Transcript Formal Symbolic Logic - Missouri Western State University

Introduction to Counting
Discrete Structures
A Multiplication Principle
Example 3
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Suppose you're buying equipment for a
home office. You wish to purchase a
computer, a scanner, and a printer
("3 decisions to make").
If you have narrowed your choices to 3
models of computers, 4 scanners, and 2
printers, how many different overall
outcomes are possible?
The Count
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By the multiplication principle, the product
3
x
4
x
2
= 24
computer
scanner
printer
tells us 24 different systems are possible.
“ a branch of a tree”
Decision Tree: 24 branches
Total of
24
different
systems
computer
scanner
printer
Example 4
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Consider a license plate consisting of any 3 single digit
numbers followed by any 3 letters.
Examples of such license plates include
533 ATZ, 285 VCC, etc.
There are 6 decisions to make.
For each digit, we have the 10 choices 0, 1,..., 9
and for each letter, we have 26 choices a,b,...,z.
The Plates
The total number of different plates is given by
10 x 10 x 10 x 26 x 26 x 26
digit 1 digit 2 digit 3 letter 1 letter 2 letter 3
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A total of 17,576,000 different plates!
Example 5
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In a deli, suppose we may choose from 4 types
of bread, 6 types of meat, and 3 types of
cheese.
Consider the sandwiches which include one
type of meat plus one type of cheese. How
many such sandwiches are possible?
There are three decisions to make
(bread, meat, cheese)
The Sandwiches
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There are three decisions to make
4
bread
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x
6
meat 1
x
3
= 72
cheese
A total of 72 different sandwiches.
Cardinality of a Set
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The number of elements in a set A is
called the “cardinality” of A , denoted | A|.
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By the multiplication principle,
if | A| = n and | B| = m, then | A x B| = nm.
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By the multiplication principle, if | A| = n,
then
n
( A)  2  2  2
n times
22
No Overlap?
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In this special case, count the total
elements by counting each set
separately.
| A B |  | A|  | B |
In this case,
the sets are said
to be "disjoint".
Multiply and Add
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Consider the license plate example again. This
time allow either 3 digits followed by 3 letters or just
6 digits. These two sets of plates are disjoint (no
overlap).
A total of 18,576,000 license plates
Inclusion/Exclusion
and Combinations
Discrete Structures
How many elements?
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Recall our sets A = {2,4,5,8,10} and
B = { 2,3,5,7,8,9}.
Determine the cardinality, | A  B |
| A B |  | A|  | B |
Because the sets
are not disjoint.
Avoid Double Counting
| A B |  A  B  A B
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When adding the 5
elements from A with the
6 elements from B, the 3
elements which lie in
both A and B must be
counted only once.
Addition Principle
For any two sets A and B,
| A B |  A  B  A B
In particular, if A and B are disjoint sets,
then
| A B |  A  B
Playing Cards
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The 13 cards in each category or "suit"
include a 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack,
Queen, King, and an Ace. The Jack,
Queen, and King are called "face cards".
Aces or Spades
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In a deck of playing cards, how many of
the cards are Aces or Spades?
That's a total of 16 distinct cards.
Using Addition Rule
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By counting the set of Aces and set of
Spades individually and subtracting the
overlap…
| Aces  Spades |
 Aces  Spades  Aces  Spades
 4  13  1  16
Solving for other term
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Given any 3 of the values in the addition rule,
we may solve for the remaining unknown
value.
Find the number of elements in B, given
| A  B | 140, A  90, and A  B  35
Substitute in the known values…
| A B |  A  B  A B
140  90  B  35, and so B  85
Elements in B
The 85 elements of B includes the 35 in the
overlap with A. These regions contain the
total of 140 elements.
Example
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“In a survey of 1000 students, 700 indicate
they are enrolled in a math or english class.
