Counting Principles and Pigeonhole Principle

LECTURE 15
Chapter 5.1
Counting
Prepared by Khairun Nahar,Assistant Professor,
Department of CSE, Comilla University

Basic Counting Principles
We will present two basic counting principles, the product rule
and the sum rule. Then we will show how they can be used to
solve many different counting problems.
THE PRODUCT RULE: Suppose that a procedure can be
broken down into a sequence of two tasks. If there are 𝑛1 ways
to do the first task and for each of these ways of doing the first
task, there are 𝑛2 ways to do the second task, then there are
𝑛1𝑛2 ways to do the procedure.
The product rule applies when a procedure is made up of
separate tasks.

Example-1
Example-2

Example-3
Example-4

Example-5
Example-6
How many functions are there from a set with m elements to a set with n elements?

Example-7

N=5. n-k+1 way=5-3+1=3
M=n
A1->n1
A2->n2
A3->n3
A4->n4
A4->n5

THE SUM RULE If a task can be done either in one of 𝑛1 ways or in
one of 𝑛2 ways, where none of the set of 𝑛1 ways is the same as any of
the set of 𝑛2 ways, then there are 𝑛1 + 𝑛2 ways to do the task.
Example 11: Suppose that either a member of the mathematics faculty
or a student who is a mathematics major is chosen as a representative
to a university committee. How many different choices are there for
this representative if there are 37members of the mathematics faculty
and 83 mathematics majors and no one is both a faculty member and a
student?
Solution: There are 37 ways to choose a member of the mathematics
faculty and there are 83 ways to choose a student who is a mathematics
major. Choosing a member of the mathematics faculty is never the
same as choosing a student who is a mathematics major because no one
is both a faculty member and a student. By the sum rule it follows that
there are 37 +83 = 120 possible ways to pick this representative.

EXAMPLE 14
In a version of the computer language BASIC, the name of a variable is a string of
one or two alphanumeric characters, where uppercase and lowercase letters are not
distinguished. (An alphanumeric character is either one of the 26 English letters or
one of the 10 digits.) Moreover, a variable name must begin with a letter and must
be different from the five strings of two characters that are reserved for
programming use. How many different variable names are there in this version of
BASIC?
Solution: Let V equal the number of different variable names in this
version of BASIC. Let V1 be the number of these that are one character long
and V2 be the number of these that are two characters long. Then by the sum
rule, V = V1 + V2. Note that V1 = 26,because a one-character variable name
must be a letter. Furthermore, by the product rule there are 26 .36 strings of
length two that begin with a letter and end with an alphanumeric character.
However, five of these are excluded, so V2 = 26 . 36 - 5 = 931. Hence, there
are V = V1 + V2 = 26 +931 = 957 different names for variables in this version
of BASIC.

Assignment
1. There are 18 mathematics majors and 325 computer science majors at a
college.
a) How many ways are there to pick two representatives so that one is a
mathematics major and the other is a computer science major?
b) How many ways are there to pick one representative who is either a
mathematics major or a computer science major?
2. An office building contains 27 floors and has 37 offices on each floor.
How many offices are in the building?
3. A multiple-choice test contains 10 questions. There are four possible
answers for each question.
a) How many ways can a student answer the questions on the test if the
student answers every question?
b) How many ways can a student answer the questions on the test if the
student can leave answers blank?

4. How many bit strings of length n , where n is a positive
integer, start and end with 1s?
5. How many strings of eight English letters are there
a) that contain no vowels, if letters can be repeated?
b) that contain no vowels, if letters cannot be repeated?
c) that start with a vowel, if letters can be repeated?
d) that start with a vowel, if letters cannot be repeated?
e) that contain at least one vowel, if letters can be repeated?
f) that contain exactly one vowel, if letters can be repeated?
g) that start with X and contain at least one vowel, if
letters can be repeated?
h) that start and end with X and contain at least one vowel, if
letters can be repeated?

