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Sets and Counting

Module by: Rupinder Sekhon. E-mail the author

Summary: This chapter covers principles of sets and counting. After completing this chapter students should be able to: use set theory and venn diagrams to solve counting problems; use the multiplication axiom to solve counting problems; use permutations to solve counting problems; use combinations to solve counting problems; and use the binomial theorem to expand x+y^n.

Chapter Overview

In this chapter, you will learn to:

  1. Use set theory and Venn diagrams to solve counting problems.
  2. Use the Multiplication Axiom to solve counting problems.
  3. Use Permutations to solve counting problems.
  4. Use Combinations to solve counting problems.
  5. Use the Binomial Theorem to expand x+ynx+yn size 12{ left (x+y right ) rSup { size 8{n} } } {}.

Sets

In this section, we will familiarize ourselves with set operations and notations, so that we can apply these concepts to both counting and probability problems. We begin by defining some terms.

A set is a collection of objects, and its members are called the elements of the set. We name the set by using capital letters, and enclose its members in braces. Suppose we need to list the members of the chess club. We use the following set notation.

C = Ken, Bob, Tran, Shanti, Eric C = Ken, Bob, Tran, Shanti, Eric size 12{C= left lbrace "Ken, Bob, Tran, Shanti, Eric" right rbrace } {}
(1)

A set that has no members is called an empty set. The empty set is denoted by the symbol Ø.

Two sets are equal if they have the same elements.

A set AA size 12{A} {} is a subset of a set BB size 12{B} {} if every member of AA size 12{A} {} is also a member of BB size 12{B} {}.

Suppose C=Al, Bob, Chris, David, EdC=Al, Bob, Chris, David, Ed size 12{C= left lbrace "Al, Bob, Chris, David, Ed" right rbrace } {} and A=Bob, DavidA=Bob, David size 12{A= left lbrace "Bob, David" right rbrace } {}. Then AA size 12{A} {} is a subset of CC size 12{C} {}, written as ACAC size 12{A subseteq C} {}.

Every set is a subset of itself, and the empty set is a subset of every set.

Union Of Two Sets

Let AA size 12{A} {} and BB size 12{B} {} be two sets, then the union of AA size 12{A} {} and BB size 12{B} {}, written as ABAB size 12{A union B} {}, is the set of all elements that are either in AA size 12{A} {} or in BB size 12{B} {}, or in both AA size 12{A} {} and BB size 12{B} {}.

Intersection Of Two Sets

Let AA size 12{A} {} and BB size 12{B} {} be two sets, then the intersection of AA size 12{A} {} and BB size 12{B} {}, written as ABAB size 12{A intersection B} {}, is the set of all elements that are common to both sets AA size 12{A} {} and BB size 12{B} {}.

A universal set UU size 12{U} {} is the set consisting of all elements under consideration.

Complement of a Set

Let AA size 12{A} {} be any set, then the complement of set AA size 12{A} {}, written as AˉAˉ size 12{ { bar {A}}} {}, is the set consisting of elements in the universal set UU size 12{U} {} that are not in AA size 12{A} {}.

Disjoint Sets

Two sets AA size 12{A} {} and BB size 12{B} {} are called disjoint sets if their intersection is an empty set.

Example 1

Problem 1

List all the subsets of the set of primary colors red, yellow, bluered, yellow, blue size 12{ left lbrace "red, yellow, blue" right rbrace } {}.

Solution

The subsets are ∅, redred size 12{ left lbrace "red" right rbrace } {}, yellowyellow size 12{ left lbrace "yellow" right rbrace } {}, blueblue size 12{ left lbrace "blue" right rbrace } {}, red, yellowred, yellow size 12{ left lbrace "red, yellow" right rbrace } {}, red, bluered, blue size 12{ left lbrace "red, blue" right rbrace } {}, yellow, blueyellow, blue size 12{ left lbrace "yellow, blue" right rbrace } {}, red, yellow, bluered, yellow, blue size 12{ left lbrace "red, yellow, blue" right rbrace } {}

Note that the empty set is a subset of every set, and a set is a subset of itself.

