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Unformatted text preview: Discrete Mathematics
Dr. Fred Phelps
Lecture 1 Counting and
Set Theory Announcements
• • • Please register for discrete math on the
site http://dl.iitu.kz. The main textbook is
on that site.
The syllabus is also on that site.
And a copy of the main textbook:
“Discrete Mathematics” by Lovasz,
Pelikan and Veztergombi. LPV 1.1 Counting Problems
from a Birthday Party
• • • Alice (♀) has a birthday party.
Invites Bob (♂), Carl (♂), Diane (♀), Eve
(♀), Frank (♂), George (♂)
Total 7 people – four boys, three girls LPV 1.1 Counting Problems from a
Birthday Party
Problem 1  Handshakes
Everyone shakes everyone else’s hand.
How many handshakes does this make?
7 people each shake 6 hands so 7·6=42.
But, each handshake was double
counted
7 ×6 2 = 21. LPV 1.1 Counting Problems from a
Birthday Party
Problem 2 – Seating Arrangements
How many ways can the guests be seated at the seven
chairs at the table if Alice (the birthday girl) stays at the
head of the table?
Alice’s chair is fixed – only one possibility.
Any one of six people (anyone besides Alice) can sit in
the next (2nd) chair – six possibilities.
Any one of five (anyone besides Alice or the occupant
of the first chair) can sit in the third chair – five
possibilities. LPV 1.1 Counting Problems from a
Birthday Party
Problem 2 – Seating Arrangements
And so the total number of possible
seating arrangements is:
1·6·5·4·3·2·1=6!=720. LPV 1.1 Counting Problems from a
Birthday Party
Problem 3 – Dance Pairings
How many different sets of dancing pairs (boys
dance with girls only) can be formed from the 4
boys and 3 girls?
The textbook says this is easy.
But the textbook’s solution is wrong!
Notice that one boy will not have a partner. So
lets start with the girls. LPV 1.1 Counting Problems from a
Birthday Party
Problem 3 – Dance Pairings
1st girl has four choices of partner
2nd girl has three choices
3rd girl has two choices
So the number of possible dance pairs =
4x3x2=24. LPV 1.1 Counting Problems from a
Birthday Party
Problem 3 – Dance Pairings
Now let’s try to solve this starting with the
boys.
First we choose which three of the four boys
will dance. One must sit out. There are 4 ways
to do this:
Bob, Carl, Frank, but not George
Bob, Carl, George, but not Frank
Bob, Frank, George, but not Carl
Carl, Frank, George, but not Bob LPV 1.1 Counting Problems from a
Birthday Party
Problem 3 – Dance Pairings
This is called the number of combinations of 4
things taken 3 at a time. We will see the general
formula for combinations later.
Next we order the boys (by age or height or
alphabetical order of their names). Thus we can
uniquely refer to “the first boy” without
knowing who he is.
1st boy has three choices of partner
2nd boy has two choices
3rd boy is stuck with the last girl LPV 1.1 Counting Problems from a
Birthday Party
Problem 3 – Dance Pairings
So, for every combination of three boys, there
are 3·2·1=6 dancing pairs.
Given the four choices for which boys dance,
there are 4·6=24 different ways for the friends
to “pair off” to dance.
Its good that we get the same answer either
way! LPV 1.1 Counting Problems from a
Birthday Party
Problem 4 – Lottery Tickets
A lottery ticket consists of the first ninety
positive integers {1, 2, 3, …, 88, 89, 90}. To play
the lottery, you chose and mark five of the
numbers. How many different ways to mark the
lottery ticket are there?
There are 90 choices for the first number
There are 89 choices for the second number
And so on
89
87 86
And so the90 × ×88 × × =of 273,912,160
total number 5, ways to mark the
ticket is LPV 1.1 Counting Problems from a
Birthday Party
Problem 4 – Lottery Tickets
But winning does not depend on the order you mark
the ticket, (i.e. if you choose {2}, {7}, {49}, {32}, {14} it is
the same as if you choose {2}, {49}, {14}, {7}, {32}).
Each set of five numbers can be chosen in 5·4·3·2·1=5!
orders.
So there9are × × ×
0 × 88 87 86
89
5 ×4 × ×2 ×
31 5, 273,912,160
=
= 43,949, 268
120 ways to mark the lottery card. In order to win, you need LPV 1.1 Counting Problems from a
Birthday Party
Problem 4 – Lottery Tickets
Note – the number of ways to mark the lottery ticket is
the number of ways to choose 5 numbers out of a set
of 90 or the number of combinations of 90 things taken
5 at a time.
90!
.
(90 − 5)!×
5! You can see that the answer is LPV 1.1 Counting Problems from a
Birthday Party
Problem 5 – Bridge Hands
How many different bridge (i. e. the
card game) hands are there? (A bridge
hand is 13 cards chosen from a deck of
52).
The first card is any one of 52
•
The second is any one of 51
•
etc.
