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SECTION 10-2
• Using the Fundamental
Counting Principle
Slide 10-2-1
USING THE FUNDAMENTAL COUNTING
PRINCIPLE
• Uniformity and the Fundamental Counting
Principle
• Factorials
• Arrangements of Objects
Slide 10-2-2
UNIFORMITY CRITERION FOR
MULTIPLE-PART TASKS
A multiple-part task is said to satisfy the
uniformity criterion if the number of choices for
any particular part is the same no matter which
choices were selected for the previous parts.
Slide 10-2-3
FUNDAMENTAL COUNTING PRINCIPLE
When a task consists of k separate parts and satisfies
the uniformity criterion, if the first part can be done
in n1 ways, the second part can be done in n2 ways,
and so on through the k th part, which can be done in
nk ways, then the total number of ways to complete
the task is given by the product
n1  n2  n3 
 nk .
Slide 10-2-4
EXAMPLE: TWO-DIGIT NUMBERS
How many two-digit numbers can be made from the
set {0, 1, 2, 3, 4, 5}? (numbers can’t start with 0.)
Solution
Part of Task
Number of ways
Select first digit
5
(0 can’t be used)
Select second
digit
6
There are 5(6) = 30 two-digit numbers.
Slide 10-2-5
EXAMPLE: TWO-DIGIT NUMBERS WITH
RESTRICTIONS
How many two-digit numbers that do not contain
repeated digits can be made from the set
{0, 1, 2, 3, 4, 5} ?
Solution
Part of
Task
Number of
ways
Select first Select second digit
digit
6
5
(repeated digits not allowed)
There are 6(5) = 30 two-digit numbers.
Slide 10-2-6
EXAMPLE: TWO-DIGIT NUMBERS WITH
RESTRICTIONS
How many ways can you select two letters followed
by three digits for an ID?
Solution
Part of
Task
Number
of ways
First
letter
26
Second Digit
letter
26
10
Digit
Digit
10
10
There are 26(26)(10)(10)(10) = 676,000 IDs possible.
Slide 10-2-7
FACTORIALS
For any counting number n, the product of all
counting numbers from n down through 1 is
called n factorial, and is denoted n!.
Slide 10-2-8
FACTORIAL FORMULA
For any counting number n, the quantity
n factorial is given by
n !  n(n  1)(n  2)
2 1.
Slide 10-2-9
EXAMPLE:
Evaluate each expression.
a) 4!
b) (4 – 1)!
5!
c)
3!
Solution
a) 4!  4  3  2 1  24
b) (4  1)!  3  2 1  6
5! 5  4  3!
c)

 5  4  20
3!
3!
Slide 10-2-10
DEFINITION OF ZERO FACTORIAL
0!  1
Slide 10-2-11
ARRANGEMENTS OF OBJECTS
When finding the total number of ways to
arrange a given number of distinct objects,
we can use a factorial.
Slide 10-2-12
ARRANGEMENTS OF N DISTINCT
OBJECTS
The total number of different ways to arrange
n distinct objects is n!.
Slide 10-2-13
EXAMPLE: ARRANGING BOOKS
How many ways can you line up 6 different
books on a shelf?
Solution
The number of ways to arrange 6 distinct objects
is 6! = 720.
Slide 10-2-14
ARRANGEMENTS OF N OBJECTS
CONTAINING LOOK-ALIKES
The number of distinguishable arrangements of n
objects, where one or more subsets consist of lookalikes (say n1 are of one kind, n2 are of another kind,
…, and nk are of yet another kind), is given by
n!
.
n1 !n2 ! nk !
Slide 10-2-15
EXAMPLE: DISTINGUISHABLE
ARRANGEMENTS
Determine the number of distinguishable
arrangements of the letters of the word INITIALLY.
Solution
9 letters total
3 I’s and 2 L’s
9!
 30240.
3!2!
Slide 10-2-16