Of these students, 400 are enrolled in a math
class and 650 are enrolled in an english
class. How may are enrolled in math and
english classes?”
| M  E | 700, M  400, and E  650
Find Intersection
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Using the addition principle, setup and solve
for the intersection.
|M E|  M + E  M E
700  400  650  M  E
M  E  350
M
50
350
300
E
Extended to 3 Sets?
| A B C |  A  B  C
 A B  AC  B C
 A B C
May generalize further
for any n sets.
Problem 8, page 203
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Among 150 students
83 own a car,
97 own a bicycle,
28 own a motorcycle,
53 own car and bicycle,
14 own car and motorcycle,
7 own bicycle and motorcycle,
2 own all three items
How many own only a bicycle?
How many don’t own any of these items?
Permutations and
Combinations
Discrete Structures
Arranging Letters
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When we consider 3 letters on a license plate,
the order in which the letters appear is
significant. That is, the sequence of letters
PHT is different than TPH, even though the
same letters are used.
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Using the “multiplication principle”,
there are 26 x 25 x 24 = 15600 ways
to pick and arrange 3 distinct letters
(not using the same letter twice).
“Permutations”
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When we wish to consider the many different
arrangements of the various choices, as with
letters on license plates, we use the term
"permutations".
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“Permutations of n objects, taken r at a time”
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Consider 26 letters, choose and arrange 3.
How many ways may this be done?
Permutation Formula
When repetitions are not allowed,
the number of ways to choose and arrange
any r objects chosen from a set of n available
objects is denoted Pn, r .
Using the Formula
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For the arrangements of 3 letters
recall by “multiplication principle”,
there are 26 x 25 x 24 = 15600 ways
Same as before!
A convenient notation, but why not
just use the multiplication principle?
Gold, Silver, Bronze
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Consider the top 3 winners in a race with 8
contestants. How many results are possible?
Or equivalently,
Calculate it
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Calculators have a built-in feature for these
computations (labeled as nPr ).
Use the MATH button and PRB submenu.
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To compute the value P8,3 = 336
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we simply enter:
8 nPr 3
Compare 2 Cases:
Consider a club with 16 members:
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Case 1:
If a president, VP, and
treasurer are elected, how
many outcomes are
possible?
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(a choice of 3 members,
order is not important)
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Since we don't count the
different arrangements,
this total should be less.
(select and arrange 3,
order is important)
16 x 15 x 14 = 3360
pres. VP treas.
Case 2:
If a group of 3 members is
chosen, how many groups
are possible ?
Case 2 “Combinations”
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We’re interested in the members of the group, not
all the possible arrangements.
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Think of the group as
one "choice of 3” from the 16 members.
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The number of different combinations is denoted
by C16, 3 .
“Combinations of 16 objects, taken 3 at a time”
Combinations Formula
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number of combinations of r items chosen
from n available choices, is given by
“Combinations of n objects, taken r at a time”
Often read as “n, choose r"
Calculate “16, choose 3”
simplifies as
Or we may calculate the value C16,3 = 560
by simply entering:
16 nCr 3
Our Comparison
Case 1:
Case 2:
Given one group of 3 members, such as Joe,
Bob, and Sue, 6 arrangements are possible:
( Joe, Bob, Sue), ( Joe, Sue, Bob), ( Bob, Joe, Sue)
( Bob, Sue, Joe), ( Sue, Joe, Bob), ( Sue, Bob, Joe)
Each group gets counted 6 times for permutations.
Divide by 6 to “remove this redundancy”.
Arranging Letters
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How many distinct ways can the letters
“MISSOURI” be arranged?
There are 8 letters, so there are
8! permutations. But not all distinct!
RUIS1 MS2 OI is equal to RUIS2 MS1 OI
so don’t count these twice.
Same consideration for the letter “I’s”.
8!, reduce by half, and reduce by half again.
Perhaps an easier way?
Rearranging Missouri
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How many distinct ways can the letters
“MISSOURI” be arranged?
8!
 10080
(2!)(2!)
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Alternate, choose 2 slots for the M’s,
choose 2 slots for the I’s,
then arrange the other 4 letters.