Example 2

Problem 1

Let F=Aikman, Jackson, Rice, Sanders, YoungF=Aikman, Jackson, Rice, Sanders, Young size 12{F= left lbrace "Aikman, Jackson, Rice, Sanders, Young" right rbrace } {}, and B=Griffey, Jackson, Sanders, ThomasB=Griffey, Jackson, Sanders, Thomas size 12{B= left lbrace "Griffey, Jackson, Sanders, Thomas" right rbrace } {}. Find the intersection of the sets FF size 12{F} {} and BB size 12{B} {}.

Solution

The intersection of the two sets is the set whose elements belong to both sets. Therefore,

FB=Jackson, SandersFB=Jackson, Sanders size 12{F intersection B= left lbrace "Jackson, Sanders" right rbrace } {}
(2)

Example 3

Problem 1

Find the union of the sets FF size 12{F} {} and BB size 12{B} {} given as follows.

F=Aikman, Jackson, Rice, Sanders, YoungB=Griffey, Jackson, Sanders, ThomasF=Aikman, Jackson, Rice, Sanders, Young size 12{F= left lbrace "Aikman, Jackson, Rice, Sanders, Young" right rbrace } {}B=Griffey, Jackson, Sanders, Thomas size 12{B= left lbrace "Griffey, Jackson, Sanders, Thomas" right rbrace } {}
(3)

Solution

The union of two sets is the set whose elements are either in AA size 12{A} {} or in BB size 12{B} {} or in both AA size 12{A} {} and BB size 12{B} {}. Therefore

FB=Aikman, Griffey, Jackson, Rice, Sanders, Thomas, YoungFB=Aikman, Griffey, Jackson, Rice, Sanders, Thomas, Young size 12{F union B= left lbrace "Aikman, Griffey, Jackson, Rice, Sanders, Thomas, Young" right rbrace } {}
(4)

Observe that when writing the union of two sets, the repetitions are avoided.

Example 4

Problem 1

Let the universal set U=red, orange, yellow, green, blue, indigo, violetU=red, orange, yellow, green, blue, indigo, violet size 12{U= left lbrace "red, orange, yellow, green, blue, indigo, violet" right rbrace } {}, and P=red, yellow, blueP=red, yellow, blue size 12{P= left lbrace "red, yellow, blue" right rbrace } {}. Find the complement of PP size 12{P} {}.

Solution

The complement of a set PP size 12{P} {} is the set consisting of elements in the universal set UU size 12{U} {} that are not in PP size 12{P} {}. Therefore,

Pˉ=orange, green, indigo, violetPˉ=orange, green, indigo, violet size 12{ { bar {P}}= left lbrace "orange, green, indigo, violet" right rbrace } {}
(5)

To achieve a better understanding, let us suppose that the universal set UU size 12{U} {} represents the colors of the spectrum, and PP size 12{P} {} the primary colors, then PˉPˉ size 12{ { bar {P}}} {} represents those colors of the spectrum that are not primary colors.

Example 5

Problem 1

Let U=red, orange, yellow, green, blue, indigo, violetU=red, orange, yellow, green, blue, indigo, violet size 12{U= left lbrace "red, orange, yellow, green, blue, indigo, violet" right rbrace } {}, P=red, yellow, blueP=red, yellow, blue size 12{P= left lbrace "red, yellow, blue" right rbrace } {}, Q=red, greenQ=red, green size 12{Q= left lbrace "red, green" right rbrace } {}, and R=orange, green, indigoR=orange, green, indigo size 12{R= left lbrace "orange, green, indigo" right rbrace } {}. Find PQ¯RˉPQ¯Rˉ size 12{ {overline {P union Q}} intersection { bar {R}}} {}.

Solution

We do the problems in steps.

PQ=red, yellow, blue, greenPQ¯=orange, indigo, violetRˉ=red, yellow, blue, violetPQ¯Rˉ=violetPQ=red, yellow, blue, greenPQ¯=orange, indigo, violetRˉ=red, yellow, blue, violetPQ¯Rˉ=violet size 12{ matrix { P union Q= left lbrace "red, yellow, blue, green" right rbrace {} ## {overline {P union Q}} = left lbrace "orange, indigo, violet" right rbrace {} ## { bar {R}}= left lbrace "red, yellow, blue, violet" right rbrace {} ## {overline {P union Q}} intersection { bar {R}}= left lbrace "violet" right rbrace } } {}
(6)

We now use Venn diagrams to illustrate the relations between sets. In the late 1800s, an English logician named John Venn developed a method to represent relationship between sets. He represented these relationships using diagrams, which are now known as Venn diagrams. A Venn diagram represents a set as the interior of a circle. Often two or more circles are enclosed in a rectangle where the rectangle represents the universal set. To visualize an intersection or union of a set is easy. In this section, we will mainly use Venn diagrams to sort various populations and count objects.

