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3
Chapter 5
Probability
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Section 5.5
Counting Techniques
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5-2
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5-3
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EXAMPLE
Counting the Number of Possible Meals
For each choice of appetizer, we have 4 choices of
entrée, and that, for each of these 2 • 4 = 8 choices,
there are 2 choices for dessert. A total of
2 • 4 • 2 = 16
different meals can be ordered.
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5-5
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5-6
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5-7
A permutation is an ordered
arrangement in which r objects are
chosen from n distinct (different)
objects and repetition is not allowed.
The symbol nPr represents the
number of permutations of r objects
selected from n objects.
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5-8
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5-9
EXAMPLE
Betting on the Trifecta
In how many ways can horses in a 10-horse race finish
first, second, and third?
The 10 horses are distinct. Once a horse crosses the
finish line, that horse will not cross the finish line again,
and, in a race, order is important. We have a
permutation of 10 objects taken 3 at a time.
The top three horses can finish a 10-horse race in
10!
10! 10  9  8  7!


 10  9  8  720 ways
10 P3 
7!
10  3! 7!
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5-10
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A combination is a collection, without
regard to order, of n distinct objects
without repetition. The symbol nCr
represents the number of
combinations of n distinct objects
taken r at a time.
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5-12
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5-13
EXAMPLE
Simple Random Samples
How many different simple random samples of size 4
can be obtained from a population whose size is 20?
The 20 individuals in the population are distinct. In
addition, the order in which individuals are selected is
unimportant. Thus, the number of simple random
samples of size 4 from a population of size 20 is a
combination of 20 objects taken 4 at a time.
Use Formula (2) with n = 20 and r = 4:
20!
20! 20 19 18 17 16! 116, 280



 4, 845
20 C 4 
4!20  4 ! 4!16!
4  3 2 116!
24
There are 4,845 different simple random samples of
size 4 from a population whose size is 20.
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5-14
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5-15
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5-16
EXAMPLE
Arranging Flags
How many different vertical arrangements are there of
10 flags if 5 are white, 3 are blue, and 2 are red?
We seek the number of permutations of 10 objects, of
which 5 are of one kind (white), 3 are of a second kind
(blue), and 2 are of a third kind (red).
Using Formula (3), we find that there are
10!
10  9  8  7  6  5!

 2, 520 different
5!  3!  2!
5!  3!  2!
vertical
arrangements
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5-18
EXAMPLE
Winning the Lottery
In the Illinois Lottery, an urn contains balls numbered 1
to 52. From this urn, six balls are randomly chosen
without replacement. For a $1 bet, a player chooses
two sets of six numbers. To win, all six numbers must
match those chosen from the urn. The order in which
the balls are selected does not matter. What is the
probability of winning the lottery?
The probability of winning is given by the number of
ways a ticket could win divided by the size of the
sample space. Each ticket has two sets of six numbers,
so there are two chances of winning for each ticket. The
sample space S is the number of ways that 6 objects
can be selected from 52 objects without replacement
and without regard to order, so N(S) = 52C6.
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5-19
EXAMPLE
Winning the Lottery
The size of the sample space is
52!
52  51 50  49  48  47  46!
N S   52 C6 

 20, 358, 520
6! 52  6 !
6!  46!
Each ticket has two sets of 6 numbers, so a player has
two chances of winning for52 6each $1. If E is the
event “winning ticket,” then N1E2 = 2
2
P E  
 0.000000098
20, 358, 520
There is about a 1 in 10,000,000 chance of winning the
Illinois Lottery!
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