The Counting Rules

6
CHAPTER
The Counting Rules
Introduction
Since probability problems require knowing the total number of ways one or
more events can occur, it is necessary to have a way to compute the number
of outcomes in the sample spaces for a probability experiment. This is
especially true when the number of outcomes is large. For example, when
finding the probability of a specific poker hand, it is necessary to know the
number of different possible ways five cards can be dealt from a 52-card deck.
(This computation will be shown later in this chapter.)
In order to do the computation, we use the fundamental counting rule, the
permutation rules, and the combination rule. The rules then can be used to
compute the probability for events such as winning lotteries, getting a specific
hand in poker, etc.
94
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The Fundamental Counting Rule
The first rule is called the Fundamental Counting Rule.
For a sequence of n events in which the first event can occur in k
1
ways
and the second event can occur in k
2
ways and the third event can occur in
k
3
ways, and so on, the total number of ways the sequence can occur is
k
1
Á k

2
Á k
3
...k
n
:
EXAMPLE: In order to paint a room, a person has a choice of four colors:
white, light blue, yellow, and light green; two types of paint: oil or latex; and
three types of texture: flat, semi-glass, or satin. How many different selections
can be made?
SOLUTION:
There are four colors, two types of paint, and three textures, so the total
number of ways a paint can be selected is 4 Á 2 Á 3 ¼ 24 ways.
EXAMPLE: There are four blood types A, B, AB, and O. Blood can be Rh
þ
or Rh
À
. Finally, a donor can be male or female. How many different
classifications can be made?
SOLUTION:
4 Á 2 Á 2 ¼ 16
When determining the number of different ways a sequence of events can
occur, it is necessary to know whether or not repetitions are permitted. The
next two examples show the difference between the two situations.
EXAMPLE: The employees of a company are given a 4-digit identification
number. How many different numbers are available if repetitions are
permitted?
SOLUTION:
There are 10 digits (zero through nine), so each of the four digits can be
selected in ten different ways since repetitions are permitted. Hence the

total number of identification numbers is 10 Á 10 Á 10 Á 10 ¼ 10
4
¼ 10,000.
CHAPTER 6 The Counting Rules
95
EXAMPLE: The employees of a company are given 4-digit identification
numbers; however, repetitions are not allowed. How many different numbers
are available?
SOLUTION:
In this case, there are 10 ways to select the first digit, 9 ways to select the
second digit, 8 ways to select the third digit, and 7 ways to select the fourth
digit, so the total number of ways is 10 Á 9 Á 8 Á 7 ¼ 5040.
PRACTICE
1. A person can select eight different colors for an automobile body,
five different colors for the interior, and white or black sidewall tires.
How many different color combinations are there for the automobile?
2. A person can select one of five different colors for brick borders, one
type of six different ground coverings, and one of three different types
of shrubbery. How many different types of landscape designs are
there?
3. How many different types of identification cards consisting of 4 letters
can be made from the first five letters of the alphabet if repetitions are
allowed?
4. How many different types of identification cards consisting of 4 letters
can be made from the first 5 letters of the alphabet if repetitions are
not allowed?
5. A license plate consists of 2 letters and 3 digits. How many different
plates can be made if repetitions are permitted? How many can be
made if repetitions are not permitted?
ANSWERS

