Chapter 4: Sets and Counting
4.1 Sets Sets are among the most basic of mathematical objects. For this reason it is not possible to describe sets in terms of other, more primitive concepts. This is just as well, for by this time you probably have an intuitive understanding of a set anyway as simply a collection of objects. What is most important when we are talking about sets is that we all have a common understanding as to exactly what the elements of the sets under discussion are. For this reason, we need to talk briefly about the notation that is used to represent sets and to describe the elements that belong to a set. One common way to represent small sets is simply to list all the elements. For example, if we want to talk about the set which consists of the numbers 1,2,3,4, and 5, we will represent this set by simply listing the elements as follows: {1,2,3,4,5}. It is useful to have “names” for sets in order to be able to refer to them easily, and for this purpose the names that are commonly used are capital letters. For instance, if we say A = {1,2,3,4,5}, this says that we are talking about the set whose elements are the integers from 1 to 5 and that we are using the name “A” to identify this set. As another example, we might want to think of a group of four people named Bob, Mary, Sue, and Ellen as constituting a set. If we want to name this set “B”, then we can describe the set by writing something like B = {Bob, Mary, Sue, Ellen}. It is important to keep in mind that there is no order associated with the elements of a set. Saying B = {Bob, Mary, Sue, Ellen} and saying B = {Mary, Ellen, Bob, Sue} are completely equivalent. In each case the set is described as consisting of the same four people. Another way of describing a set is to describe a condition that must be met in order for something to be an element of the set. For example, we could describe the set A = {1,2,3,4,5} by saying something like this: The set A consists of all integers which are greater than 0 and less than 6. Sometimes this method is used out of necessity. For example, we might want to speak of the set of all positive numbers. In this case it would be impossible to list all the elements because there are infinitely many of them. When we wish to describe a set by indicating a condition that must be satisfied in order for an element to belong to the set, we do so by writing something like this: A = { x | x is an integer and 0 < x < 6}.
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The symbol “x” is being used here in a generic way to represent a potential element of the set A, and the “rule” to the right of the vertical bar “|” describes what conditions “x” must satisfy in order to be admitted as a member of the set A. Example 4.1. Here are some sets described in each of the two ways described above: A = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday} or A = { x | x is a day of the week} B = {–2, 2} or B = { x | x2 = 4} C = {a,e,i,o,u} or C = {x | x is a vowel in the alphabet} The sets in Example 4.1 are each described using both of the methods for defining sets, but often in practice one method will be more natural than the other. For instance, suppose D = {9, 14, –37, 522}. There’s not an obvious rule we can use to describe a condition that must be satisfied in order for a number to belong to the set D. On the other hand, large sets such as E = { x | x is an integer and 1 ≤ x ≤ 100} are not conveniently described by listing the elements because the list is too long. In talking about sets, there are some special symbols that are very useful. The symbol “ ∈ ” for example is used to mean “is an element of.” For instance we write 5 ∈ A if we want to indicate that the number 5 is an element of the set A. Similarly we write 3 ∉ A if we want to indicate that the number 3 is not an element of the set A. We speak of two sets as being equal only if the sets consist of exactly the same elements. In this case we write A = B. So if A = {a,b,c} and B = {c,a,b} then A = B. However, if A = {a,b,c} and B = {a,b} then we write A ≠ B. Another important symbol is the subset symbol “! ”. We write A ! B to indicate that “A is a subset of B.” This means that every element of the set A is also an element of the set B. Example 4.2. If A = {a,b,c,d,e} and B = {d,b}, then B ! A because both d and b (the elements of B) are also elements of A. Notice that every set is automatically a subset of itself, i.e. A ! A no matter what the set A is. This is no more peculiar than the fact that n ≤ n no matter what the number n happens to be. What if A and B are sets for which A ! B and B ! A? This says that every element of A is an element of B, and every element of B is also an element of A. These two statements together indicate that A and B contain exactly the same elements, so A = B. This is much like saying that if a and b are numbers for which a ≤ b and b ≤ a, then a = b.
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There is a special name for the set consisting of no elements whatsoever. It is called the empty set and is denoted by Ø. This set has the peculiar property that it is a subset of every set. No matter what the set A might be, Ø ! A. Every element of Ø is an element of A simply because Ø doesn’t have any elements. This is a bit like saying that “all unicorns are blue” is true because there are not any unicorns that are not blue. While this may seem a bit strange to your customary way of thinking, it’s logically correct. Example 4.3. List the subsets of the set A = {1,2,3} Solution:
{1,2,3} {1,2} {1,3} {2,3} {1} {2} {3} Ø The first subset listed here is A itself. There is one subset with 3 elements, there are three subsets with 2 elements, three with one element, and one (the empty set) with no elements. Example 4.4. If A = {0,1,2,3,4,5} and B = {2,3,4}, indicate which of the following statements are true and which are false: 1. B ! A 2. 0 ∈ A 3. 0 ! A 4. {0} ! A 5. Ø ∈ A 6. Ø ! A 7. Ø ! B Solution: (1) True, because every element of B is an element of A. (2) True, because the number 0 belongs to the set A. (3) False. 0 is a number. It is not a set, so it can’t possibly be a subset of something. (4) True. {0} is the set consisting of the single number 0. Since every element of this set is an element of A, it is true that {0} ! A. (5) False. The five elements of A are all numbers. Ø is not a number. So it can’t be an element of A. (6) True, the empty set is a subset of every set. (7) True, the empty set is a subset of every set. Most discussions of sets take place within the context of some large set lurking somewhere in the background. For example, if we are talking about sets of numbers then every set is a subset of the set of all numbers. If we are talking about different groups of students at the university, then all these sets are subsets of the set of all students at the university. If we are talking about groups of voters in North Carolina, then all sets under discussion are subsets of the set of all registered voters in North Carolina. The large set in the background which sets the context for a particular discussion is called the universal set. It should always be clear from context what the universal set is when a particular collection of sets are being discussed. The complement of a set A is made up of all elements of the universal set that do not belong to A. It is denoted by Ac.
Section 4.1: Sets
129 Ac = { x | x is in the universal set and x ∉ A}
For example, the complement of the set of all positive numbers would consist of zero and all negative numbers. The complement of the set of male students at the university would be the set of female students at the university. If the universal set is the set of all books in the library, then the complement of the set of all math books would be the set of all books that are not math books. If the universal set consists of the letters of the English alphabet, then the complement of the set of letters from a through m would be the set of letters from n through z. Example 4.5. Suppose the universal set is U = {1,2,3,4,5,6,7,8,9,10} and A = {1,7,8}. Then Ac = {2,3,4,5,6,9,10}, Øc = U, and Uc = Ø.
Notation introduced in this section 1. x ∈ A means “x belongs to A” 2. x ∉ A means “x does not belong to A” 3. A ! B means that A is a subset of B, i.e. every element of A is also an element of B. 4. Ø denotes the empty set, i.e. the set with no elements. 5. Ø ! A for every set A. 6. Ac = the set of elements in the universal set that are not in A.
Problems 1.
For each of the following sets, give a rule that describes the criteria that an element must satisfy in order to be a member of the set. For example, the first set could be written as A = { x | x is an integer and 1 ≤ x ≤ 3} A = {1,2,3} B = {2,4,6} C = {a,e,i,o,u} D = {Colorado, Connecticut, California}
2.
List all the elements of the following sets: (a) { x | x is an integer and 3 < x < 9} (b) { x | x is a letter in the word “escalate”} (c) { x | x2 + 9 = 25}
3.
Which of the following sets are equal? (a) A = {2, –2} (b) B = { x | x + 1 = 3}
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Chapter 4: Sets and Counting (c) C = { x | x2 + 1 = 5} (d) D = { x | x is an integer and 1 < x < 3} (e} E = {–2, 2}
4.
List all the subsets of the set U = {Bob, Ed, Sue}.
5.
List all the subsets of the set U = {Bob, Ed, Sue, Carol}.
6.
Can you guess how many subsets there are for the set U = {Bob, Ed, Sue, Carol, Alice} without listing them all?
7.
List all the 3-element subsets of the set U = {a,b,c,d,e}.
8.
List all the 2-element subsets of the set U = {a,b,c,d,e,f}.
9.