a) How many strings of eight English letters are there that
contain no vowels, if letters can be repeated?
=>There are 5 vowels . Then any of 21 letters can appear in any
of the 8 positions, so the answer is 218=37822859361
b) How many strings of eight English letters are there that
contain no vowels, if letters cannot be repeated?
=>The first can be any of 21 letters, the next can’t be the same as
the first, so there are 20 possible letters, the third can be any of
19, etc.
Therefore, the answer is 21×20×19×18×17×16×15×14
=8204716800
that is 21!/13! =8204716800
Solution of Problem (5)

c) How many strings of eight English letters are there that start
with a vowel, if letters can be repeated?
=> Since there are 5 possibilities for the first letter and 26 for
each of the rest. That’s 5 · 267=40159050880
d) How many strings of eight English letters are there that start
with a vowel, if letters cannot be repeated?
=>There are 5 possibilities for the first letter, 25 for the second,
24 for the third, etc. That gives 5·25·24.23.22.21.20.19.18
=12113640000 or 5·25!/17!= 12113640000
e) How many strings of eight English letters are there that
contain at least one vowel, if letters can be repeated?
=>There are 268
strings in all, but only 218
that don’t have a
vowel. Therefore, there are 268
− 218
=171004205215 that have
at least one vowel.

f) How many strings of eight English letters are there that that
contain exactly one vowel, if letters can be repeated?
The vowel (any of 5 possibilities) can occur in any of 8 places,
and the other 7 places each can have any of 21 possible letters.
That gives 5 · 8 · 217
= 72043541640
g) How many strings of eight English letters are there that that
start with X and contain at least one vowel, if letters can be
repeated?
=>There are 267 strings that start with an X, and 217 that don’t
have any vowel, so the answer is 267 − 217=6230721635.
h) How many strings of eight English letters are there that that
start and end with X and contain at least one vowel, if letters
can be repeated?
=>There are 266 strings that start and end with an X, and 216 of
those that don’t have any vowel, so the answer is 266 −
216=223149655.

Chapter 5.2
The Pigeonhole Principle

THE PIGEONHOLE PRINCIPLE

THE PIGEONHOLE PRINCIPLE If k is a positive integer
and k+ 1 or more objects are placed into k boxes, then there is at
least one box containing two or more of the objects.

THEOREM 2
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.
Proof: We will use a proof by contradiction. Suppose that
none of the boxes contains more than ⌈
𝑁
𝐾
⌉- 1 objects. Then,
the total number of objects is at most
where the inequality ⌈
𝑁
𝐾
⌉ < (
𝑁
𝐾
)+ 1 has been used. This is a
contradiction because there are a total of N objects.
N=6,k=5=> 5x(ceil(1.2)-1)< 5x ((1.2+1)-1);
=> 5x(2-1)< 5.2 But N=6

Exercise-1: Show that in any set of six classes, each meeting
regularly once a week on a particular day of the week, there must be
two that meet on the same day, assuming that no classes are held on
weekends.
Solution: Here are six classes (pigeons), but only five weekdays
(pigeonholes). Therefore, by the pigeonhole principle, at least (⌈
6
5
⌉ )
= 2 classes must be held on the same day.
Exercise-2: Show that if there are 30 students in a class, then at least
two have last names that begin with the same letter.
Solution: We have 26 letters, and 30 students so using Pigeonhole
Principle we have (⌈
30
26
⌉) = 2.

Chapter 5.3
Permutations and Combinations

Permutations
A permutation of a set of distinct objects is an ordered
arrangement of these objects. We also are interested in ordered
arrangements of some of the elements of a set. The word
"permutation" also refers to the act or process of changing the
linear order of an ordered set. An ordered arrangement of r
elements of a set is called an r-permutation.
The number of r-permutations of a set with n elements is denoted
by 𝑷(𝒏 , 𝒓).
Example: Let S = {1 , 2, 3} . The ordered arrangement {3, 1, 2}
is a permutation of S. The ordered arrangement {3, 2} is a 2-
permutation of S.
3-permutation of S are: {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1},
{3, 1, 2}, and {3, 2, 1}.

Combinations
The combination is a way of selecting items from a collection,
such that (unlike permutations) the order of selection does not
matter. An 𝑟 −combination of elements of a set is an unordered
selection of r elements from the set. Thus, a 𝑟 −combination is
simply a subset of the set with r elements.
The number of r-combinations of a set with n distinct elements is
denoted by 𝐶(𝑛, 𝑟 ).
Note that 𝐶(𝑛 , 𝑟) is also denoted by (𝑛
𝑟
) and is called a binomial
coefficient.

Permutation and Combination
S1={a,b,c} S2={Saimon, Rahim, Karim} n=3
S1={a,b},{b,c},{c,a},{a,c},{b,a},{c,b}=6 way= permutation
S2={Saimon,Rahim} =3 way= combination
{Saimon, Karim}
{Rahim,Karim}
{Rahim, Saimon}
{Karim, Saimon}
{Karim, Rahim}

Relation between Permutation and Combination
P(n , r) = C (n, r)× r !
Permutation Formula Combination Formula

Counting Principles and Pigeonhole Principle

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