1. 8 Á 5 Á 2 ¼ 80 color combinations
2. 5 Á 6 Á 3 ¼ 90 types
3. 5 Á 5 Á 5 Á 5 ¼ 5
4
¼ 625 cards
4. 5 Á 4 Á 3 Á 2 ¼ 120 cards
5. Repetitions permitted: 26 Á 26 Á 10 Á 10 Á 10 ¼ 676,000
Repetitions not permitted 26 Á 25 Á 10 Á 9 Á 8 ¼ 468,000
CHAPTER 6 The Counting Rules
96
Factorials
In mathematics there is a notation called factorial notation, which uses the
exclamation point. Some examples of factorial notation are
6! ¼ 6 Á 5 Á 4 Á 3 Á 2 Á 1 ¼ 720
3! ¼ 3 Á 2 Á 1 ¼ 6
5! ¼ 5 Á 4 Á 3 Á 2 Á 1 ¼ 120
1! ¼ 1
Notice that factorial notation means to start with the number and find its
product with all of the whole numbers less than the number and stopping at
one. Formally defined,
n! ¼ n Áðn À 1ÞÁðn À 2Þ ...3 Á 2 Á 1
Factorial notation can be stopped at any time. For example,
6! ¼ 6 Á 5! ¼ 6 Á 5 Á 4!
10! ¼ 10 Á 9! ¼ 10 Á 9 Á 8!
In order to use the formulas in the rest of the chapter, it is necessary to
know how to multiply and divide factorials. In order to multiply factorials,
it is necessary to multiply them out and then multiply the products. For
example,
3! Á 4! ¼ 3 Á 2 Á 1 Á 4 Á 3 Á 2 Á 1 ¼ 144
Notice 3! Á 4! 6¼ 12! Since 12! ¼ 479,001,600

EXAMPLE: Find the product of 5! Á 4!
SOLUTION:
5! Á 4! ¼ 5 Á 4 Á 3 Á 2 Á 1 Á 4 Á 3 Á 2 Á 1 ¼ 2880
Division of factorials is somewhat tricky. You can always multiply them
out and then divide the top number by the bottom number. For example,
8!
6!
¼
8 Á 7 Á 6 Á 5 Á 4 Á 3 Á 2 Á 1
6 Á 5 Á 4 Á 3 Á 2 Á 1
¼
40;320
720
¼ 56
CHAPTER 6 The Counting Rules
97
or
you can cancel out, as shown:
8!
6!
¼
8 Á 7 Á 6!
6!
¼ 8 Á 7 ¼ 56
You cannot divide factorials directly:
8!
4!
6¼ 2! since 8! ¼ 40,320 and 4! ¼ 24, then
40;320
24

¼ 1680
EXAMPLE: Find the quotient
7!
3!
SOLUTION:
7!
3!
¼
7 Á 6 Á 5 Á 4 Á 3!
3!
¼ 7 Á 6 Á 5 Á 4 ¼ 840
Most scientific calculators have a factorial key. It is the key with ‘‘!’’. Also
0! ¼ 1 by definition.
PRACTICE
Find the value of each
1. 2!
2. 7!
3. 9!
4. 4!
5. 6! Á 3!
6. 4! Á 8!
7. 7! Á 2!
8.
10!
8!
9.
5!
2!
10.
6!

3!
CHAPTER 6 The Counting Rules
98
SOLUTIONS
1. 2! ¼ 2 Á 1 ¼ 2
2. 7! ¼ 7 Á 6 Á 5 Á 4 Á 3 Á 2 Á 1 ¼ 5040
3. 9! ¼ 9 Á 8 Á 7 Á 6 Á 5 Á 4 Á 3 Á 2 Á 1 ¼ 362,880
4. 4! ¼ 4 Á 3 Á 2 Á 1 ¼ 24
5. 6! Á 3!¼6 Á 5 Á 4 Á 3 Á 2 Á 1 Á 3 Á 2 Á 1 ¼ 4320
6. 4! Á 8! ¼ 4 Á 3 Á 2 Á 1 Á 8 Á 7 Á 6 Á 5 Á 4 Á 3 Á 2 Á 1 ¼ 967,680
7. 7! Á 2!¼7 Á 6 Á 5 Á 4 Á 3 Á 2 Á 1 Á 2 Á 1 ¼ 10,080
8.
10!
8!
¼
10 Á 9 Á 8!
8!
¼ 10 Á 9 ¼ 90
9.
5!
2!
¼
5 Á 4 Á 3 Á 2!
2!
¼ 5 Á 4 Á 3 ¼ 60
10.
6!
3!
¼
6 Á 5 Á 4 Á 3!