If A = {1,2,4}, B = {2,4,5}, and the universal set U is given by U = {1,2,3,4,5,6}, what are the sets Ac and Bc?
10. If A = {0,1,2,3,4,5}, B = {0,2,3}, and C = {1,4,5}, indicate whether each of the following statements is true or false: (a) 0 ∈ A (b) 0 ! A (c) B ! A (d) {1,2} ∈ A (e) 1 ∉ B (f) {0} ! C (g) {0,2,3} ! B (h) {4,5,1} = C (i) {0} = Ø (j) Ø ∈ {0} (k) 0 ∈ Ø (l) 0 = Ø 11. If S = {a,b,c,d,e}, tell whether each of the following statements is true or false: (a) b ∈ S (e) {c,b} ! S (b) 2 ∉ S (f) 0 ∈ S (c) {b,c} ∈ S (g) 0 ! S (d) S ∈ S (h) {0} ! S 12. Give the complement of each subset listed in Problem 7. 13. Give the complement for each subset listed in Problem 8.
4.2 Set Operations and Venn Diagrams Arithmetic with numbers is described in terms of two basic operations which are addition and multiplication. [Subtraction and division can be described in terms of addition and multiplication if we know about negatives and inverses. For example a – b = a + (–b), and a/b = a × (1/b).] The two basic operations on sets are intersection and union. Also there are many parallels between the negative of a number and the complement of a set.
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The intersection of two sets consists of everything that the two sets have in common, or the overlap. The symbol for intersection is “∩”. The union is the set that is produced by taking everything in both the sets and lumping it together into one big set, and the symbol for union is “∪”. The important difference in A ∩ B and A ∪ B is in the criteria for membership: to qualify for membership in A ∩ B, an element must belong to both of the sets A and B, whereas to get into the set A ∪ B all that is required is to belong to at least one of the sets A or B. It might help you to think of intersection as corresponding to and and union as corresponding to or. To belong to A ∩ B an element must belong to A and B. To belong to A ∪ B it must belong to A or B.
Definition: (union and intersection) The intersection of two sets A and B is the set A ∩ B defined by A ∩ B = { x | x ∈ A and x ∈ B}. The union of two sets A and B is the set A ∪ B defined by A ∪ B = { x | x ∈ A or x ∈ B}.
Example 4.7. Suppose A = {1,3,5,7} and B = {1,2,3,4}. Then A ∩ B = {1,3} while the union of the two is given by A ∪ B = {1,2,3,4,5,7}. Notice that in listing the elements of the union, elements are listed only once even though they may belong to both sets. Venn diagrams provide a useful aid for visualizing the relation between sets. In a Venn diagram, each set is represented by a circle. A Venn diagram that would help in visualizing the relation between the sets A and B of Example 4.7 might look something like this: A
B
The left circle represents the set A, and the right one 5 1 represents the set B. The overlap or intersection of A and 2 B is the football shaped region in the middle, and we 3 7 show the elements 1 and 3 in this region because they are 4 the elements of A ∩ B. The numbers 5 and 7 are shown inside the A circle because they belong to A, but outside the B circle because they do not belong to B. Similarly, the numbers 2 and 4 are shown inside the B circle but outside the A circle. Often we will put a box around all the circles to indicate the universal set. In the Venn diagram that follows, the box represents the universal set U and the circles represent sets A and B that are subsets of the universal set. Notice that each region in the Venn diagram represents one of the four possibilities: A ∩ B includes the elements that are in A and in B. A ∩ Bc consists of elements that are
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in A and are not in B. Ac ∩ B consists of elements that are in B and are not in A. And finally Ac ∩ Bc consists of the elements that are not in A and not in B. In the Venn diagram this corresponds to the region that lies outside of both circles.
A
B c
c
A! B A! B A! B c
c
A! B
Venn diagrams are also useful when three sets are involved. For example, the Venn diagram below shows three circles labeled A, B, and C which represent three sets A, B, and C. The shaded region shown in the figure is the region that represents the set which is A ∩ B ∩ Cc .
B
A
C Venn diagrams are useful in helping to understand some of the important ways that the various set operations relate to each other. The table below catalogs the most important properties of the set operations.
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Properties of unions, intersections, and complements. A∪B=B∪A A∩B=B∩A
commutative law for unions commutative law for intersections A ∪ (B ∪ C) = (A ∪ B) ∪ C associative law for unions A ∩ (B ∩ C) = (A ∩ B) ∩ C associative law for intersections A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) distributive law for unions A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) distributive law for intersections c c c (A ∪ B) = A ∩ B DeMorganʼs law c c c (A ∩ B) = A ∪ B DeMorganʼs law
Venn diagrams can help you to understand why the above laws are true. We’ll illustrate this with the distributive law for intersections from the table above. The idea is to construct a Venn diagram for the set on each side of the equation, and then to compare them and see that, in fact, the two sets represent exactly the same regions in the Venn diagrams. This means that the two sets, though they may be written in quite different forms, represent sets that are logically the same. Example 4.8. Verify the distributive law for intersections using Venn diagrams. Solution: Let’s start out by working on a Venn diagram for A ∩ (B ∪ C). First we have to locate the region corresponding to B ∪ C. It is shown below:
B
A
C Now we have to visualize what we will have left when we intersect this with A. It leaves us with
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B
A
C So this finishes up the Venn diagram representing A ∩ (B ∪ C) in the sense that the shaded region in our picture is the region that represents the set A ∩ (B ∪ C). Now what about the other side of the equation which is (A ∩ B) ∪ (A ∩ C)? First check that A ∩ B and A ∩ C are represented by the regions shown in the following two Venn diagrams. B
A
B
A
C
C
Finally, we must form the union of the sets corresponding to these two regions. It should be apparent that when you do that you wind up with the same Venn diagram we obtained above, namely
B
A
C
.
Example 4.9. Suppose U = {a,b,c,d,e,f,g,h}, A = {a,b,e,f,g}, B = {a,c,e,h}, and C = {c,d,e}. Find each of the following sets: (a) A ∩ (B ∪ C)c (b) (Ac ∩ B) ∪ Cc (c) (Ac ∪ C) ∩ Bc.
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Solution: (a) B ∪ C = {a,c,d,e,h} so (B ∪ C)c = {b,f,g}. And in this case it is also true that A ∩ (B ∪ C)c = {b,f,g}. (b) Ac = {c,d,h} and so Ac ∩ B = {c,h}. And Cc = {a,b,f,g,h}. So (Ac ∩ B) ∪ Cc is just the union of these two sets, or {a,b,c,f,g,h} (c) Ac = {c,d,h}, so Ac ∪ C = {c,d,e,h}. Bc = {b,d,f,g}. When the intersection of these is formed we get (Ac ∪ C) ∩ Bc = {d}. Example 4.10. Let’s denote by M and B the students in a particular university that are studying mathematics and business. Write down the set that describes each of the following groups of students: (a) students studying math but not business (b) students studying both math and business (c) students studying either math or business (d) students who study neither math nor business (e) students who don’t study math and who don’t study business Solution: (a) M ∩ Bc, (b) M ∩ B, (c) M ∪ B, (d) (M ∪ B)c or Mc ∩ Bc (see the first of DeMorgan’s Laws), (e) same as (d). Example 4.11. This example is similar to the previous one except that three sets are involved. Let’s consider three categories of farmers: those who grow wheat, those who grow soybeans, and those who grow corn. These three sets of farmers will be denoted by W, S, and C respectively. Describe in terms of these three sets each of the following sets of farmers: (a) those who grow all three crops (b) those who grow at least one of the three crops (c) those who grow corn but don’t grow wheat or soybeans (d) those who grow wheat and one of the other two crops but not both (e) those who grow exactly two of the three crops Solution: (a) W ∩ S ∩ C, (b) W ∪ S ∪ C, (c) C ∩ Wc ∩ Sc or C ∩ (W ∪ S)c Parts (d) and (e) can be done in a variety of ways, none of which is very short. For example, the farmers who grow wheat and corn only are W ∩ C ∩ Sc, whereas those who grow wheat and soybeans only are W ∩ S ∩ Cc. What we want is the union of these two, so one possibility is just to write down exactly that: (W ∩ C ∩ Sc) ∪ (W ∩ S ∩ Cc). Another way to say the same thing would be to say W ∩ [(C ∩ Sc) ∪ (S ∩ Cc)]. An easy to understand but not particularly pretty way to write down the needed set in part (e) is simply (W ∩ S ∩ Cc) ∪ (Wc ∩ S ∩ C) ∪ (W ∩ Sc ∩ C).