• So there are 52 × × × × ways to receive 13 cards
51 50 ... 40
But order does not matter. So we need to divide by 13! LPV 1.1 Counting Problems from a
Birthday Party
Problem 5 – Bridge Hands
So the number of bridge hands is
52 × ×
51 ...40
52!
=
= 635, 013,559, 600.
13!
39!×
13! This is the number of combinations of 52 things taken 13 at a
time. LPV 1.1 Counting Problems from a Birthday
Party
Problem 6 – Chess Pairings
How many different pairs can play chess? (Alice will only
watch.)
Six people can be seated in six chairs (at three tables) in
6! = 720 ways.
Each set of three pairs can choose their tables 3x2=6
different ways:
White Table
Black
White Table
Black
White
Table
Black
1 2 1 2 3 4 3 4 5 6 1 2 5 6 3 4 5 6 3 4 5 6 5 6 5 6 1 2 3 4 1 2 3 4 1 2 LPV 1.1 Counting Problems from a Birthday
Party
Problem 6 – Chess Pairings With both the pairs and the tables fixed there are still
23=8 seating arrangements.
White Table Black White Table Black White Table Black White Table Black 1 2 1 2 1 2 1 2 3 4 3 4 4 3 4 3 5 6 6 5 5 6 6 5 2 1 2 1 2 1 2 1 3 4 3 4 4 3 4 3 5 6 6 5 5 6 6 5 So each set of three pairs appears 6x8=48 times in the
720 possible seating arrangements and the number of
pairs is
6!/ (6 × = 15.
8) LPV 1.1 Counting Problems from a
Birthday Party
Problem 6 – Chess Pairings
An easier way to do the previous problem:
Bob is youngest. Let him choose his partner first. He has
five choices.
Whoever is next (the youngest among the rest) will have
three choices.
The last pair will be determined because there are only
two nonselected people left.
This gives 5x3x1=15 pairs. LPV 1.2 Sets
A set is a collection of things (elements). If A is a set and b is an element of A, we write b ∈ A.
Set Elements deck of cards each card is an element guests at the
party
real numbers Alice, Bob, Carl, Diane, Eve, Frank, P
George
ℝ
number for every point on the
number line
ℤ
…, 2, 1, 0, 1, 2, 3,… integers
nonnegative
integers
empty set Symbo
l
D 0, 1, 2, 3, … Z+ none ∅ LPV 1.2 Sets
The cardinality of a set A is the number of
its elements and is denoted by A.
P = 7 Ø = 0 ¢= ∞ Ways to define a set:
We can define a set by listing elements between
braces:
P = {Alice, Bob, Carl, Diane, Eve, Frank, George}
2.
We can use a verbal description:
{the boys at Alice’s party}
3.
Use properties inside brackets after a colon:
1. ¢ + = { x ∈ ¢ : x ≥ 0} LPV 1.2 Subsets
A set A is called a subset of set B if every
element of A is also an element of B.
We write this as: A ⊆ B. If two sets A and B have exactly the same
elements they are said to be equal.
We write this as:
A = B. Example of subsets
∅⊆¥ ⊆¢ + ⊆¢ ⊆¤ ⊆¡
• In plain English, the empty set is a subset
of the natural numbers (1, 2, 3, …) which
is a subset of the nonnegative integers (0,
1, 2, …), which is a subset of the integers
(…2, 1, 0, 1, 2…) which is a subset of
the rational numbers (all numbers which
can be written as a fraction of integers),
which is a subset of the real numbers. intersection of two sets
• • The intersection of two sets A and B is
the set consisting of those elements which
are common to both sets.
A
We write this as: I B. If A I B = ∅ then A and B are said to be "disjoint". union of two sets
• • The union of two sets A and B is defined
as the set consisting of elements which
are in at least one of A and B.
A
We write this as: U B. universe
• In most cases the sets naturally come
from some “universe” of possible sets. For
example, let S be the set of all sequences
of capital (i.e. big) English letters with
length 4 which do not have any A’s. universe complement
• • The complement of a
set G in universe U is
all of the members of U
which are not in G.
People write this as:
C
G or G′ or G . difference of A and B A− B B− A Notice that A − B ∩ B − A = ∅. notation Examples of notation:
A1 ≠ A2 means that ∃a ∋: a ∈ A1 ∪ A2 and a ∉ A1 I A2 . So, ∃i ∈ {1, 2} ∋: a ∉ Ai . Symmetric Difference:
• • • The symmetric difference of sets A and B
is the set of all elements of A or B which
are not in both A and B.
We write this as A Δ B.
A Δ B = {x(x ∈ A ∧ x ∉ B) ∨ (x ∉ A ∧ x ∈
B)} DeMorgan’s Laws Venn Diagram “proof” of
DeMorgan Law Proof of 2nd DeMorgan Law Goodbye! ?? ...
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This note was uploaded on 10/15/2011 for the course MATHS 100 taught by Professor Fredphelps during the Spring '11 term at Jordan University of Science & Tech.
 Spring '11
 FredPhelps
 Math, Set Theory, Counting

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