3!
¼ 6 Á 5 Á 4 ¼ 120
The Permutation Rules
The second way to determine the number of outcomes of an event is to use
the permutation rules. An arrangement of n distinct objects in a specific order
is called a permutation. For example, if an art dealer had 3 paintings, say A,
B, and C, to arrange in a row on a wall, there would be 6 distinct ways to
display the paintings. They are
ABC BAC CAB
ACB BCA CBA
The total number of different ways can be found using the fundamental
counting rule. There are 3 ways to select the first object, 2 ways to select the
second object, and 1 way to select the third object. Hence, there are
3 Á 2 Á 1 ¼ 6 different ways to arrange three objects in a row on a shelf.
Another way to solve this kind of problem is to use permutations. The
number of permutations of n objects using all the objects is n!.
CHAPTER 6 The Counting Rules
99
EXAMPLE: In how many different ways can 6 people be arranged in a row
for a photograph?
SOLUTION:
This is a permutation of 6 objects. Hence 6! ¼ 6 Á 5 Á 4 Á 3 Á 2 Á 1 ¼ 720 ways.
In the previous example, all the objects were used; however, in many
situations only some of the objects are used. In this case, the permutation rule
can be used.
The arrangement of n objects in a specific order using r objects at a time is
called a permutation of n objects taking r objects at a time. It is written as
n
P
r

and the formula is
n
P
r
¼
n!
ðn À rÞ!
EXAMPLE: In how many different ways can 3 people be arranged in a row
for a photograph if they are selected from a group of 5 people?
SOLUTION:
Since 3 people are being selected from 5 people and arranged in a specific
order, n ¼ 5, r ¼ 3. Hence, there are
5
P
3
¼
5!
ð5 À 3Þ!
¼
5!
2!
¼
5 Á 4 Á 3 Á 2!
2!
¼ 5 Á 4 Á 3 ¼ 60 ways
EXAMPLE: In how many different ways can a chairperson and secretary be
selected from a committee of 9 people?
SOLUTION:
In this case, n ¼ 9 and r ¼ 2. Hence, there are
9

P
2
ways of selecting two people
to fill the two positions.
9
P
2
¼
9!
ð9 À 2Þ!
¼
9!
7!
¼
9 Á 8 Á 7!
7!
¼ 72 ways
CHAPTER 6 The Counting Rules
100
EXAMPLE: How many different signals can be made from seven different
flags if four flags are displayed in a row?
SOLUTION:
Hence n ¼ 7 and r ¼ 4, so
7
P
4
¼
7!
ð7 À 4Þ!
¼

7!
3!
¼
7 Á 6 Á 5 Á 4 Á 3!
3!
¼ 7 Á 6 Á 5 Á 4 ¼ 840
In the preceding examples, all the objects were different, but when some of
the objects are identical, the second permutation rule can be used.
The number of permutations of n objects when r
1
objects are identical,
r
2
objects are identical, etc. is
n!
r
1
!r
2
!...r
p
!
where r
1
þ r
2
þ ...þ r
p
¼ n
EXAMPLE: How many different permutations can be made from the letters

of the word Mississippi?
SOLUTION:
There are 4 s, 4 i, 2 p, and 1 m; hence, n ¼ 11, r
1
¼ 4, r
2
¼ 4, r
3
¼ 2, and r
4
¼ 1
11!
4! Á 4! Á 2! Á 1!
¼
11 Á 10 Á 9 Á 8 Á 7 Á 6 Á 5 Á 4!
4! Á 4 Á 3 Á 2 Á 1 Á 2 Á 1 Á 1
¼
1,663,200
48
¼ 34,650
EXAMPLE: An automobile dealer has 3 Fords, 2 Buicks, and 4 Dodges to
place in the front row of his car lot. In how many different ways by make of
car can he display the automobiles?
SOLUTION:
Let n ¼ 3 þ 2 þ 4 ¼ 9 automobiles; r
1
¼ 3 Fords, r
2
¼ 2 Buicks, and r
3

¼ 4
Dodges; then there are
9!
3!Á2!Á4!
¼
9Á8Á7Á6Á5Á 4!
3Á2Á1Á2Á1Á
4!
¼ 1260 ways to display the
automobiles.
CHAPTER 6 The Counting Rules
101