Problems 1.
If U = {a,b,c,d,e,f,g,h,i,j}, A = {a,c,e,g,i}, B = {c,d,e,f}, and C = {a,b,c,d,e}, draw a Venn diagram and properly locate each element of the universal set U in the
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Chapter 4: Sets and Counting appropriate region of the Venn diagram.
2.
A = {1,2,3,4}, B = {1,3,5,7}, and C = {7,9,3}, and the universal set U is given by U = {1,2,3,4,5,6,7,8,9}. Locate all this information appropriately in a Venn diagram.
3.
Verify the two distributive laws for the sets A = {1,2,3,4}, B = {1,3,5,7}, and C = {7,9,3}.
4.
Verify that DeMorgan’s laws are true for the case in which A = {1,2,3,4}, B = {1,3,5,7}, and the universal set U = {1,2,3,4,5,6,7,8,9}.
5.
A = {1,2,3}, B = {1,3,5,7}, and C = {1,5,8,9}. These are subsets of the universal set U = {0,1,2,3,4,5,6,7,8,9}. List the elements in each of the following sets: (a) Ac ∪ (B ∩ Cc) (b) [A ∪ (B ∩ Cc)]c (c) Ac ∪ (Bc ∩ C)
6.
For each of the sets below, draw a Venn diagram with overlapping circles representing the sets A, B, and C and shade the region in the Venn diagram that represents the given set: (a) (A ∪ B) ∩ Cc (b) (A ∪ Bc) ∩ C (c) (Ac ∩ B) ∪ Cc
7.
In the Venn Diagram that follows, find all the points in each of these sets. (a) A ∪ (B ∩ Cc) (b) [A ∪ (B ∩ Cc)]c (c) Ac ∪ (B ∩ C)
A
t
a p
n
e
B
y
r
w k
g f
C 8.
Use Venn diagrams to determine whether each of the following set equations is true or false: (a) A ∩ (B ∪ Cc) = (A ∩ B) ∪ (Cc ∩ A) (b) (A ∩ Bc)c = A ∩ B (c) (A ∩ Bc)c = Ac ∩ B (d) (A ∪ B ∪ C)c = Ac ∩ Bc ∩ Cc
9.
Answer TRUE or FALSE. (a) A ∪ B ! B (c) A ∩ B ! A (e) B ! A ∩ B
(b) (A ∩ Bc) ∩ B = Ø (d) A ! A ∪ B (f) (A ∩ Bc) ∪ (A ∩ B) = A
10. Verify the distributive law for unions by using Venn diagrams.
Section 4.2: Set Operations and Venn Diagrams
137
11. Verify DeMorgan’s laws by using Venn diagrams. 12. A symposium on campus is attended by faculty, students, and guests. Among those in attendance, let M = males F = females S = students P = professors U = universal set of all people attending the symposium Write (in terms of these four sets) the set that describes each of the following: (a) the female students. (b) the males attending who are not students. (c) the guests (not students or professors). (d) females who are not professors. (e) people associated with the university (students or professors). 13. A church has 3 Bible study groups. Group A is studying Isaiah, Group B is studying the synoptic gospels, and Group C is studying the book of Romans. Some people are participating in more than one group. Using the letters A, B, and C to represent the three sets of people involved in the 3 groups, give the set that represents each of the following: (a) people studying Romans and the synoptic gospels. (b) people studying Romans but not Isaiah. (c) people that are in at least one of the groups. (d) people that are in at least two of the groups. (e) people that are in all three groups. (f) people that are in exactly two of the groups.
4.3 Counting techniques using Venn diagrams Often we are not so much interested in the specific elements that make up a set as in how many elements there are in a set. For example, a political candidate is more interested in how many people vote for him or her than in exactly who the people casting the votes are. In fact, a list of all the people voting might be so big as to be unmanageable, whereas the number giving the vote totals is simple to express. Similarly, a girl scout selling cookies might be more interested in the number of boxes she sells than in exactly who her customers are. And in computing your income tax, it is the number of dependents you have that is important in doing the computations, not what their names are.
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In this section we will look at some simple ways to use Venn diagrams as a way of helping to classify information pertaining to the number of items in a set.
Number of elements in a set The number of items in a set E will be denoted by n(E).
Example 4.12. Let’s denote by M the students studying math and by B the students studying business, and let’s suppose that there are 30 students studying math, 43 studying business, and 16 studying both math and business. All this information can be located in a Venn diagram as follows:
M
B 14
16
27
Since there are 30 studying math, and 16 of these study business, that leaves us with 14 who study math and who don’t study business. Similarly there are 27 who study business and not math. Altogether there are 14 + 16 + 27 = 57 students. Notice that 57 = 30 + 43 – 16, i.e. n(M ∪ B) = n(M) + n(B) – n(M ∩ B). These reason for this is that when we add n(M) and n(B) we have counted the students in the intersection of M and B twice, so if we subtract off n(M ∩ B) we will then have the total number of students in the two categories combined, i.e. n(M ∪ B).
Number of elements in the union of two sets. n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Example 4.13. In a group of 100 students who are majoring in business or accounting, 63 will be getting a degree in business and 47 will be getting a degree in accounting. How many will be getting a double major in both? Solution: If A = accounting majors and B = business majors, then n(A) = 47 and n(B) = 63, whereas n(A ∪ B) = 100. Therefore
Section 4.4: The Multiplication Principle and Tree Diagrams
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n(A ∩ B) = n(A) + n(B) – n(A ∪ B) = 47 + 63 – 100 = 10. The next several examples will illustrate the manner in which Venn diagrams may be used to categorize numerical information. Example 4.14. A survey of 500 households in Raleigh turned up the following information with regard to the readership of local newspapers: 342 households subscribe to the News & Observer 219 subscribe to the Raleigh Times 48 subscribe to the N.C. Independent 142 subscribe to both the News & Observer and the Times 41 subscribe to the News & Observer and the Independent 17 subscribe to the Times and the Independent 11 subscribe to all three Based on this data, answer the following questions: 1. 2. 3. 4.
How many households don’t subscribe to any of the three papers? How many households subscribe to exactly one paper? How many households subscribe to exactly two papers? How many households subscribe to the Times but do not subscribe to the News & Observer?
Solution: The key idea is simply to place all the data in a Venn diagram. Let’s use N, T, and I for the households that receive the News & Observer, the Times, and the Independent. The form in which the data is presented requires that we start processing the data at the bottom. The reason is that we initially can’t do anything with the information at the top. For instance, of the 342 households that receive the News & Observer, we don’t know to start with how many of these also read the Times or the Independent, so we don’t know how to divide these 342 up in the Venn diagram. If we start at the bottom, however, it’s clear where the 11 who read all three go, because they form the intersection of the three sets. The completed Venn diagram appears below:
T
N 170
131
71
11 30
6 1
80
! We obtain this picture by reasoning as follows. Since 17 households receive the Independent and Times, and 11 receive all three, that means that 11 of the 17 receive the
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News & Observer also and 6 receive the Independent and Times but do not receive the News & Observer. So we place the 6 in the figure as shown. Similarly, since 41 receive the News & Observer and Independent, when we subtract the 11 who receive all three that leaves 30 that receive the Independent and News & Observer and who do not receive the Times. Finally the 1 in the figure is obtained from the 48 households that subscribe to the Independent via the computation 48 – 30 – 11 – 6 = 1. The 131, 170, and 71 are determined similarly. And the 80 that is outside all three circles is obtained by adding up all the other numbers and subtracting from the 500 households. After all this information is tabulated in the Venn diagram, everything else is easy. The answers to the questions are obtained like this: 1. 80 2. 170 + 71 + 1 = 242 3. 131 + 30 + 6 = 167 4. 71 + 6 = 77 Example 4.15. A survey of a group of 200 people determined that 42 listen only to public radio stations, 74 listen only to privately owned radio stations, and 51 listen to both public and privately owned radio stations. Display this information in a Venn diagram. How many people who were included in the survey don’t listen to radio at all? Solution: This example is deliberately worded so as to be different from the last one. The key difference in this problem and the last is that in this example we are told that 42 listen only to public radio and 74 listen only to privately owned stations. So 42 and 74 are not the totals for the two categories of public and private. Rather 42 is the number who listen to public stations but not privately owned stations. And similarly 74 is the number who listen to privately owned stations but not to public ones. The correct Venn diagram is this one: Public
Private 42
51
74 33
If we think of A as the number of people surveyed who listen to public radio and B as the number of people surveyed who listen to privately owned radio, then the information we are being given is that n(A ∩ Bc) = 42 and n(B ∩ Ac) = 74. The number who don’t listen to radio at all is obtained by subtraction: 200 – 42 – 51 – 74 = 33.
