Transcript Probability - Count with Kellogg

Probability
Part 2 – Factorial and other
Counting Rules
Probability
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Warm-up
Canadian “zip” codes are as follows
NLN – LNL, where N is a number and L is
a letter. How many “zip” codes are possible?
Probability
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Agenda
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Warm-up
Homework Review
Objective: To understand and apply concepts
related to the factorial counting rule,
permutations, and combinations.
Summary
Homework
Factorial Counting Rule
If you have n items that occur in a sequence
of events and no repetitions are permitted,
then the number of ways the sequence can
occur is n!
Where n is the number of items.
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Probability
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Example
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You have five items you wish to arrange items on
a shelf. How many different ways can they be
arranged?
5! = 5 x 4 x 3 x 2 x 1 = 120
Partitions Rule
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Determines the number of different ways, you
can partition the elements of a set of n
elements into k groups consisting of n1, n2,
…, nk objects respectively
n!
where
n1!n2!nk !
k
n
i 1
i
n
Partitions Example
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You have 10 children to help around the house.
You want to assign three to clean up the yard, four
to help paint the downstairs and three to wash the
family car. In how many different ways can you
group your children?
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k = 3 (different areas)
n1= 3, n2=4, n3=3 (group sizes)
n!
10!
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 4,200
n1!n2 ! nk ! 3!4!3!
Permutations
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Consider the possible arrangements of the letters
a, b, and c.
The possible arrangements are: abc, acb, bac,
bca, cab, cba.
If the order of the arrangement is important then
we say that each arrangement is a permutation
of the three letters. Thus there are six
permutations of the three letters.
Permutations
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An arrangement of n distinct objects in
a specific order is called a permutation
of the objects.
Note: To determine the number of
possibilities mathematically, one can
use the multiplication rule to get:
3  2  1 = 6 permutations.
Permutations
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Permutation Rule : The arrangement of n
objects in a specific order using r objects
at a time is called a permutation of n
objects taken r objects at a time. It is
written as nPr and the formula is given by
nPr = n! / (n – r)!.
Permutations - Example
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How many different ways can a chairperson
and an assistant chairperson be selected for
a research project if there are seven
scientists available?
Solution: Number of ways
= 7P2 = 7! / (7 – 2)! = 7!/5! = 42.
Permutations - Example
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How many different ways can four
different books be arranged in a specific
location on a shelf if they can be
selected from nine different books?
Solution: Number of ways
=9P4 = 9! / (9 – 4)! = 9!/5! = 3024.
Combinations
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Consider the possible arrangements of the letters
a, b, and c.
The possible arrangements are: abc, acb, bac,
bca, cab, cba.
If the order of the arrangement is not important
then we say that each arrangement is the same.
We say there is one combination of the three
letters.
Combinations
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Combination Rule : The number of
combinations of r objects from n
objects is denoted by nCr and the
formula is given by nCr = n! / [(n – r)!r!] .
Combinations - Example
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How many combinations of four objects
are there taken two at a time?
Solution: Number of combinations:
4C2 = 4! / [(4 – 2)! 2!] = 4!/[2!2!] = 6.
Combinations - Example
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In order to survey the opinions of customers at
local malls, a researcher decides to select 5
malls from a total of 12 malls in a specific
geographic area. How many different ways can
the selection be made?
Solution: Number of combinations:
12C5 = 12! / [(12 – 5)! 5!] = 12!/[7!5!] = 792.
Combinations - Example
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In a club there are 7 women and 5 men. A
committee of 3 women and 2 men is to be
chosen. How many different possibilities are
there?
Solution: Number of possibilities: (number of
ways of selecting 3 women from 7) (number of
ways of selecting 2 men from 5) = 7C3 5C2 =
(35)(10) = 350.
Combinations - Example
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A committee of 5 people must be
selected from 5 men and 8 women.
How many ways can the selection be
made if there are at least 3 women on
the committee?
Combinations - Example
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Solution: The committee can consist of
3 women and 2 men, or 4 women and 1
man, or 5 women. To find the different
possibilities, find each separately and
then add them:
8C3 5C2 + 8C4 5C1
+ 8C5 5C0= (56)(10) + (70)(5) + (56)(1)
= 966.
Probability – Counting Rules
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1) A television news director wishes to use three news stories on an
evening show. One story will be the lead story, one will be the second story,
and the last will be a closing story. If the director has a total of eight stories
to choose from, how many possible ways can the program be set up? 336
2) How many different ways can a chairperson and an assistant chairperson
be selected for a research project if there are seven scientists available? 42
3) A bicycle shop owner has 12 mountain bicycles in the showroom. The
owner wishes to select 5 of them to display at a bicycle show. How many
different ways can a group of 5 be selected? 792
4) In a club there are 7 women and 5 men. A committee of 3 women and 2
men is to be chosen. How many different possibilities are there? 350
Probability
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Summary
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Factorial Counting Rule
Permutations
Combinations
Probability
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Homework
Probability Practice Sheet 2