Example 6

Problem 1

Suppose a survey of car enthusiasts showed that over a certain time period, 30 drove cars with automatic transmissions, 20 drove cars with standard transmissions, and 12 drove cars of both types. If every one in the survey drove cars with one of these transmissions, how many people participated in the survey?

Solution

We will use Venn diagrams to solve this problem.

Let the set AA size 12{A} {} represent those car enthusiasts who drove cars with automatic transmissions, and set SS size 12{S} {} represent the car enthusiasts who drove the cars with standard transmissions. Now we use Venn diagrams to sort out the information given in this problem.

Since 12 people drove both cars, we place the number 12 in the region common to both sets.

Figure 1
(a) (b) (c)
The Venn diagrams show that 12 people drove both automatic and standard transmission. While only 18 drove automatic and 8 drove standard. The Venn diagrams show that 12 people drove both automatic and standard transmission. While only 18 drove automatic and 8 drove standard. The Venn diagrams show that 12 people drove both automatic and standard transmission. While only 18 drove automatic and 8 drove standard.

Because 30 people drove cars with automatic transmissions, the circle A must contain 30 elements. This means x+12=30x+12=30, or x=18x=18. Similarly, since 20 people drove cars with standard transmissions, the circle B must contain 20 elements, or y+12=20y+12=20 which in turn makes y=8y=8.

Now that all the information is sorted out, it is easy to read from the diagram that 18 people drove cars with automatic transmissions only, 12 people drove both types of cars, and 8 drove cars with standard transmissions only. Therefore, 18+12+8=3818+12+8=38 people took part in the survey.

Example 7

Problem 1

A survey of 100 people in California indicates that 60 people have visited Disneyland, 15 have visited Knott's Berry Farm, and 6 have visited both. How many people have visited neither place?

Solution

The problem is similar to the one in Example 6.

Let the set DD size 12{D} {} represent the people who have visited Disneyland, and KK size 12{K} {} the set of people who have visited Knott's Berry Farm.

Figure 2
(a) (b)
The figure shows that everything in the box is equals U. The Venn diagram shows that D has 54 and that K has 9, while D and K have 6. While x is equal to 31 and is not included in D or K.The figure shows that everything in the box is equals U. The Venn diagram shows that D has 54 and that K has 9, while D and K have 6. While x is equal to 31 and is not included in D or K.

We fill the three regions associated with the sets D and K in the same manner as before. Since 100 people participated in the survey, the rectangle representing the universal set U must contain 100 objects. Let x represent those people in the universal set that are neither in the set D nor in K. This means 54+6+9+x=10054+6+9+x=100, or x=31x=31.

Therefore, there are 31 people in the survey who have visited neither place.

Example 8

Problem 1

A survey of 100 exercise conscious people resulted in the following information:

  • 50 jog, 30 swim, and 35 cycle
  • 14 jog and swim
  • 7 swim and cycle
  • 9 jog and cycle
  • 3 people take part in all three activities
  1. How many jog but do not swim or cycle?
  2. How many take part in only one of the activities?
  3. How many do not take part in any of these activities?
Solution

Let JJ size 12{J} {} represent the set of people who jog, SS size 12{S} {} the set of people who swim, and CC size 12{C} {} who cycle.

In using Venn diagrams, our ultimate aim is to assign a number to each region. We always begin by first assigning the number to the innermost region and then working our way out.

Figure 3
(a) (b) (c)
The figure shows that everything in the square is equal to U. The Venn diagram shows that S,J and C have three in common, and J and C have 6 in common. While S and J have 11 in common and S and C have 4. Where S only has 12, C only has 22, and J only has 30. The remaining 12 have no relation to S, J, or C. The figure shows that everything in the square is equal to U. The Venn diagram shows that S,J and C have three in common, and J and C have 6 in common. While S and J have 11 in common and S and C have 4. Where S only has 12, C only has 22, and J only has 30. The remaining 12 have no relation to S, J, or C. The figure shows that everything in the square is equal to U. The Venn diagram shows that S,J and C have three in common, and J and C have 6 in common. While S and J have 11 in common and S and C have 4. Where S only has 12, C only has 22, and J only has 30. The remaining 12 have no relation to S, J, or C.