Problems
Section 4.4: The Multiplication Principle and Tree Diagrams
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1.
A company produces 100 different types of cookies. Of these 63 require vanilla, 39 require chocolate, and 72 require nuts. Furthermore, 44 require both vanilla and nuts, 25 require both chocolate and nuts, and 20 require both vanilla and chocolate. 13 of the types of cookies require all 3 ingredients. Let V = types of cookies requiring vanilla C = types of cookies requiring chocolate N = types of cookies requiring nuts (a) Find n(C ∪ V) (b) Find n[Cc ∩ (Nc ∪ V)] (c) Find n[N ∪ (C ∪ V)c]
2.
In a group of 100 senior citizens, 56 are female, 24 are Methodists, and 17 of the females are Methodists. How many men are there in the group who are not Methodists?
3.
100 workers were asked if they were college graduates and if they belong to a union. 60 were not college graduates, 23 were nonunion college graduates, and 30 were union members. How many of the workers were neither college graduates nor union members?
4.
In a group of 100 people, 55 are college-educated. 25 of the college-educated watch news, 15 watch sports, and 10 watch both. 25 who did not go to college watch sports, 25 who did not go to college watch news, and 15 of them watch both. (a) How many who went to college do not watch either news or sports? (b) How many watch news but not sports? (c) How many who did not go to college watch news but not sports?
5.
In a group of 150 adults it is determined that 53 smoke cigarettes, 64 drink alcohol, and 83 watch TV. Furthermore, 34 both smoke and drink, 45 drink and watch TV, 42 watch TV and smoke, and 28 engage in all three vices (smoking, drinking, and TV). (a) Draw a Venn diagram to represent this data. (b) How many people watch TV but do not smoke? (c) How many neither smoke nor drink? (d) How many engage in exactly one of the three activities? (e) How many engage in exactly two of the three activities?
6.
A college has 700 students. A survey has determined that 200 read French 100 read German 175 read Spanish 30 read French and German 22 read French and Spanish 14 read German and Spanish 6 read all three languages (a) How many students read exactly 2 of these 3 languages? (b) How many read at least 1 of the 3 languages?
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7.
A survey of people on campus determined that the following numbers of people surveyed ate these meals at Baxley’s last Friday: 130 students ate breakfast 181 students ate lunch 270 students ate dinner 70 students ate breakfast and lunch 110 students ate breakfast and dinner 62 students ate lunch and dinner 58 students ate all three meals (a) How many ate at least one meal at Baxley's? (b) How many ate exactly one meal at Baxley's? (c) How many ate exactly two meals at Baxley's? (d) How many ate breakfast but not lunch?
8.
In a pollution study of 600 U.S. rivers, the following data was obtained: 190 were polluted with heavy metals 160 were polluted with phosphates 88 were polluted with crude oil 52 were polluted with crude oil and heavy metals 48 were polluted with crude oil and phosphates 100 were polluted with phosphates and heavy metals 22 were polluted with all three (a) Put this data into a Venn diagram. (b) How many rivers are polluted by at least one of the three compounds? (c) How many rivers are polluted by at exactly two of the three compounds? (d) How many rivers are not polluted by any of them?
9.
Sue conducted a survey among 100 students to determine their recreational interests. She learned that 52 enjoy hiking 39 enjoy swimming 42 enjoy jogging 19 enjoy all 3 activities 33 enjoy hiking and jogging 22 enjoy hiking and swimming 10 enjoy only swimming (a) Draw a Venn diagram to represent all this information (b) How many enjoy exactly one of the three activities? (c) How many enjoy at least one of the three activities? (d) How many enjoy exactly two activities?
10. In this section we found a formula for n(A ∪ B). See if you can find a formula for n(A ∪ B ∪ C).
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4.4 The Multiplication Principle and Tree Diagrams Example 4.16. Ed’s three clean shirts are white, blue, and yellow. He has two pairs of pants, one black and one gray. Considering only these articles of clothing, how many different ways can Ed dress? Solution: A complete listing of all the possible ways in which he can dress includes white shirt & black pants blue shirt & black pants yellow shirt & black pants white shirt & gray pants blue shirt & gray pants yellow shirt & gray pants So there are six different ways that Ed can dress. Why six ways? Because no matter which pairs of pants he wears, there are three possible choices for which shirt is worn, and 2 × 3 = 6. Alternatively, for any choice of shirt, either of the two pairs of pants may be worn, and 3 × 2 = 6. Another good way to illustrate the possibilities is with a tree diagram. A tree diagram that illustrates all the possible ways that Ed can dress might look as follows: black
gray black
white
blue
gray
black
gray
yellow
The reason this is called a tree diagram is that it bears a clear physical resemblance to a tree. The root at the bottom is where things start. At the first level in the tree we have listed all the possibilities for the choice of shirt. Then for each shirt, the choices for which pants may be worn continue at the next level in the tree. The six circles at the top are the leaves of the tree and correspond to the six possible ways of dressing. For example, the circle at the top left represents the possibility of wearing a white shirt and black pants. Similarly, the circle at the top right represents the choice of a yellow shirt and gray pants. The fact that three choices for one item and two choices for the other leads to 3 × 2 = 6 possibilities altogether is a special case of the multiplication principle.
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Multiplication Principle If one operation can be performed in r different ways and another operation can be performed in s different ways, then the number of different ways in which the two combined operations can be performed is r × s.
Example 4.17. In purchasing a car, Kim is choosing between 5 different models. In addition there are 4 banks that are willing to finance the purchase, and 4 insurance companies that offer insurance. If she chooses to purchase a car, finance it through one of the banks, and insure it with one of the insurance companies, how many different ways are there in which she might do all these things? Solution: 5 × 4 × 4 = 80. Example 4.18. A fast food restaurant offers hamburgers with the following trimmings: lettuce, tomato, pickles, onions, mayonnaise, mustard, and cheese. How many different styles of hamburger is it possible to order? Solution: What do you have to decide about lettuce? You either want it on your hamburger or you don’t. This represents two possibilities for lettuce. Similarly you choose to have tomato or else not to have tomato, two possibilities. This is true for each of seven options listed. Since for each of the seven there are two possibilities, the total number of options is 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128. There are 128 possible ways you can have your hamburger. Presumably all hamburgers include the hamburger itself and the bun. We haven’t listed those as options. But the 128 possibilities we have counted include all the possibilities as far as the optional ingredients are concerned. Some of these possibilities might taste pretty terrible. For example, one of the 128 possibilities would be a hamburger and bun with mayonnaise and mustard and nothing else. Example 4.19. The Reds and the Cubs have a series of three baseball games coming up. Draw a tree diagram to show all the possible ways that the series can go. Solution:
Section 4.4: The Multiplication Principle and Tree Diagrams
R
C
R
C R
R
C
C
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R
C C
R
R
C
The lowest level of branches in the tree represents the first game of the series. “R” means that the Reds win the game and “C” means the Cubs win. The second level of branches represents the second game, and the third level represents the third and final game. The eight circles represent the eight “possible outcomes” of the entire series. For example, the circle at the top left of the tree represents the outcome in which the Reds sweep the series, winning all three games. And the third circle (moving from left to right) represents the outcome in which the Reds win the first game, the Cubs win the second, and the Reds win the third. How could we see that the series has eight possible outcomes without drawing a tree? Simply think about the multiplication principle. There are to be three games, and for each game there are two possible winners. So the number of possibilities is 2 × 2 × 2 = 8. Example 4.20. Kim and Susan are going to play a tennis match. The winner of the match will be the first person to win two sets. Draw a tree diagram to represent all the possible outcomes for the tennis match. Solution: S
K S
K
K
S S
K
K
S
Notice that the fundamental difference in this situation and the previous example is that in the case of the tennis match we don’t know in advance exactly how many sets the match will last. There may be three sets played, or there may be only two (if the same player wins the first two sets). In the case of the baseball series we knew for certain at the beginning that exactly three games were going to be played. Whenever a set is played, one of the two players will win. But because in this problem we don’t know to start with exactly how many sets will be played, there is no obvious way to use the multiplication principle to see that the number of possible outcomes is six. From the tree diagram we see this simply by counting the six circles. Example 4.21. A restaurant has the following menu:
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entrees chicken beef pork
vegetables squash carrots potatoes corn
desserts pie cake ice cream
If a customer chooses one item from each category, how many different possible meals are there? Solution: This situation is ready-made for the multiplication principle. Three choices of entree times four choices of vegetable times three choices of dessert means that the total number of different possible meals is 3 × 4 × 3 = 36. It should be quite apparent to you how you might go about drawing a tree diagram to represent the 36 different meals. This would be a big tree, having 36 circles at the top. If all you want to know is the number of different possibilities, it is much simpler in this example to use the multiplication principle than to draw a tree.
