations | Permutation and Combination

Here you will learn how to find total number of combinations and distribution of distinct objects and alike objects.

Let’s begin –

Total Number of Combinations

(a)  Given n different objects, the number of ways of selecting at least one of them is,

\(^{n}C_1\) + \(^{n}C_2\) + \(^{n}C_3\) +…….+ \(^{n}C_n\)

This can also be stated as the total number of combinations of n distinct things.

(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 + 1)(q + 1)(r + 1)……….-1

(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 + 1)(q + 1)(r + 1)\(2^n\) – 1

Example : 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-
(i) one book of each subject?
(ii) at least one book of each subject?
(iii) at least one book of english?

Solution : (i) \(^{3}C_1\times ^{4}C_1\times ^{5}C_1\) = 60

(ii) \((2^3 – 1)(2^4 – 1)(2^5 -1)\) = \(7\times 15\times 31\) = 3255

(iii) \((2^5 – 1)(2^3)(2^4)\) = \(31\times 128\) = 3968

Distribution of Distinct Objects and Alike Objects

(a)  Distribution of distinct objects

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 :


(b)  Distribution of alike objects

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


Example : 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 ?

Solution : 5 different mangoes can be distributed by following ways among 3 children such that each gets at least 1 :

Total number of ways : (\(5!\over 3! 1! 1! 2!\) + \(5!\over 2! 2! 2!\))\(\times 3!\)

Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children = \(3^7\) (as each fruit has three options).

\(\therefore\)    Total no. of ways = (\(5!\over 3! 2!\) + \(5!\over {(2!)^3}\))\(\times 3!\times 3^7\)

Leave a Comment

Your email address will not be published. Required fields are marked *