#### Transcript Section_04_04 - it

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SECTION 4.4
COUNTING
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Objectives
1.
2.
3.
Count the number of ways a sequence of operations can
be performed
Count the number of permutations
Count the number of combinations
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Objective 1
Count the number of ways a sequence of
operations can be performed
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The Fundamental Principle of Counting
If an operation can be performed in m ways, and a second
operation can be performed in n ways, then the total
number of ways to perform the sequence of two operations
is mn.
If a sequence of several operations is to be performed, the
number of ways to perform the sequence is found by
multiplying together the numbers of ways to perform each
of the operations.
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Example
A certain make of automobile is available in
any of three colors: red, blue, or green, and
comes with either a large or small engine. In
how many ways can a buyer choose a car?
Solution:
There are 3 choices of color and 2 choices of engine. The total
number of choices is 3·2 = 6
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Example
License plates in a certain state contain three letters followed by
Solution:
There are six operations in all; choosing three letters and
choosing three digits. There are 26 ways to choose each letter
and 10 ways to choose each digit. The total number of license
plates is therefore 26·26·26·10·10·10 = 17,576,000.
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Objective 2
Count the number of permutations
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Definition: Permutation
The word “permutation” is another word for “ordering.”
When we count the number of permutations, we are counting
the number of different ways that a group of items can be
ordered.
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Permutation of n Objects
The number of permutations of n objects is n!
Definition of n!
For any positive integer n, the number n! is pronounced “n
factorial” and is equal to the product of all the integers from
n down to 1.
n! = n(n − 1)(n –2)∙ ∙ ∙ 3(2)(1)
By definition, 0! = 1.
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Example
Five runners run a race. One of them will finish first, another
will finish second, and so on. In how many different orders
can they finish?
Solution:
The number of different orders in which the runners can
finish is 5! = 5·4·3·2·1 = 120.
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Example
Ten runners run a race. The first-place finisher will win a gold
medal, the second-place finisher will win a silver medal, and the
third-place finisher will win a bronze medal. In how many
different ways can the medals be awarded?
Solution:
We use the Fundamental Principle of Counting. There are 10
possible choices for the gold medal winner. Once the gold medal
winner is determined, there are nine remaining choices for the
silver medal. Finally, there are eight choices for the bronze
medal. The total number of ways the medals can be awarded
is10·9·8 = 720.
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Permutation of r Objects Chosen from n
In the last example, three runners were chosen from a group of ten, then
ordered as first, second, and third. This is referred to as a permutation of
three items chosen from ten.
In general, a permutation of r items chosen from n items is an ordering of the r
items. It is obtained by choosing r items from a group of n items, then choosing
an order for the r items. The number of permutations of r items chosen from n
is denoted nPr.
The number of permutations of r objects chosen from n is
n!
P

n

(
n

1)



(
n

r

1)

n r
(n  r )!
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Example
Five lifeguards are available for duty one Saturday afternoon.
There are three lifeguard stations. In how many ways can three
lifeguards be chosen and ordered among the stations?
Solution:
We are choosing three items from a group of five and ordering
them. The number of ways to do this is
5 P3 
5!
5! 5  4  3  2  1


 5  4  3  60
(5  3)! 2!
2 1
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Objective 3
Count the number of combinations
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Combinations
In some cases, when choosing a set of objects from a larger set, we don’t care
about the ordering of the chosen objects; we care only which objects are
chosen. For example, we may not care which lifeguard occupies which station;
we might care only which three lifeguards are chosen. Each distinct group of
objects that can be selected, without regard to order, is called a combination.
The number of combinations of r objects chosen from a group of n objects is
n Cr 
n!
r !(n  r )!
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Example
At a certain event, 30 people attend, and 5 will be chosen at random to
receive prizes. The prizes are all the same, so the order in which the
people are chosen does not matter. How many different groups of 5
people can be chosen?
Solution:
Since the order of the 5 chosen people does not matter, we need to
compute the number of combinations of 5 chosen from 30. This is
30 C5 
30!
30  29  28  27  26

 142,506
5!(30  5)!
5  4  3  2 1
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Permutations and Combinations on the TI-84 PLUS
Calculator commands for permutations
and combinations are accessed by pressing
MATH and scrolling to the PRB menu.
The format for entering these commands is to enter the value of n into the
calculator, select either nPr (permutations) or nCr (combinations) from the PRB
menu, and then enter the value of r. To run the command, press Enter.
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Do You Know
•
•
•
How to use the Fundamental Principle of Counting to count
the number of ways a sequence of operations can be
performed?
How to count the number of permutations?
How to count the number of combinations?
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