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Example 4.22. Three good light bulbs and two defective ones have become mixed in a box. In order to find which are which, bulbs are going to be withdrawn from the box one at a time and tested until both bad ones or all three good ones have been found. Draw a tree diagram to show all the possible things that can happen when this process is carried out. Solution: B
G
G
B
G
G
G
B
G
B G
B
B
B
G G
B
B
Here “G” represents a good bulb and “B” represents a bad or defective bulb. The lowest level of branches represents the first bulb tested, etc. Each circle represents a possible outcome for the experiment. What outcome is represented, for example, by the circle farthest to the right at the top level (the fourth level) in the tree? It represents the outcome in which the first bulb tested is bad, the next two are good, and the fourth tested is bad. At this point the testing process stops because both defective bulbs have been found. What does the circle farthest to the right at the third level represent? It represents the outcome in which a bad bulb is found first, then a good bulb, then another bad one. The testing stops at this point because both bad bulbs have been found. At each stage in drawing the tree you should ask yourself whether the “stopping condition” has been reached. Remember that you stop only when all three good bulbs or both defective ones have been found. The ten circles represent the ten possible outcomes when the testing process is carried out. We have no way of knowing ahead of time (before doing the testing) exactly how many bulbs we will have to test. So again (as in Example 4.20) there is no obvious way to use the multiplication principle to see that there will be ten possibilities. Example 4.23. A pair of dice are rolled, one red and one green. How many different outcomes are there for this experiment? Solution: Since there are 6 possible numbers that might appear on the red die, as well as 6 numbers that might appear on the green die, this gives 6 × 6 = 36 possible results for the two dice. This method of counting assumes that we are keeping track of not only what numbers appear, but which dice they appear on. For instance, a “5” on red and a “3” on green is being counted as different from a “3” on red and a “5” on green. If
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we want to list the 36 possibilities, we might represent each outcome as an ordered pair of numbers, with the first number giving the number on the red die and the second the number on the green die. Then the outcome “(5,3)” would mean “5” on red and “3” on green. To make sure we understand what the 36 possibilities look like when written as an ordered pair of numbers, let’s list them all. {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} This example will reappear in the next chapter.
Problems 1.
An appliance manufacturer has 4 models of automatic washers and 3 models of automatic dryers. How many washer-dryer combinations can be bought?
2.
Diners at Brothers may select their pizzas from 3 varieties of crust, 2 varieties of sauces, and 15 varieties of toppings. How many pizzas are possible consisting of crust, sauce, and one topping?
3.
A coin is tossed 4 times and the sequence of heads and tails is recorded. (a) Determine the number of outcomes to this activity. (b) Exhibit all sequences by means of a tree diagram.
4.
How many 7-digit telephone numbers are possible (a) if the first digit must not be 0 or 1? (b) if the first digit can’t be 0 or 1 and the last 4 digits must all be odd digits.
5.
There are 4 buses and 3 vans departing from campus to the Vet School in the morning, and there are 3 buses and 2 vans operating on the return trip in the evening. In how many ways can a commuter make a round trip?
6.
In an election being held by a student organization, there are 6 candidates for President, 4 for Vice President, 5 for Secretary, and 6 for Treasurer. How many different outcomes are possible for the election?
7.
How many ways are there to fill out a multiple choice test of 5 questions if each question has 4 possible answers?
8.
A truck license plate consists of 2 letters of the alphabet followed by 4 digits. How
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many different license plates can be printed? 9.
How many different social security numbers are there in which the first digit is different from zero?
10. How many different 5-digit numbers can be formed from the digits 2, 3, 4, 5, 6, and 7 (a) if multiple “copies” of a digit may be used, as in the 5-digit number 37337? (b) if no digit may be used more than once? (c) How many of the possibilities you counted in part (a) are 5-digit numbers less than 50,000? (d) How many of the possibilities you counted in part (b) are 5-digit numbers less than 50,000? 11. The Dodgers and Giants are playing a best 3 out of 5 baseball series. (The series ends when someone has won 3 games.) (a) Draw a tree diagram to show all possible outcomes for the series. (b) How many possible outcomes are there? (c) How many of these possible outcomes show the series lasting 5 games? 12. How many different possible answer sheets could be turned in for a test consisting of 8 “true-false” questions? 13. A coin is tossed until a “head” is obtained or until 4 consecutive “tails” are obtained. (a) Draw a tree diagram to show all possible outcomes for this experiment. (b) How many outcomes are there? (c) How many of these outcomes show at least 2 “tails” being observed? 14. How many different answer sheets could there be for a multiple choice test of 6 questions in which (a) each question has four possible answers, and every question must be answered? (b) each question has four possible answers, and it is possible to leave questions unanswered (because wrong answers count against you more than unanswered questions do)? 15. A committee of 4 men and 3 women are meeting in a room. As the meeting ends, the people start filing out of the room one at a time. You stand and watch them come out of the room until (1) 2 consecutive people of the same sex leave the room, or (2) all 3 women have come out of the room. (a) Draw a tree diagram to represent all the possible things you could observe. (b) How many possibilities does your tree show? (c) How many possibilities are there in which you see at least 2 women leave the room? (d) How many possibilities are there in which you see at least 3 men leave the room?
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16. Ed has $2 and is going to play roulette. Each time he plays he either wins $1 or loses $1. He will stop playing when he goes bankrupt (he has $0), when he has won $2 (so he has $4), or when he has played a total of 5 times. Draw a tree diagram to show all the possible scenarios. (One good way to do this is to just keep track of how much money he has. Label the root of your tree $2, and then show branches going up to $1 and $3 showing his possible totals after he wins or loses the first round. Remember to stop whenever you reach $0, $4, or a total of 5 rounds.) 17. A meeting is to be addressed by people named Smith, Jones, Watt, Evans, and Fudd. In how many ways can the order of the speakers be arranged if Fudd insists on speaking before Watt? 18. Sarah has 2 nickels, 1 dime, and 1 quarter in her pocket. She pulls out coins one at a time until she has a total of at least 20¢. (a) Draw a tree diagram. (b) How many possibilities does your tree show in which at least 3 coins are removed from her pocket? (c) How many outcomes have her taking at least 35¢ from her pocket? 19. How many house designs are possible if a contractor offers 5 choices of roof, 3 choices of window design, and 7 choices of brick? 20. Kathy’s grandmother took Kathy shopping and they found 3 dresses, 4 sweaters, and 2 blouses that Kathy likes. (a) If her grandmother decides to buy her one of each (one dress, one sweater, and one blouse) for her birthday, how many possibilities are there for the articles of clothing her grandmother buys? (b) If her grandmother decides to buy only one article of clothing from among the things Kathy likes, then how many choices does the grandmother have? 21. A student has the following course requirements: (1) She must take either MA 103 or MA 114. (2) She must take 2 out of a selection of 3 humanities courses. (3) She must take 1 of 4 available language courses. How many different possibilities are there for the selection of courses she chooses?
