6
SETS AND
COUNTING
Copyright © Cengage Learning. All rights reserved.
6.2
The Number of Elements
in a Finite Set
Copyright © Cengage Learning. All rights reserved.
Counting the Elements in a Set
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Counting the Elements in a Set
The solution to some problems in mathematics calls for
finding the number of elements in a set.
Such problems are called counting problems and
constitute a field of study known as combinatorics.
The number of elements in a finite set is determined by
simply counting the elements in the set.
If A is a set, then n(A) denotes the number of elements
in A.
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Counting the Elements in a Set
For example, if
A = {1, 2, 3, ..., 20}
B = {a, b}
C = {8}
then
n(A) = 20, n(B) = 2, n(C) = 1.
The empty set has no elements in it, so n() = 0.
Another result that is easily seen to be true is the following:
If A and B are disjoint sets, then
n(A  B) = n(A) + n(B)
(3)
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Example 1
If A = {a, c, d} and B = {b, e, f, g}, then n(A) = 3 and
n(B) = 4, so
n(A) + n(B) = 7.
Moreover, A  B = {a, b, c, d, e, f, g} and
n(A  B) = 7.
Thus, Equation (3) holds true in this case.
Note that A  B = .
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Counting the Elements in a Set
In the general case, A and B need not be disjoint, which
leads us to the formula
n(A  B) = n(A) + n(B) – n(A  B)
(4)
To see this, we observe that the set A  B may be viewed
as the union of three mutually disjoint sets with x, y, and z
elements, respectively (Figure 8).
n(A  B) = x + y + z
Figure 8
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Counting the Elements in a Set
This figure shows that
n(A  B) = x + y + z
Also,
n(A) = x + y
and
n(B) = y + z
So
n(A) + n(B) = (x + y) + (y + z)
= (x + y + z) + y
= n(A  B) + n(A  B)
n(A  B) = y
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Counting the Elements in a Set
Solving for n(A  B), we obtain
n(A  B) = n(A) + n(B) – n(A  B)
which is the desired result.
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Example 2
Let A = {a, b, c, d, e}, and let B = {b, d, f, h}.
Verify n(A  B) = n(A) + n(B) – n(A  B) directly.
Solution:
A  B = {a, b, c, d, e, f, h}
so
n(A  B) = 7
A  B = {b, d}
so
n(A  B) = 2
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Example 2 – Solution
Furthermore,
n(A) = 5
So
and
cont’d
n(B) = 4
n(A) + n(B) – n(A  B) = 5 + 4 – 2
=7
= n(A  B)
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Counting the Elements in a Set
An equation similar to Equation (4) can be derived for the
case that involves any finite number of finite sets.
For example, a relationship involving the number of
elements in the sets A, B, and C is given by
n(A  B  C) = n(A) + n(B) + n(C) – n(A  B)
– n(A  C) – n(B  C) + n(A  B  C)
(5)
As useful as equations such as Equation (5) are, in practice
it is often easier to attack a problem directly with the aid of
Venn diagrams, as shown by the next example.
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Applied Example 4 – Marketing Surveys
A leading cosmetics manufacturer advertises its products in
three magazines: Allure, Cosmopolitan, and the Ladies
Home Journal.
A survey of 500 customers by the manufacturer reveals the
following information:
180 learned of its products from Allure.
200 learned of its products from Cosmopolitan.
192 learned of its products from the Ladies Home Journal.
84 learned of its products from Allure and Cosmopolitan.
52 learned of its products from Allure and the Ladies Home
Journal.
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Applied Example 4 – Marketing Surveys
cont’d
64 learned of its products from Cosmopolitan and the
Ladies Home Journal.
38 learned of its products from all three magazines.
How many of the customers saw the manufacturer’s
advertisement in
a. At least one magazine?
b. Exactly one magazine?
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Applied Example 4 – Solution
Let U denote the set of all customers surveyed, and let
A = {x  U | x learned of the products from Allure}
C = {x  U | x learned of the products from Cosmopolitan}
L = {x  U | x learned of the products from Ladies Home
Journal}
The result that 38 customers
learned of the products from
all three magazines translates
into n(A  C  L) = 38
(Figure 9a).
All three magazines
Figure 9(a)
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Applied Example 4 – Solution
cont’d
Next, the result that 64 learned of the products from
Cosmopolitan and the Ladies Home Journal translates into
n(C  L) = 64.
This leaves
64 – 38 = 26
who learned of the products
from only Cosmopolitan and
the Ladies Home Journal
(Figure 9b).
Two or more magazines
Figure 9(b)
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Applied Example 4 – Solution
cont’d
Similarly, n(A  L) = 52,
so
52 – 38 = 14
learned of the products from only Allure and the Ladies
Home Journal, and n(A  C) = 84, so
84 – 38 = 46
learned of the products from only Allure and Cosmopolitan.
These numbers appear in the appropriate regions in
Figure 9b.
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Applied Example 4 – Solution
cont’d
Continuing, we have n(L) = 192, so the number who
learned of the products from the Ladies Home Journal only
is given by
192 – 14 – 38 – 26 = 114
(Figure 10).
The completed Venn diagram
Figure 10
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Applied Example 4 – Solution
cont’d
Similarly, n(C) = 200,
so
200 – 46 – 38 – 26 = 90
learned of the products from only Cosmopolitan, and
n(A) = 180,
so
180 – 14 – 38 – 46 = 82
learned of the products from only Allure.
Finally,
500 – (90 + 26 + 114 + 14 + 82 + 46 + 38) = 90
learned of the products from other sources.
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Applied Example 4 – Solution
cont’d
We are now in a position to answer questions (a) and (b).
a. Referring to Figure 10, we see that the number of
customers who learned of the products from at least one
magazine is given by
n(A  C  L) = 500 – 90
= 410
The completed Venn diagram
Figure 10
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Practice
p. 336 Self-Check Exercises #1 & 2
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