Transcript Mathematical Ideas - Millersville University of Pennsylvania

Chapter 11
Counting
Methods
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All rights reserved
Chapter 11: Counting Methods
11.1
11.2
11.3
11.4
11.5
Counting by Systematic Listing
Using the Fundamental Counting Principle
Using Permutations and Combinations
Using Pascal’s Triangle
Counting Problems Involving “Not” and
“Or”
11-2-2
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Chapter 1
Section 11-2
Using the Fundamental Counting
Principle
11-2-3
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Using the Fundamental Counting
Principle
• Uniformity and the Fundamental Counting
Principle
• Factorials
• Arrangements of Objects
11-2-4
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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.
11-2-5
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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 kth 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 .
11-2-6
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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
Select first digit Select second
digit
Number of ways
5
6
(0 can’t be used)
There are 5(6) = 30 two-digit numbers.
11-2-7
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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
Select second digit
first digit
5
5
(repeated digits not allowed)
There are 5(5) = 25 two-digit numbers.
11-2-8
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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.
11-2-9
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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!.
11-2-10
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Factorial Formula
For any counting number n, the quantity n
factorial is given by
n!  n(n 1)(n  2)
2 1.
11-2-11
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Example:
Evaluate each expression.
a) 4!
b) (4 – 1)!
Solution
5!
c)
3!
a) 4!  4  3  2 1  24
b) (4  1)!  3  2 1  6
5! 5  4  3!
c)

 5  4  20
3!
3!
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11-2-12
Definition of Zero Factorial
0!  1
11-2-13
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Arrangements of Objects
When finding the total number of ways to
arrange a given number of distinct
objects, we can use a factorial.
11-2-14
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Arrangements of n Distinct Objects
The total number of different ways to
arrange n distinct objects is n!.
11-2-15
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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.
11-2-16
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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 !
11-2-17
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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!
11-2-18
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