4.5 Permutations and Combinations Permutations John, Mary, and Sue have bought tickets for three seats together for a basketball game. In how many different possible ways could they arrange themselves in the three seats? One possible seating arrangement is represented in the picture:
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John
Mary
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Sue
Of course, you don’t need to draw a fancy picture to represent a possible seating arrangement. If you like, you can think of the seats as a left seat, a middle seat, and a right seat, and then a seating arrangement simply means specifying who sits in each seat. Or you could think of the seats as numbered 1, 2, and 3, and an arrangement then amounts to assigning a seat number to each person. However you think of the situation, there are six possible seating arrangements. left John John Mary Mary Sue Sue
middle Mary Sue John Sue John Mary
right Sue Mary Sue John Mary John
How can you tell without listing the possibilities that there are six? Think of the seats from left to right. Any of the three might choose the left seat. But then one of the other two gets the middle seat. And the person left over gets the right seat. Proceeding in this way, the multiplication principle says that there are 3 choices (left seat) × 2 choices (middle seat) × 1 choice (right seat) = 6 choices. What if instead of 3 seats for 3 people there were 5 seats and 5 people? In this case there would be 5 × 4 × 3 × 2 × 1 = 120 possibilities.
Factorial If n is a positive integer, then n! is defined to be n! = n (n – 1) (n – 2) ... 1. Also, 0! is defined to be equal to 1.
“Factorials” give a simple way of writing some expressions that appear frequently. We have just seen that the number of different ways of arranging three people in a row is 3!, and the number of ways of arranging five in a row is 5!.
Definition: Permutations A permutation of a set of objects is a listing of the objects in some specified order. The number of different permutations of n different objects is given by n!.
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Example 4.24. A baseball team has nine players. How many different possible batting orders are there once it has been decided who the starting players will be? How many batting orders are possible if the pitcher is going to bat last? Solution: A “batting order” is simply a listing of the nine starting players in some specified order. So the number of possible “batting orders” is simply the number of different permutations of the nine players. There are therefore 9! = 362,880 different batting orders. If the pitcher is to bat last, then the question becomes, “How many different ways are there in which the other eight players can be ordered.” The answer is 8! = 40,320. Example 4.25. A baseball coach has nine players who can play “infield” (third base, shortstop, and second base). How many different ways are there in which he could choose his starting infield for an upcoming game? Solution: Suppose he chooses his third baseman first. Since there are 9 players to choose from, this gives him 9 possible choices. After choosing the third baseman, there are then 8 left to choose from in picking the shortstop. And then 7 remain when the second baseman is chosen. So, according to the multiplication principle, there are 9 × 8 × 7 = 504 ways to fill out his infield. Notice in this problem that what we are doing is considering ordered arrangements of some of the nine players but not all of them.
P(n,r) Given a collection of n objects, the number of different ways in which r of them can be lined up in a row is P(n,r) = n (n – 1) (n – 2) ... (n – r + 1). Such an ordered arrangement of r objects chosen from n objects is called a permutation of r objects chosen from n objects.
Combinations In many contexts a group of objects is being selected, but the order of selection is immaterial. For instance, suppose you are joining a tape club, and part of the incentive to join is that you get to select 4 tapes for only $1.98. If there are 20 tapes to choose from, then that means your selection can consist of any 4 of the 20 tapes being offered. In this case the only thing that is important to you is which 4 tapes you select, and there is no reason to think of the 4 tapes selected as forming any kind of ordered arrangement. Example 4.26. If you have five clean shirts and are going to pack two of them to go on a weekend trip, how many possibilities are there for the two that you select?
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Solution: Let’s identify the five shirts as A,B,C,D, and E. The question is this: In how many different ways could we choose 2 of these 5 items? A complete list of all the possible choices includes the following: {A,B} , {A,C} , {A,D} , {A,E} , {B,C} , {B,D} , {B,E} , {C,D} , {C,E} , {D,E} So there are 10 possibilities for the two shirts you pick. Notice that we have not listed {A,B} and {B,A} as two different possibilities. The reason is that the set {A,B} and the set {B,A} are identical. Sets do not have any order associated with their elements, so whether we write {A,B} or {B,A} makes no difference. Both notations represent the same set of shirts. Also notice that the number of possibilities listed above (which is 10) is not the same thing as P(5,2), and in fact P(5,2) = 20 rather than 10. P(5,2) would represent the number of different ordered arrangements of 2 objects that can be chosen from 5 objects, and P(5,2) = 5 × 4 because you could think of there being 5 choices for the first object chosen and 4 remaining choices for the second. If, for instance, we were counting how many ways there would be of choosing one clean shirt for Saturday and another for Sunday, we could reason that there are 5 to choose from for the Saturday shirt, and then 4 remaining ones to choose from for the Sunday shirt, giving 5 × 4 = 20 possibilities. In the original question, however, no distinction is made between the two shirts being chosen. The question is simply, “How many different ways are there in which a set of 2 objects can be chosen from an available collection of 5 objects?”
Definition: Combinations A set of r objects chosen from a set of n objects is called a combination of r objects chosen from n objects. The number of different combinations of r objects that may be chosen from n given objects is n! C(n,r) = r! (n – r)! So C(n,r) represents the number of different unordered sets of r objects that could be chosen from a set of n objects.
5! Notice that C(5,2) = 2! 3! = 10. This confirms the listing of the 10 possible choices for 2 shirts from 5 shirts in the previous example. Example 4.27. Jerry has seven compact discs that Michelle would like to borrow for a party. He has agreed to let her take four of them. In how many different ways could Michelle make her choice? Solution: This is simply a matter of choosing 4 objects from 7, so the number of possible ways of picking the compact discs would be
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Example 4.28. Michelle has 8 compact discs for her party. Before the party starts, she wants to decide which compact disc to play first, which to play second, and which to play third. In how many ways can she make these choices? Solution: Do you see why this is a permutations rather than a combinations question? She is not just choosing 3 compact discs from the 8. She is specifically choosing a first compact disc, a second compact disc, and a third compact disc. So the correct answer is not C(8,3) but instead is P(8,3) = 8 × 7 × 6 = 336. Example 4.29. A class consists of 14 boys and 17 girls. Four students from the class are to be selected to go on a trip. (a) How many different possibilities are there for the 4 students selected to make the trip? (b) If it has been decided that 2 boys and 2 girls will make the trip, then in how many different ways could the 4 students be selected? Solution: (a) Altogether there are 31 students. Since 4 are to be chosen, we should think of this question as asking how many possible ways there are to choose 4 items from 31 items. The answer is C(31,4) = 31,465. (b) Since there are 14 boys from which 2 will be selected for the trip, there are C(14,2) = 91 possibilities for which two boys are chosen. Similarly, since there are 17 girls there are C(17,2) = 136 ways in which the girls could be chosen. Where do we go from here? We use the multiplication principle. Since there are 91 ways to choose the boys for the trip and 136 ways to choose the girls, the number of ways to choose the boys and the girls for the trip is 91 × 136 = 12,376. Notice in this last example that we used two different ideas in part (b). We used the multiplication principle as well as the combinations formula. You should think of the multiplication principle and the combinations and permutations formulas as tools that can be used in a variety of situations. You have to think about the situation and decide which tools are applicable. And often more than one tool will be required to do the complete job. Example 4.30. A poker hand consists of 5 cards dealt from an ordinary 52-card deck. How many different poker hands are there? How many are there that give a “full house”? (A full house is a hand that contains 3 cards of one rank and 2 cards of some other rank, for example 3 aces and 2 sevens.) Solution: The first question is easy. There are 52 cards, and a 5-card hand consists of 5 of the 52 cards. The number of possibilities is C(52,5) = 2,598,960. For the second question we have to proceed more carefully. Let’s first consider an easier question. How many hands would there be that contain 3 aces and 2 sevens? The 3 aces would have to come from the 4 aces in the deck, and so there are C(4,3) = 4
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possibilities for the 3 aces. The 2 sevens would have to come from the 4 sevens in the deck, so there are C(4,2) = 6 possibilities. Then for the hand to have 3 aces and 2 sevens we have 4 possibilities for the aces and 6 possibilities for the sevens, so the multiplication principle says we have 4 × 6 = 24 ways to get 3 aces and 2 sevens. Now what about a full house? There are 13 different ranks of cards (twos, threes, ..., kings, aces). Once we specify the rank from which the 3 of a kind come from, that leaves 12 possibilities for the rank that the two of a kind will come from. Then the 3 of a kind can be chosen in C(4,3) = 4 ways from the cards of that rank, and the 2 of a kind can be chosen in C(4,2) = 6 ways from the cards of the second rank. The multiplication principle says then that the whole process has 13 × 12 × 4 × 6 = 3744 possibilities. So there are 3744 different poker hands that give a full house. Summary: There are 13 × 12 × 4 × 6 ways to get a full house because there are 13 possibilities for the rank from which the 3 of a kind are dealt, 12 possibilities then for the rank from which the 2 of a kind come, 4 possibilities for the 3 of a kind (once the rank is established), and 6 possibilities for the 2 of a kind (once the rank is established). Example 4.31. A concert orchestra knows 17 different pieces. For an evening program they will choose 3 of the pieces to perform. How many different programs are possible? Solution: We will interpret this as meaning that two programs are different if they consist of the same pieces performed in a different order. For instance, a program starting with Handel’s Water Music, followed by Brahms’ Serenade in D and concluding with Beethoven’s Fifth Symphony would be considered different from a program which began with the Serenade in D and was followed by Handel’s Water Music and then Beethoven’s Fifth Symphony. With this interpretation the answer to the question is simply P(17,3) = 4080, since we are counting ordered listings of 3 items from 17 available items. Suppose, however, that the conductor first decides to choose which 3 pieces to perform, and then decides what the order of performance should be. The 3 pieces can be chosen (without regard to order) in C(17,3) = 680 different ways. Then after they are chosen, the order of performance for the 3 pieces can be determined in P(3,3) = 3! = 6 different ways. The multiplication principle says then that the number of different programs would be C(17,3) × P(3,3) = 4080. Example 4.32. How many visibly different ways are there to arrange (in a row) the letters of the word “HANNA”? Solution: If we had a five different letters in the word, the answer would simply be 5! = 120. How to we compensate for the fact that the two N’s are indistinguishable, and similarly for the two A’s? Think about lining the five letters up in the five positions below: .