C8,2C6,2 (4!)  (28)(15)(24)  10080
Rearranging Mississippi
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How many distinct ways can the letters
“mississippi” be arranged?
11!
 34650
(4!)(4!)(2!)
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Alternate, choose 4 slots for the S’s,
4 slots for the I’s, 2 slots for the P’s,
then only one slot left to place the M.
 C  C  C  (1)  (330)(35)(3)(1)  34650
11,4
7,4
3,2
More Combinations
Discrete Structures
5 card hands?
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When a hand of cards is drawn from a deck,
which cards we receive is important, not the
arrangement.
How many different 5-card hands are possible ?
Consider all combinations of 5 cards, taken from the
52 cards. That is, “52, choose 5”.
Series of Choices
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Try combining our new counting formulas with our
previous counting principles.
How many 5-card hands include exactly 3 kings
and 2 aces?
Here we select kings and select aces. It’s a series
of decisions, so we apply the multiplication
principle:
?
choose 3 of
the 4 kings
x
?
choose 2 of
the 4 aces
= ???
3 Kings, 2 Aces
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“4 kings, choose 3”;
there are C4,3 = 4 possible outcomes.
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“4 aces, choose 2”;
there are C4,2 = 6 possible outcomes.
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C4, 3
choose 3 of
the 4 kings
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x
C4, 2
= (4)(6) = 24
choose 2 of
the 4 aces
There are 24 hands including 3 kings and 2 aces.
Committee
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Consider a group of 20 juniors and 25 seniors.
Question 1: How many ways can a committee of 4 of
these students be selected?
"45 available students, choose 4“
C45,4 = 148,995 possible committees.
Question 2: How many ways can a committee with 2
juniors and 2 seniors be selected?
Select 2 of the 20 juniors, select 2 of 25 seniors:
C20, 2 x C25, 2 = (190)(300) = 57,000 committtees.
Continued…
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Question 3:
How many ways can a committee of 4 students be
selected, including at least 3 seniors?
Adding the two cases yields a total of
(20)(2300) + 12650 = 58,650 committees.
Defective, or not?
Among a collection of 20 clocks, 5 are defective.
 Question 1: How many ways can 3 of the clocks
be selected?
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"20 available clocks, choose any 3“
C20,3 = 1140 possible selections.
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Question 2: How many ways can 3 of the clocks
be selected such that none are defective?
Select 3 from the 15 non-defective clocks
C15,3 = 455 selections don't involve defective clocks
Partition as Disjoint Sets
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All 5-card hands categorized as:
no face cards:
C40, 5 = 658,008
1 face card: C12,1 C40, 4 = 1,096,680
2 face cards: C12,2 C40, 3 = 652,080
3 face cards: C12,3 C40, 2 = 171,600
4 face cards: C12,4 C40, 1 = 19,800
all 5 face cards: C12,5 =
792
Total 5-card hands:
2,598,960
Throw out the rest
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Sometimes instead of counting the objects
we’re interested in…
…it’s easier to count “all objects” and
subtract out the objects we don’t want.
Consider a subset A of a “universal” set S
S
A
A
A  S  A
The LONG Way…
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To count the 5-card hands with at least one
face card…
hands with 1 face card + hands with 2 face cards
+ hands with 3 face cards + hands with 4 face cards
+ hands with 5 face cards
C12,1C40,4  C12,2C40,3  C12,3C40,2  C12,4C40,1  C12,5
 1,940,952 ?
The Easy Way!
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But, to count the 5-card hands with at least
one face card…
…we may instead count all 5-card hands and
subtract the hands with no face cards.
one or more
face cards
no face
cards
all 5-card hands
C52,5  C40,5
 2,598,960  658, 008
 1,940,952
Subtract the complement
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From a group of 15 juniors and 15 seniors…
…how many ways is it possible to choose a
team of 11 students including at least 2
seniors?
C30,11  C15,11  C15,1C15,10
two or more
seniors
0 or 1
senior
all 11-student teams
 54,627,300  1,365  45,045
 54,580,890