Tips on Solving Aptitude Questions Based On Permutations

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Tips on Solving Aptitude Type Questions on
Permutation
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Understanding Permutation Combination
Suppose we have 4 objects A, B, C and D and we
are required to choose 3 from them and then
arrange them on a shelf. This can be done in the
following ways
Selecting objects Arranging the objects
ABC ABD ACD BCD ABC, ACB, BCA, BAC, CAB, CBA ABD, ADB, BDA, BAD, DBA, DAB ACD, ADC, CAD, CDA, DAC, DCA BCD, BDC, CBD, CDB, DBC, DCB

• Thus, there are 4 ways of choosing 3 objects from
  4 and there are 6 ways of arranging the chosen
  objects. The process of selecting things is
  called combination and that of arranging things
  is called permutation.
• Examples of problems relating to combination
• Formation of a team from a number of players.
• Formation of a particular committee from a number
  of players.
• Examples of problems relating to permutation
• Arrangement of books on a shelf.
• Formation of numbers with the given digits.
• Formation of words with the given letters.

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The Sum Rule
The Sum Rule If A and B are 2 disjoint events (A
or B can occur but they never occur together)
such that A occurs in m ways and B in n ways,
then A or B can occur in m n ways. Question
How many odd numbers less than 1000 can be
formed using the digits 1,2,5,7,8 if repetition
of digits is allowed? Solution Total no. of
digits 5, No. of odd digits 3. One-digit
numbers Only 3 1-digit numbers are possible
1,5,7 Two-digit numbers No. of possible ways 5
x 3 15 Three-digit numbers No. of possible
ways 5 x 5 x 3 75 Thus, there are (31575)
93 ways of forming odd numbers less 1000 using
the given digits.
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The Product Rule
The Product Rule If an operation can be
performed in m ways, and if corresponding to each
of these m ways of performing this operation,
there are n ways of performing a second
operation, then the number of ways of performing
the two operations together is m x n. Question
There are 10 buses running between 2 towns X and
Y. In how many ways can a man go from X to Y
using a specific bus and return by a different
bus? Solution A man can go from X to Y in 10
ways and return in (10 1) ways 9 ways. Thus,
number of ways of doing the journey 10 x 9 90.
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Formula for Permutations
Question What is the number of ways in which 5
persons can occupy 3 vacant seats? Solution
Assume that any one of the 5 persons can sit in
Seat 1. That would imply that the choice for
filling Seat 1 can be done in 5 ways. We now have
4 persons left that can sit in Seat 2. That
implies that the choice for filling Seat 2 can be
done in 4 ways. With 3 persons left after Seat 1
2 are filled, we can fill Seat 3 in 3
ways. Applying product rule, no of ways of
filling 3 seats with 5 persons 5 x 4 x 3
60. Formula for Permutations No of ways of
choosing and arranging r elements from a total
of n given elements is nPr or P(n, r)
n! (n r)!
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5 persons
3 persons
4 persons
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Restricted Permutations Illustration 1
Question In how many ways can a shelf for 4
books be formed out of 10 books such that a
particular book is always (i) included (ii) left
out? Solution (i) If a book is always included,
then it can come in the first, second, third or
fourth positions (i.e.) its position can be
selected in 4 ways. The other 3 books in the
shelf can be selected from the remaining 9 books
in P(9,3) ways Total Number of ways P(9,3) x
4 9! / 6! x 4 9 x 8 x 7 x 4 2016 (ii) If a
book is always excluded, effectively we only have
9 books to fill in 4 positions. Total Number of
ways P(9,4) 9! / 5! 9 x 8 x 7 x 6 3024
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Restricted Permutations Types and Formulae
Restricted Permutations Illustration 2
Question In how many ways can 6 boys and 4
girls be arranged in a straight line such that no
two girls are ever together? Solution First,
fixing the positions of boys, no. of
permutations 6! Since two girls cannot sit
together, a girl can sit either between two boys
or at the ends resulting in 7 possible seats as
shown below Fixing the positions of girls
P(7,4) Thus, no. of ways of arranging 6 boys and
4 girls such that no two girls are together
6! x P(7,4) 7x 6 x 5 x 4 x 6 x 5 x 4 x 3 x 2 x
1 604800.
G
B2
B1
G
G
B3
G
B4
G
B5
G
B6
G
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Restricted Permutations Illustration 3
Question How many 5-digit numbers can be formed
out of the digits 0, 1, 2,.., 9 if each number
starts with 35 and no digit is repeated? Solution
Since the first two positions are defined and
no digit is to be repeated, the remaining 3
positions have to be filled with digits
0,1,2,4,6,7,8,9, i.e., 8 digits. Number of
ways of forming required number P(8,3) 8! /
5! 8 x 7 x 6 336.
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5
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8 digits
6 digits
7 digits
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Permutations of Alike Things
The number of permutations (x) of n things taken
all at a time where p things are alike of one
kind, q things are alike of a second kind, r
things are alike of a third king and so on,
is, x n! p! q! r! Question
There are 3 copies each of 4 different books.
Find the number of ways of arranging them on a
shelf. Solution Total no. of books 3 x 4
12 Required no. of ways of arranging them
12! 369600 3! 3! 3! 3!
Question Find the number of arrangements of
the letters of the words BANANA such that the 2
Ns do not appear together. Solution There are
3 As and 2 Ns. Total no. of ways of arranging
the letters 6! / (3! 2!) 60 No. of
arrangements in which Ns appear together 5! /
3! 20. We assume that the two Ns are combined
to form a single character Thus, no. of
arrangements in which the 2 Ns do not appear
together 60 20 40.
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Permutations of Repeated Things
The number of permutations of n different things
taken r at a time, when each thing may occur any
number of times is nr . Question 8 different
letters of the alphabet are given. Words of 4
letters are formed. Find the no. of such words
with at least one letter repeated. Solution
If any letter can be used any no. of times, no.
of letters that can be formed 84 4096. No. of
words with no letter repeated P(8,4) 8 x 7 x 6
x 5 1680. No. of words with at least 1 letter
repeated 4096 1680 2416. Question How many
3-digit numbers can be formed with the digits
1,2,3,4,5 when the digits may be
repeated? Solution Required no. of 3-digit
numbers that can be formed 53 125.
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8 letters
8 letters
8 letters
5 letters
8 letters
8 letters
7 letters
6 letters
When letters cannot be repeated
When letters can be repeated
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Circular Permutations

• Consider A, B and C to be arranged in a circular
  fashion.
• 3 linear arrangements ABC, BCA and CAB are result
  in the same circular arrangement
• No of circular arrangements with n elements No
  of linear arrangements with n elements/ n
• n! / n (n 1)!
• The number of ways in which n objects taken r at
  a time can form a ring nPr / r
• If clock-wise and counter clock-wise arrangements
  are equivalent, divide the number of ways by 2
• Question In how many ways can 10 boys and 5
  girls sit around a circular table such that no 2
  girls sit together?
• Solution 10 boys can be seated around a table in
  9! ways.
• There are 10 spaces between the boys which can be
  filled up by the 5 girls in P(10,5) ways.
• Thus, total no. of ways of arranging the boys and
  girls 9! x 10P5 (9! 10! ) 5!
• Question Find the no. of ways in which 10
  different flowers can form a garland so that 4
  particular flowers are never separated.

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