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Chapter 4: Sets and Counting The N’s might, for instance, occupy the following positions: N
N.
In fact, since there are 5 positions, there are C(5,2) = 10 different possibilities for which 2 positions the N’s will occupy. Now what about the two A’s? After the N’s are placed, there remain 3 unfilled blanks. The A’s will fill 2 of these. AN
AN.
The two positions to be occupied by the A’s can be picked in C(3,2) = 3 ways. Then the H gets the last vacant spot, so there is only C(1,1) = 1 way to position the H. The number of ways to arrange the letters therefore is C(5,2) × C(3,2) × C(1,1) = 10 × 3 × 1 = 30. Convince yourself that the order in which you consider the letters is irrelevant. For instance, if you first locate the H, there are C(5,1) possible ways of positioning it: H . Then if you position the 2 A’s, there are C(4,2) possibilities: AAH . And that leaves C(2,2) ways to position the N’s. NAAHN. The number of arrangements is therefore C(5,1) × C(4,2) × C(2,2) = 5 × 6 × 1 = 30. Both the expressions C(5,2) × C(3,2) × C(1,1) or C(5,1) × C(4,2) × C(2,2) can be simplified to 5! 2! 2! 1! . In fact, another way to understand why this is correct is to think of the 5! in the numerator as the number of ways to arrange 5 different items. But here we have to divide by 2! to compensate for the fact that the two N’s are indistinguishable, and again by 2! to compensate for the fact that the two A’s are indistinguishable. The 1! in the denominator corresponds to the one H and is irrelevant to the computation. Example 4.33. How many ways are there to arrange the letters in each of these words? (a) HANNAH (b) BARBARA (c) MISSISSIPPI
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6! Solution: (a) 2! 2! 2! = 90 because there are 2 H’s, 2 A’s, and 2 N’s. 7! (b) 2! 3! 2! = 210 because there are 2 B’s, 3 A’s, and 2 R’s. 11! (c) 4! 4! 2! = 34,650 because there are 4 I’s, 4 S’s, and 2 P’s. Each part of this question could, of course, have been answered using a strategy like the one used in Example 4.32. For instance, if in part (b) you first consider the B’s, then the A’s, and then the R’s, you get the answer C(7,2) × C(5,3) × C(2,2) = 21 × 10 × 1 = 210.
Problems 1.
Evaluate each of the following: (a) P(8,5)(b) P(52,50) (c) P(7,7) (d) C(8,5) (e) C(52,50) (f) C(7,7)
2.
Evaluate these: (a) C(20,0)
3.
A course has enrollment capacity of 30 students. Suppose 35 people want to enroll. How many different classes are possible?
4.
A quiz team of 5 students is to be chosen from 15 girls and 10 boys. How many teams can be selected (a) without regard to sex? (b) with 3 girls and 2 boys? (c) with at least 3 girls?
5.
A student is required to answer any 2 questions out of 5 questions on a math test. In how many ways can the student choose the 2 questions?
6.
How many different committees of 3 could be formed from 8 people? If Jane is one of the 8 people, how many different committees could be formed with Jane as a committee member?
7.
A baseball team has 6 outfielders. (a) In how many different ways can the manager make up a starting lineup for his outfield from this group? (In other words, in how many different ways can he pick a left fielder, a center fielder, and a right fielder?) (b) How many possibilities are there if one of the players can play only center field?
8.
The manager of a baseball team has chosen the 9 players for his starting lineup.
(b) C(0,0)
(c) C(1000,1000)
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Chapter 4: Sets and Counting How many different ways can he arrange his batting order if (a) either the pitcher or catcher bats last? (b) the pitcher bats last and the shortstop bats first? (c) the pitcher bats last, the shortstop bats first, and the 3 outfielders all bat consecutively.
9.
A conference of political leaders is attended by 2 people from Alabama, 3 from Georgia, 3 from New Jersey, 4 from California, and 2 from Michigan. A steering committee of 4 people is to be chosen from those attending. How many different steering committees are possible if (a) there are no restrictions? (b) exactly one member of the steering committee is to be from the South? (c) at most one member of the steering committee is to be from the South? (d) no member of the steering committee is to be from California?
10. The people attending the conference in Problem #9 have decided to elect a chairman and a vice-chairman. How many different results are possible for these elections if (a) there are no restrictions? (b) exactly one of the two elected is to be from the South? (c) neither of the two is to be from California? [In all parts of this question assume that the two offices are to be held by two different people.] 11. A supermarket chain is going to choose 3 of 5 possible sites to build stores, and 3 of 8 candidates to be managers of the stores. In how many ways can this be done? (Assume that the order of selection doesn’t matter, and that it doesn’t matter among the 3 managers chosen which of the 3 new stores they are assigned to. In other words, we are interested only in which sites are chosen and which people are chosen.) 12. How many different trios can be formed in a music class with 10 students? How many quartets can be formed? 13. A sociologist finds 20 high-income families, 30 middle-income families, and 15 low-income families in Isothermal, N.C. He decides to interview 3 families from each income level. In how many different ways can the families to be interviewed be chosen? 14. A basketball team has 5 centers, 5 forwards, and 6 guards. For a road game the coach will select 3 forwards, 4 guards, and 2 centers to make the trip. In how many different ways can the coach select the traveling team? 15. Four boys and three girls have bought tickets for 7 seats together in a theater. (a) In how many different ways can the 7 people seat themselves in the seven seats? (b) How many arrangements are possible if the boys all sit together and the girls all sit together?
