Basic Concepts

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King ABDUL AZIZ UniversityFaculty Of Computing
and Information Technology
CPCS 222 Discrete Structures I Counting
Dr. Eng. Farag Elnagahy farahelnagahy_at_hotmail.com
Office Phone 67967
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The Basics of counting

• Combinatorics is the mathematics of counting and
  arranging objects.
• Counting of objects with certain properties
  (enumeration) is required to solve many different
  types of problems. For example, counting is used
  to
• Determine number of ordered or unordered
• arrangement of objects.
• Generate all the arrangements of a specified
  kind
• which is important in computer simulations.
• Compute probabilities of events.
• Analyze the chance of winning games, lotteries
  etc.
• Determine the complexity of algorithms.

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The Basics of counting

• Two basic counting principles
• The Sum Rule
• The Product Rule
• Let us consider two tasks
• m is the number of ways to do task1
• n is the number of ways to do task2
• Performing task1 does not accomplish task2 and
  vice versa (task1 and task2 are independent of
  each other).
• Sum rule the number of ways that either task1
  or task2 can be done, but not both, is mn.
• Product rule the number of ways that both
  task1 and task2 can be done in mn.

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The Basics of counting (Examples)
Example A student can choose a computer project
from one of three lists. The three lists contain
23, 15, and 19 possible projects respectively.
How many possible projects are there to choose
from? (23151957) Example The chairs of an
auditorium are to be labeled with a letter and a
positive integer not to exceed 100. What is the
largest number of chairs that can be labeled
differently? (26x1002600)
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The Basics of counting (Set version)

• If A is the set of ways to do task1, and B the
  set of ways to do task2, and if A and B are
  disjoint, then
• The ways to do either task1 or task2 are
• A?B, and A?BAB
• The ways to do both task1 and task2 are
• A?B, and A?BAB
• The number of different subsets of a finite
  set(s) is ?
• 000000000000
• 100000000000
• 010000000000
  2s
• .
• 111111111111

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The Basics of counting (Examples)
What is the value of k after the following code
has been executed?
K0 for i11 to n1 K K 1 for i21 to
n2 K K 1 for i31 to n3 K K 1 K n1 n2
n3
K0 for i11 to n1 for i21 to n2 for
i31 to n3 K K 1 K n1x n2x n3
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The Basics of counting (Examples)
Count the number of print statements in this
algorithm The total number of print
statements executed is n (nn) 2n2.
for i 1 to n begin for j 1 to n
print hello for k 1 to n
print hello end
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The Basics of counting (Examples)
Count the number of print statements in this
algorithm for each i, the number of
print statements executed is i in the j loop plus
n-i in the k loop. Therefore, for each i, the
number of print statements is i (n-i)
n. Therefore the total number of print statements
executed is n n n2.
for i 1 to n begin for j 1 to i
print hello for k i 1 to n
print hello end
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The Basics of counting (Examples)

• In a computer language
• The name of a variable is a string of one or two
  alphanumeric characters.
• uppercase and lowercase letters are not
  distinguished.
• 26 English letter , 10 digits.
• the variable name must begin with letter.
• there are five strings of two characters that
  are reserved for programming use.
• How many different variable names are there ?
• V1 string (one character) ,V2 string (two
  characters)
• VV1V2
• V126
• V226x(2610) 5 26x36 5 931
• V26931957

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The Basics of counting (Examples)
How many different license plates are available
if each plate contains a sequence of three
letters followed by three digits? L1
L2 L3 D1 D2 D3 Each of the three letters can be
written in 26 different ways, and each of the
three digits can be written in 10 different
ways. Hence, by the product rule, there is a
total of 26 ? 26 ? 26 ? 10 ? 10 ? 10
17,576,000 different license plates possible.
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The Basics of counting (Examples)
Each user on a computer system has a password,
which is six to eight characters long, where each
character is an uppercase letter or digit. Each
password must contain at least one digit. How
many passwords are there? (string includes
Letters Digits)- (string with no
digits) PP6P7P8 P6366-2661 867 866
560 P7367-26770 332 353 920 P8368-2682 612
282 842 880 PP6P7P82 684 483 063 360
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The Basics of counting (Examples)

• In version 4 of the Internet Protocol (IPv4)
• The internet address is a string of 32 bits as
  follows
• Class A address 0 -netid(7 bits) hostid (24
  bits)
• Class B address 10 -netid(14 bits) hostid (16
  bits)
• Class C address 110 -netid(21 bits) hostid (8
  bits)
• Class D address 1110 multicast address (28
  bits)
• Class E address 11110 - address (27 bits)
• Where, Network number(netid) -host
  number(hostid)
• Restrictions
• 1111111 is unavailable in netid
• All 0s and all 1s are unavailable in hostid
• The computer on the Internet has either class A
  or B or C addresses. How many different IPv4
  addresses are available for computers on the
  Internet?
• (27-1)(224-2)(214-1)(216-2)(221-1)(28-2)

