Download Combinations & Permutations

Permutations
and
Combinations
Mathematics and
Statistics
Permuntation & Combination
1
• In this section, techniques will be
introduced for counting
•
the unordered selections of
distinct objects and
•
the ordered arrangements of
objects
• of a finite set.
Mathematics and
Statistics
Permuntation & Combination
2
6.2.1 Arrangements
• The number of ways of arranging n
unlike objects in a line is n !.
• Note: n ! = n (n-1) (n-2) ···3 x 2 x 1
Mathematics and
Statistics
Permuntation & Combination
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Example
• It is known that the password on a
computer system contain
• the three letters A, B and C
• followed by the six digits 1, 2, 3, 4, 5,
6.
• Find the number of possible
passwords.
Mathematics and
Statistics
Permuntation & Combination
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Solution
• There are 3! ways of arranging the
letters A, B and C, and
• 6! ways of arranging the digits 1, 2, 3,
4, 5, 6.
• Therefore the total number of
possible passwords is
• 3! x 6! = 4320.
• i.e. 4320 different passwords can be
formed.
Mathematics and
Statistics
Permuntation & Combination
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Like Objects
• The number of ways of arranging in a
line
• n objects,
• of which p are alike, is
n!
p!
Mathematics and
Statistics
Permuntation & Combination
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The result can be extended
as follows:
• The number of ways of arranging in a
line n objects
• of which p of one type are alike,
• q of a second type are alike,
• r of a third type are alike, and so on,
is
n!
p!q!r!  
Mathematics and
Statistics
Permuntation & Combination
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Example
• Find the number of ways that the
letters of the word
•
STATISTICS
• can be arranged.
Mathematics and
Statistics
Permuntation & Combination
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Solution
•
•
•
•
The word STATISTICS contains
10 letters, in which
S occurs 3 times,
T occurs 3 times and
• I occurs twice.
Mathematics and
Statistics
Permuntation & Combination
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• Therefore the number of ways is
10!
 50400
3!3!2!
• That is, there are 50400 ways of
arranging the letter in the word
STATISTICS.
Mathematics and
Statistics
Permuntation & Combination
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Example
• A six-digit number is formed from
the digits
• 1, 1, 2, 2, 2, 5 and
• repetitions are not allowed.
• How many these six-digit numbers
are divisible by 5?
Mathematics and
Statistics
Permuntation & Combination
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Solution
• If the number is divisible by 5 then it
must end with the digit
• 5.
• Therefore the number of these sixdigit numbers which are divisible by 5
is equal to the number of ways of
arranging the digits
• 1, 1, 2, 2, 2.
Mathematics and
Statistics
Permuntation & Combination
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• Then, the required number is
5!
 10
2!3!
• That is, there are 10 of these sixdigit numbers are divisible by 5.
Mathematics and
Statistics
Permuntation & Combination
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6.2.2 Permutations
• A permutation of a set of distinct
objects is an ordered arrangement of
these objects.
• An ordered arrangement of r
elements of a set is called an rpermutation.
• The number of r-permutations of a
set with n distinct elements,
Mathematics and
Statistics
Permuntation & Combination
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• i.e. the number of permutations of r
objects taken from n unlike objects
is:
n!
 nn  1n  2      n  r  1
n Pr 
n  r !
• Note: 0! is defined to 1, so
n!
n!
  n!
n Pr 
n  n ! 0!
Mathematics and
Statistics
Permuntation & Combination
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Example
Find the number of ways of placing
3 of the letters A, B, C, D, E
in 3 empty spaces.
Mathematics and
Statistics
Permuntation & Combination
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Solution
•
•
•
•
•
•
The first space can be filled in
5 ways.
The second space can be filled in
4 ways.
The third space can be filled in
3 ways.
Mathematics and
Statistics
Permuntation & Combination
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• Therefore there are
• 5 x 4 x 3 ways
• of arranging 3 letters taken from 5
letters.
• This is the number of permutations
of 3 objects taken from 5 and
• it is written as 5P3
• so 5P3 = 5 x 4 x 3 = 60.
Mathematics and
Statistics
Permuntation & Combination
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• On the other hand, 5 x 4 x 3 could be
written as
5  4  3  2 1 5!
5!
 
 P5, 3
2 1
2! 5  3 !
• Notice that the order in which the
letters are arranged is important --• ABC is a different permutation from
ACB.
Mathematics and
Statistics
Permuntation & Combination
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Example
• How many different ways are there
to select
• one chairman and
• one vice chairman
• from a class of 20 students.
Mathematics and
Statistics
Permuntation & Combination
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Solution
• The answer is given by the number of
2-permutations of a set with 20
elements.
• This is
•
Mathematics and
Statistics
20P2
= 20 x 19 = 380
Permuntation & Combination
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6.2.3 Combinations
• An r-combination of elements of a set
is an unordered selection of r
elements from the set.
• Thus, an r-combination is simply a
subset of the set with r elements.
Mathematics and
Statistics
Permuntation & Combination
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• The number of r-combinations of a
set with n elements,
• where n is a positive integer and
• r is an integer with 0 <= r <= n,
• i.e. the number of combinations of r
objects from n unlike objects is
n!
n Cr 
r! n  r !
Mathematics and
Statistics
Permuntation & Combination
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Example
• How many different ways are there
to select two class representatives
from a class of 20 students?
Mathematics and
Statistics
Permuntation & Combination
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Solution
• The answer is given by the number of
2-combinations of a set with 20
elements.
• The number of such combinations is
20!
 190
20C2 
2! 18!
Mathematics and
Statistics
Permuntation & Combination
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Example
• A committee of 5 members is chosen
at random from
• 6 faculty members of the
mathematics department and
• 8 faculty members of the computer
science department.
Mathematics and
Statistics
Permuntation & Combination
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• In how many ways can the committee
be chosen if
• (a) there are no restrictions;
• (b) there must be more faculty
members of the computer science
department than the faculty
members of the mathematics
department.
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Statistics
Permuntation & Combination
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Solution
• (a) There are 14 members, from whom
5 are chosen.
•
The order in which they are
chosen is not important.
•
So the number of ways of choosing
the committee is
•
14C5= 2002.
Mathematics and
Statistics
Permuntation & Combination
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• (b)If there are to be more
•
faculty members of the computer
science department than
•
the faculty members of the
mathematics department,
•
then the following conditions must
be fulfilled.
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Statistics
Permuntation & Combination
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• (i) 5 faculty members of the
computer science department.
•
The number of ways of choosing is
•
8C5= 56.
• (ii) 4 faculty members of the
computer science department and
•
1 faculty member of the
mathematics department
Mathematics and
Statistics
Permuntation & Combination
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•
•
The number of ways of choosing is
8C4 x 6C1 = 70 x 6 = 420.
• (iii) 3 faculty members of the
computer science department and
•
2 faculty members of the
mathematics department
•
The number of ways of choosing is
•
8C3 x 6C2 = 56 x 15 = 840
Mathematics and
Statistics
Permuntation & Combination
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• Therefore the total number of ways
of choosing the committee is
• 56 + 420 + 840 = 1316.
Mathematics and
Statistics
Permuntation & Combination
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