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(c) How many arrangements are possible if every girl sits between two boys? 16. Nine people are having dinner. Someone proposes a toast, and everyone taps wine glasses with everyone else. How many “clinks” do you hear? 17. In how many ways can 7 students be assigned to sit in a row of 7 seats if (a) there are no restrictions? (b) two particular students insist on sitting next to each other? (c) two particular students refuse to sit next to each other? 18. Mary’s tape collection consists of 10 Mozart tapes, 8 Beethoven tapes, and 6 Schubert tapes. How many ways can she loan 5 tapes to her friend if (a) exactly 3 of the 5 are Beethoven tapes? (b) 2 of the 5 loaned are Mozart tapes and the other 3 are Schubert tapes? (c) all 5 are Schubert tapes? 19. How many different arrangements can be made of the letters in the word REFEREE? 20. The numbers C(n,r) are called binomial coefficients because they appear as the coefficients when a binomial expression of the form (a + b)n is expanded. For example, if you expand (a + b)4 you will find that the coefficients are C(4,0), C(4,1), C(4,2), C(4,3) and C(4,4). Carry out this expansion to see that this is true. 21. Seven points are plotted on a piece of paper, with no 3 of the points being in the same straight line. (a) If each pair of points is to be connected by a line segment, how many line segments must be drawn? (b) How many different triangles could possibly be formed having all its vertices chosen from among the 7 points? 22. S = {A, B, C, D, E, F}. (a) How many possibilities are there for a person’s initials if the first, middle, and last name all start with a letter from the set S? (b) How many possibilities are there if each initial comes from the set S and the three names start with three different letters? (c) How many possibilities are there in which one of the initials is “A” and the other two initials are two different letters (different from “A” and different from each other) from the set S? (d) How many possibilities are there if the letters are all different and are in alphabetical order (like “ACD” for example) and belong to the set S? 23. A man is going to visit 6 friends one afternoon. Three of the friends live in Durham, and three live in Raleigh. In how many different ways could he arrange the 6 visits so that he visits the three in Durham consecutively and the three in Raleigh consecutively?
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24. How many 5-card poker hands contain 2 aces, 2 kings, and a jack? 25. How many 5-card poker hands contain 2 pairs (for example, 2 aces, 2 kings, and a jack)? 26. 4 boys and 4 girls are participating in a 3-legged race. How many ways are there for the children to pair off if (a) the boys pair off with boys and the girls with girls? (b) each boy pairs off with a girl? (c) Could you answer these questions if there were 10 boys and 10 girls? 27. There are 5 permanent and 10 non-permanent members of the United Nations Security Council. A resolution is passed by the Council if it receives at least 9 favorable votes and if no permanent member votes against the measure. How many ways can a resolution be passed if (a) all members vote on the measure and there are no abstentions? (b) one particular permanent member and 3 particular non-permanent members abstain from voting? 28. Carol lives in California and she has 5 aunts who live in Asheville, Winston-Salem, Charlotte, Chicago, and Philadelphia. She plans to visit all of them. In how many different ways can she decide on the order in which she visits her aunts if (a) there are no restrictions? (b) she visits the aunt in Chicago last? (c) she visits the aunt in Chicago first and the aunt in Asheville last? (d) she visits all the aunts in North Carolina during a single trip to N.C.? (In other words, to save on air fare she decides to visit the aunts in North Carolina consecutively. For example, she doesn’t want to go to Asheville and then Philadelphia and then come back to N.C. to visit the aunt in Winston-Salem.) 29. In the Raleigh City Council elections, a voter is allowed to vote for up to two candidates in the at-large race. If 6 people are running, how many ways are there for a voter to cast a ballot, assuming that a voter has the option of voting for only one candidate or not voting for any at-large candidates? 30. A store clerk is hanging up shirts (all in the same style but different colors and sizes). There are 4 white shirts (1 small, 1 medium, 1 large, and 1 X-large), 4 red shirts (one in each of the 4 sizes), and 4 blue shirts (one of each size). How many ways are there to hang the shirts on the rack if (a) there are no restrictions? (b) shirts of the same size have to be kept together? (c) shirts of the same color are kept together? 31. Four people are going to seat themselves around a table. In how many different ways can they seat themselves? [Think of yourself as one of the four people, and
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consider how many different ways there are for the other people to position themselves relative to your seat.]
Chapter 4 Review Problems 1.
A woman has 7 skirts and 9 blouses. In how many ways can she select 3 skirts and 4 blouses to take on a trip?
2.
Construct a tree diagram to show all ways of giving 30¢ change if you have 1 quarter, 4 dimes, and 3 nickels. (Hint: In your tree let the first level of branches show the largest value coin given as change, etc.)
3.
How many five-digit numbers are there (the first digit cannot be zero) (a) in which the last digit is a 7? (b) in which the digit 7 does not appear at all? (c) in which both the first and last digits are even numbers {0,2,4,6,8}? (d) in which all the digits are even numbers? (e) in which no digit is repeated? (f) in which all the digits are even numbers and no digit is repeated?
4.
In a group of 30 students, 19 take math, 17 take music, 11 take history, 12 take math and music, 7 take math and history, 5 take music and history, and 2 take all three. (a) How many students are studying exactly 2 of the 3 subjects? (b) How many are studying no more than 1 of the 3 subjects?
5.
A club has 8 male and 6 female members. 3 members are to be selected to go on a trip. How many possibilities are there for the 3 people selected if (a) there are no restrictions? (b) 1 man and 2 women are to be selected? (c) at least 1 woman must be selected? (d) at least one person of each sex must be selected?
6.
A test consists of 3 multiple choice questions which may be answered a,b,c,d, or e. (a) How many ways can the three questions be answered? (b) How many ways can they be answered if you want to give a different answer to each question? (c) How many ways can they be answered so that at least one question is assigned the answer “b”?
7.
A pizza restaurant offers mushrooms, pepperoni, green peppers, onions, hamburger, and sausage as toppings. A customer can order as many toppings as desired. How many different pizzas can be ordered?
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8.
The outing club needs chairmen for the social, publicity, clean-up, and program committees. If 14 members are available and each is willing to head one of the committees (but not more than one), in how many different ways can the committee chairmen be chosen? [There are 2 possible interpretations to this question. Explain your interpretation by explaining exactly what things you are counting.]
9.
Five people are forming a line to buy tickets to a movie. (a) In how many different ways could they line up? (b) John and Ann are 2 of the 5 people. In how many different ways can the 5 line up if John and Ann want to be next to each other in the line? (c) How many arrangements are possible if John and Ann have had a fight and refuse to stand next to each other?
10. The numbers 1, 2, 3, and 4 are written on slips of white paper, and the numbers 5, 6, and 7 are written on slips of red paper. All the slips are dropped in a hat, and then two slips are drawn. (a) How many possibilities are there for the pair of numbers drawn? (Disregard order in all parts of this problem. We are interested only in which two slips we get, not the order in which we drew them.) (b) How many of these possibilities would consist of two white slips? (c) How many of these possibilities would consist of two red slips? (d) How many of the possibilities would consist of one slip of each color? (e) What is the relationship between the answer to part (a) and the answers to parts (b), (c), and (d)? (f) How many of the possibilities in part (a) include the slip numbered “5”? (g) How many of the possibilities in part (a) do not include the number “2”? 11. Martha is going to visit her aunts that live in Greensboro, Charlotte, Atlanta, Richmond, and Pittsburgh. (a) If there are no restrictions on the order in which she makes the visits, how many different orders are possible in which she could make the 5 visits? (b) How many ways can she plan the visits if she is going to visit her two aunts in North Carolina consecutively? (c) How many ways are possible if the aunt in Greensboro is to be the first of the 5 aunts visited and the aunt in Charlotte the last? 12. A survey of 100 families showed that 95 owned a dishwasher or a microwave or a freezer. 15 families owned all three appliances. 40 had only a dishwasher, and 14 had only a freezer. 17 had a microwave and freezer, 20 had a microwave and dishwasher, and 70 had a dishwasher. (a) How many had only one appliance? (b) How many had either a microwave or a freezer? {Read this problem carefully! The word “only” matters in each instance where it is used in the problem.}
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13. 4 math books, 3 physics books, and 2 history books are to be placed on a shelf. The math books must be kept together and must be arranged in a particular order because they are volumes 1-4 of a series. The physics books are to be kept together, but not necessarily in any particular order. The history books can go anywhere. How many ways are there in which the books can be arranged and still meet these criteria? 14. PIN numbers are 4-digit numbers. (a) If there are no restrictions, how many different PIN numbers can be formed? (b) How many different PIN numbers can be formed if none of the digits are repeated? (c) How many different PIN numbers can be formed if at least one of the digits has to be a seven? (d) How many different PIN numbers can be formed if none of the digits are repeated and the digits have to be written in decreasing order (for example, 7643)?
