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,
nC1 + nC2 + nC3 +…….+ nCn
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)2n – 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) 3C1×4C1×5C1 = 60
(ii) (23–1)(24–1)(25−1) = 7×15×31 = 3255
(iii) (25–1)(23)(24) = 31×128 = 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 :
pn
(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−1Cp−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!3!1!1!2! + 5!2!2!2!)×3!
Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children =
37 (as each fruit has three options).
∴ Total no. of ways = (5!3!2! + 5!(2!)3)×3!×37