# Total Number of Combinations | Permutation & 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 :

$$p^n$$

(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

$$^{n+p-1}C_{p-1}$$

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 = ($$5!\over 3! 2!$$ + $$5!\over {(2!)^3}$$)$$\times 3!\times 3^7$$