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Chapter
7
PERMUTATIONS AND
COMBINATIONS
7.1 Overview
The study of permutations and combinations is concerned with determining the number
of different ways of arranging and selecting objects out of a given number of objects,
without actually listing them. There are some basic counting techniques which will be
useful in determining the number of different ways of arranging or selecting objects.
The two basic counting principles are given below:
Fundamental principle of counting
7.1.1 Multiplication principle (Fundamental Principle of Counting)
Suppose an event E can occur in m different ways and associated with each way of
occurring of E, another event F can occur in n different ways, then the total number of
occurrence of the two events in the given order is m × n .
7.1.2 Addition principle
If an event E can occur in m ways and another event F can occur in n ways, and
suppose that both can not occur together, then E or F can occur in m + n ways.
7.1.3 Permutations A permutation is an arrangement of objects in a definite order.
7.1.4 Permutation of n different objects The number of permutations of n objects
taken all at a time, denoted by the symbol nPn, is given by
n
Pn n ,
... (1)
where n = n(n – 1) (n – 2) ... 3.2.1, read as factorial n, or n factorial.
The number of permutations of n objects taken r at a time, where 0 < r ≤ n,
denoted by nPr , is given by
n
Pr =
We assume that 0 1
n
n r

PERMUTATIONS AND COMBINATIONS
115
7.1.5 When repetition of objects is allowed The number of permutations of n things
taken all at a time, when repetion of objects is allowed is nn .
The number of permutations of n objects, taken r at a time, when repetition of
objects is allowed, is nr .
7.1.6 Permutations when the objects are not distinct The number of permutations
of n objects of which p1 are of one kind, p2 are of second kind, ..., p k are of kth kind and
n!
!
p1 p 2 !... pk !
7.1.7 Combinations On many occasions we are not interested in arranging but only
in selecting r objects from given n objects. A combination is a selection of some or all
of a number of different objects where the order of selection is immaterial. The number
of selections of r objects from the given n objects is denoted by nC r , and is given by
the rest if any, are of different kinds is
n
Cr =
n!
r !( n r )!
Remarks
1. Use permutations if a problem calls for the number of arrangements of objects
and different orders are to be counted.
2. Use combinations if a problem calls for the number of ways of selecting objects
and the order of selection is not to be counted.
7.1.8 Some important results
Let n and r be positive integers such that r ≤ n. Then
(i) nC r = nC n – r
(ii) nC r + nCr – 1 = n + 1C r
(iii) n n – 1C r – 1 = (n – r + 1) n Cr – 1
7.2 Solved Examples
Short Answer Type
Example 1 In a class, there are 27 boys and 14 girls. The teacher wants to select 1
boy and 1 girl to represent the class for a function. In how many ways can the teacher
make this selection?
Solution Here the teacher is to perform two operations:
(i) Selecting a boy from among the 27 boys and
(ii) Selecting a girl from among 14 girls.

116
EXEMPLAR PROBLEMS – MATHEMATICS
The first of these can be done in 27 ways and second can be performed in
14 ways. By the fundamental principle of counting, the required number of ways is
27 × 14 = 378.
Example 2
(i) How many numbers are there between 99 and 1000 having 7 in the units place?
(ii) How many numbers are there between 99 and 1000 having atleast one of their
digits 7?
Solution
(i) First note that all these numbers have three digits. 7 is in the unit’s place. The
middle digit can be any one of the 10 digits from 0 to 9. The digit in hundred’s
place can be any one of the 9 digits from 1 to 9. Therefore, by the fundamental
principle of counting, there are 10 × 9 = 90 numbers between 99 and 1000 having
7 in the unit’s place.
(ii) Total number of 3 digit numbers having atleast one of their digits as 7 = (Total
numbers of three digit numbers) – (Total number of 3 digit numbers in which 7
does not appear at all).
= (9 × 10 × 10) – (8 × 9 × 9)
= 900 – 648 = 252.
Example 3 In how many ways can this diagram be coloured subject to the following
two conditions?
(i) Each of the smaller triangle is to be painted with one of three colours: red, blue or
green.
(ii) No two adjacent regions have the same colour.
Solution These conditions are satisfied exactly when we do as follows: First paint the
central triangle in any one of the three colours. Next paint the remaining 3 triangles,
with any one of the remaining two colours.
By the fundamental principle of counting, this can be done in 3 × 2 × 2 × 2 = 24 ways.

PERMUTATIONS AND COMBINATIONS
117
Example 4 In how many ways can 5 children be arranged in a line such that (i) two
particular children of them are always together (ii) two particular children of them are
never together.
Solution
(i) We consider the arrangements by taking 2 particular children together as one
and hence the remaining 4 can be arranged in 4! = 24 ways. Again two particular
children taken together can be arranged in two ways. Therefore, there are
24 × 2 = 48 total ways of arrangement.
(ii) Among the 5! = 120 permutations of 5 children, there are 48 in which two children
are together. In the remaining 120 – 48 = 72 permutations, two particular children
are never together.
Example 5 If all permutations of the letters of the word AGAIN are arranged in the
order as in a dictionary. What is the 49th word?
Solution Starting with letter A, and arranging the other four letters, there are 4! = 24
words. These are the first 24 words. Then starting with G, and arranging A, A, I and N
in different ways, there are
4!
12 words. Next the 37th word starts with I.
2!1!1!
There are again 12 words starting with I. This accounts up to the 48 th word.
The 49th word is NAAGI.
Example 6 In how many ways 3 mathematics books, 4 history books, 3 chemistry
books and 2 biology books can be arranged on a shelf so that all books of the same
subjects are together.
Solution First we take books of a particular subject as one unit. Thus there are
4 units which can be arranged in 4! = 24 ways. Now in each of arrangements,
mathematics books can be arranged in 3! ways, history books in 4! ways,
chemistry books in 3! ways and biology books in 2! ways. Thus the total number
of ways = 4! × 3! × 4! × 3! × 2! = 41472.
Example 7 A student has to answer 10 questions, choosing atleast 4 from each of
Parts A and B. If there are 6 questions in Part A and 7 in Part B, in how many ways
can the student choose 10 questions?
Solution The possibilities are:
4 from Part A and 6 from Part B
or
5 from Part A and 5 from Part B
or
6 from Part A and 4 from Part B.
Therefore, the required number of ways is

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