Transcript Chapter 11 and 12 Slides

Chapter 11:
Counting Methods
11.1
Counting by Systematic Listing
• One-part tasks: each item in the list is
identified by one feature
• Multi-part tasks: items in the list are
identified by more than one feature (ex a
playing card has a suit and a value)
11.1
Two-part tasks: Using Tables
• Example: Determine the number of twodigit numbers that can be made using 2,4
and 7 for each digit
• Example: Determine how many ways you
can pick two people from the group of
Andria, Bridget, Connor and Dylan.
Multi-part tasks with
Tree Diagrams
• #62: Determine the number of odd, nonrepeating three digit numbers that can be
written using digits from the set {0,1,2,3}
• Emma, Finn, Garrett and Hannah have
tickets for 4 reserved seats in a row at a
concert. In how many ways can they seat
themselves so that Emma and Garrett are
not beside each other?
11.1
Uniformity Criterion
• A multi-part task is said to
satisfy the uniformity
criterion if the number of
choices for any particular part
is the same no matter which
choices were selected for
previous parts
11.2
11.2
Fundamental Counting Principle
(FCP)
• For a task that consists of k separate parts
and satisfies the uniformity criterion
• Suppose there are n1 ways to do the first
task, n2 ways to do the second, etc
• Then total number of ways to complete the
task is
n1 × n2 × … × nk
Example
11.2
Using the digits 0,1,2,3,4,5,6,7,8,9, how
many 4 digit numbers are there such that
a) 0 isn’t used for first, repetition is okay
b) 0 isn’t used for first, repetition is not okay
c) Number must begin and end with odd
digit, no repetition
d) Number is odd and greater than 5000,
repetition is okay
e) Number is inclusively between 5001 and
8000, repetition is not okay
11.2
Factorials and Arrangements
• n! = n × (n-1) × (n-2) × … 2 × 1
• 0! = 1
• The total number of different
ways to arrange n distinct
objects is n!
11.3
Permutations
• P(n,r) represents the number of
arrangements (called permutations) of n
distinct things taken r at a time
P(n,r) = n! / (n – r)!
• Permutations are applied only when
repetitions are not allowed and order
matters.
Combinations
11.3
• C(n,r) represents the number of subsets
(or combinations) of n distinct things
taken r at a time
C(n,r) = P(n,r)/r! = n!/r!(n-r)!
• Combinations are applied only when
repetitions are not allowed and order
does NOT matter
11.3
Guidelines for Choosing
Counting Method
• If selected items can be repeated, use
FCP
• If selected items cannot be repeated and
order is important, use permutations
• If selected items cannot be repeated and
order isn’t important, use combinations
Pascal’s Triangle
C(n,r) gives the number in the rth spot
of of the nth row
11.4
11.4
Binomial Theorem
• For any positive integer n,
(x + y)n = C(n,0)xn + C(n,1)xn-1y
+ C(n,2)xn-2y2 + … +
C(n,n-1)xyn-1 + C(n,n)yn
Counting and Gambling
• In a standard deck of cards, each card has
a suit (clubs, spades, hearts or diamonds)
and a rank (2,3,4,5,6,7,8,9,10, Jack,
Queen, King, Ace).
• There are 4×13 =52 cards in total
• A poker hand has 5 different cards from
the deck
Counting Problems with “Not”
• Sometimes it is easier to find the number
of ways to NOT do A, so we can use
n(A) = n(U) – n(A’)
Example: how many 5 card hands in poker
contain at least one card that is not a club?
11.5
Counting Problems with “Or”
• If A and B are any two sets then
n(A  B)= n(A) + n(B) – n(A  B)
Example: If a single card is drawn from a
standard 52 card deck, how many ways are
there to draw a card that is a face card or is
hearts?
11.5
Poker Hands
How many hands in
total? C(52,5)
How many full
houses?
How many two
pairs?
How many no pairs?
Probability
12.1
• The odds/probability that an event E will
occur is denoted by P(E) and is given by
P(E) = n(E)/n(S)
Where S denotes all possible outcomes.
Ex. Odds of getting a royal flush in poker
are 4/2,598,960 = 1 in 649,740
Poker Odds
• Odds =
1/frequency, or
in other words,
odds are 1 in
the frequency
amount
Properties of Probability
•
•
•
•
12.1
0 ≤ P(E) ≤ 1
The probability of an impossible event is 0
The probability of a certain event is 1
If E denotes an event occurring then E’
denotes the event NOT occurring
P(E) = 1 – P(E’)
• P(A or B) = P(A) + P(B) – P(A and B)