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The Basics of counting (Examples)
How many bit string of length eight either start
with a 1 bit or end with the two bits 00?
1 - - - - - - - 27128 ways
- - - - - - 0 0 2664 ways
1 - - - - - 0 0 2532 ways
128 64- 32 160
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The Basics of counting (Examples)
A computer company receives 350 applications from
computer graduates for a job. Suppose that 220 of
these people majored in CS, 147 majored in
business, and 51 majored both in CS and in
business. How many of these applicants majored
neither in CS nor in business? Let A1 be the set
of students who majored in CS Let A2 be the set
of students who majored in business The number
of students who majored either in CS or in
business (or both) is A1?A2A1A2-A1?A222
0147-51316 The number of applicants who majored
neither in CS nor in business is 350-31634
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The Basics of counting (Examples)
Exercises PP. 344-347 1-12
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The Pigeonhole principle
The Pigeonhole principle Suppose there are n
pigeons, k pigeonholes, and ngtk. If these n
pigeons fly into these k pigeonholes, then some
pigeonhole must contain at least two pigeons.
If k1 objects are assigned to k places, then
at least 1 place must be assigned 2 objects.
7 pigeons 6 pigeonholes
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The Pigeonhole principle

• In terms of the assignment function
• If fA?B and AB1, then some element of B
• has 2 pre-images under f.( f is not
  one-to-one)
• How many students must be in class to guarantee
  that at least two students receive the same score
  on the final exam, if the exam is graded on a
  scale from 0 to 100 points? Greater than 101

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The Pigeonhole principle (Examples)

• The generalized Pigeonhole principle
• If N objects are placed into k boxes, then
  there is at least one box containing at least
  ?N/K? objects.
• e.g., there are N280 students in this class.
  There are k52 weeks in the year.
• Therefore, there must be at least 1 week during
  which at least ?280/52? ?5.38?6 students in the
  class have a birthday.

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The Pigeonhole principle (Examples)

• There are 280 students in the class. Without
  knowing anybodys birthday, what is the largest
  value of n for which we can prove that at least n
  students must have been born in the same month?
• ?280/12? ?23.3? 24
• What is the minimum number of students required
  in a discrete math class to be sure that at least
  six will receive the same grade, if there are
  five possible grades, A, B, C, D, and F?

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Permutations and Combinations

• Permutations
• In how many ways can we select 3 students from a
  group of 5 students to stand in line for a
  picture?
• In how many ways can we arrange all 5 of these
  students in a line for a picture?
• Note that the order in which we select the
  students matters.
• 5 ways to select the first student
• 4 ways to select the second student
• 3 ways to select the third student
• 2 ways to select the fourth student
• 1 way to select the fifth student
• A. 5.4.360
• B. 5.4.3.2.1120

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Permutations and Combinations

• Permutations
• A permutation of a set S of objects is an ordered
  arrangement of the elements of S where each
  element appears only once
• e.g., 1 2 3, 2 1 3, 3 1 2
• An ordered arrangement of r distinct elements of
  S is called an r-permutation.
• The number of r-permutations of a set S with
  nS elements is
• P(n,r) n(n-1)(n-2) (n-r1) n!/(n-r)!

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Permutations and Combinations

• Permutations (examples)
• How many ways are there to select a first-prize
  winner, a second-prize winner,and a third-prize
  winner from 100 different people who have entered
  a contest?
• P(100,3)100.99.98970200
• How many permutations of the letters ABCDEFGH
  contain the string ABC ?
• ABC, D, E, F, G, H we have 6 objects
• Theses object can occur in any order
• 6.5.4.3.2.1
• There are 6!720 permutations

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Permutations and Combinations

• Combinations
• The number of ways of choosing r elements from S
  (order does not matter).
• S1,2,3
• e.g., 1 2 , 1 3, 2
• The number of r-combinations C(n,r) of a set with
  nS elements is

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Permutations and Combinations
Example S1,2,3, all permutations(1,2,3),(2,1,
3),(1,3,2),(2,3,1),(3,1,2),(3,2,1) all
2-permutations(1,2),(2,1),(1,3),(3,1),(2,3),(3,2
) P(3,3)3216, P(3,2)326 S1,2,3, all
2-combinations1,2,1,3,2,3 Comparing to
all 2-permutations, we see we ignore order,
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Permutations and Combinations

• Example
• How many ways are there to choose a committee of
  size five consisting of three women and two men
  from a group of ten women and seven men?
• The number of ways to choose three women is C(10,
  3)
• The number of ways to choose two men is C(7, 2).
• Using the product rule to choose three women and
  two men, the answer is
• C(10, 3) C(7, 2) 2, 520.

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Permutations and Combinations
Example A class has 20 women and 16 men. In how
many ways can you (a) Put all the students in a
row? (b) Put 7 of the students in a row? (c) Put
all the students in a row if all the women are on
the left and all the men are on the
right? Solution (a) There are 36 students. They
can be put in a row in 36! ways. (b) You need to
have an ordered arrangement of 7 out of 36
students. The number of such arrangements is
P(36, 7). (c) You need to have an ordered
arrangement of all 20 women AND and ordered
arrangement of all 16 men. By the product rule,
this can be done in 20!16! ways.
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Permutations and Combinations
Exercises PP. 360-362 1 4 5 7