Additional Counting Principles Section 3.4 June 26 Summer 2013 - Math 1040

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Additional Counting Principles
Section 3.4
Summer 2013 - Math 1040
June 26
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Roadmap
§3.4 p 168 - 173
I
Permutation Principle
I
Combination Principle
I
Application
The rules used today are extensions of the fundamental counting principle:
The ways of two events occuring in sequence is m · n where there are m
ways of occuring for event A and n ways of occuring for event B.
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Permutations
Permutations are the ordered arrangements of objects.
If we want to arrange n distinct objects, the number of permutations is n!.
Example We want to line up five relatives for a family picture. There are
a total of 5! = 5 · 4 · 3 · 2 · 1 = 120 arrangements.
A special case is 0! = 1. Then 1! = 1, 2! = 2 · 1, 3! · 3 · 2 · 1, . . .
If we want to arrange n distinct objects taken r at a time, the number of
n!
permutations is n Pr = (n−r
)! .
Example We take pictures of pairs of the five relatives, in order. Then
5!
there are a total of 5 P2 = (5−2)!
= 5!
3! ways.
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Permutations
Ordering n objects may include that some of them are the same. For
instance, we may have identical twins, same colored dice, repeated letters,
etc. Some permutations may not be distinguishable.
The number of distinguishable permutations of n total objects and
n1 -many objects of the same type, n2 -many objects of another type, ect is
n!
(n1 )! · (n1 )! · · · (nk )!
where n1 + n2 + · · · nk = n
Example How many ways can we arrange the letters of the word
BOOKKEEPER? This is a 10 letter word, with 1 B, 1 P, 1 R, 2 O’s, 2 K’s,
and 3 E’s.
10!
= 151, 200 ways.
1! · 1! · 1! · 2! · 2! · 3!
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Combinations
Combinations are the unordered arrangements of objects.
Combinations of n objects taken r at a time disregards order. The number
of ways can be donoted at n Cr , and this is
n!
(n − 1)! · (r !)
This can be thought of as the number of distingishable permutations of n
objects where are n1 objects are selected and n2 objects are not selected.
Example The number of different 5-card hands from a standard deck of
52!
. That is we take 5 cards are leave 47 behind, and we
cards is 52 C5 = 5!·47!
count the number of ways to do this.
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Probabilities
Our two definitions for a theoretical probability and statistical probability
are similar in the following way. Both are total number of outcomes in an
event, out of the total possible outcomes in a sample space. We can apply
the three counting principles: the fundamental counting principle,
permutations, and combinations.
Example A flush is a hand of cards all of one suit. Suppose we want a
flush in clubs in a 5-card hand. There are 13 C5 = 1287 many ways to do
this. The probability of such an event is then
13 C5
52 C5
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=
1287
≈ 0.0005.
2, 598, 960
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Assignments
Assignment:
1. In class examples.
2. Page 174, 1 - 53 odd.
3. Quiz over 3.3, 3.4, and the exam over 2.5 and all of chapter 3.
Vocabulary: permutation, n Pr , distinguishable permuations,
combinations, n Cr
Understand: When to use m · n, n!, n Pr ,
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n!
n1 !·n2 !···nk ! ,
and n Cr .
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