We place a 3 in the innermost region of Figure 3(a) because it represents the number of people who participate in all three activities. Next we compute x, y and z.

Since 14 people jog and swim, x+3=14x+3=14, or x=11x=11.

The fact that 9 people jog and cycle results in y+3=9y+3=9, or y=6y=6.

Since 7 people swim and cycle, z+3=7z+3=7, or z=4z=4.

This information is depicted in Figure 3(b).

Now we proceed to find the unknowns m, n and p.

Since 50 people jog, m+11+6+3=50m+11+6+3=50, or m=30m=30.

Thirty people swim, therefore, n+11+4+3=30n+11+4+3=30, or n=12n=12.

Thirty five people cycle, therefore, p+6+4+3=35p+6+4+3=35, or p=22p=22.

By adding all the entries in all three sets, we get a sum of 88. Since 100 people were surveyed, the number inside the universal set but outside of all three sets is 100 – 88, or 12.

In Figure 3(c), the information is sorted out, and the questions can readily be answered.

  1. The number of people who jog but do not swim or cycle is 30.
  2. The number who take part in only one of these activities is 30+12+22=6430+12+22=64.
  3. The number of people who do not take part in any of these activities is 12.

Tree Diagrams and the Multiplication Axiom

In this chapter, we are trying to develop counting techniques that will be used in the (Reference) to study probability. One of the most fundamental of such techniques is called the Multiplication Axiom. Before we introduce the multiplication axiom, we first look at some examples.

Example 9

Problem 1

If a woman has two blouses and three skirts, how many different outfits consisting of a blouse and a skirt can she wear?

Solution

Suppose we call the blouses b1 and  b2,  and skirts  s1,  s2,  and   s3b1 size 12{b rSub { size 8{1} } } {} and  b2 size 12{b rSub { size 8{2} } } {},  and skirts  s1 size 12{s rSub { size 8{1} } } {},  s2 size 12{s rSub { size 8{2} } } {},  and  s3 size 12{s rSub { size 8{3} } } {}.

We can have the following six outfits.

b1s1,b1s2,b1s3,b2s1,b2s2,b2s3b1s1 size 12{b rSub { size 8{1} } s rSub { size 8{1} } } {},b1s2 size 12{b rSub { size 8{1} } s rSub { size 8{2} } } {},b1s3 size 12{b rSub { size 8{1} } s rSub { size 8{3} } } {},b2s1 size 12{b rSub { size 8{2} } s rSub { size 8{1} } } {},b2s2 size 12{b rSub { size 8{2} } s rSub { size 8{2} } } {},b2s3 size 12{b rSub { size 8{2} } s rSub { size 8{3} } } {}
(7)

Alternatively, we can draw a tree diagram:

Figure 4
The tree diagram illustrates all of the six possible outfits.

The tree diagram gives us all six possibilities. The method involves two steps. First the woman chooses a blouse. She has two choices: blouse one or blouse two. If she chooses blouse one, she has three skirts to match it with; skirt one, skirt two, or skirt three. Similarly if she chooses blouse two, she can match it with each of the three skirts, again. The tree diagram helps us visualize these possibilities.

The reader should note that the process involves two steps. For the first step of choosing a blouse, there are two choices, and for each choice of a blouse, there are three choices of choosing a skirt. So altogether there are 23=623=6 possibilities.

If, in the above example, we add the shoes to the outfit, we have the following problem.

Example 10

Problem 1

If a woman has two blouses, three skirts, and two pumps, how many different outfits consisting of a blouse, a skirt, and a pair of pumps can she wear?

Solution

Suppose we call the blouses b1b1 size 12{b rSub { size 8{1} } } {} and b2b2 size 12{b rSub { size 8{2} } } {}, the skirts s1s1 size 12{s rSub { size 8{1} } } {}, s2s2 size 12{s rSub { size 8{2} } } {}, and s3s3 size 12{s rSub { size 8{3} } } {}, and the pumps p1p1 size 12{p rSub { size 8{1} } } {}, and p2p2 size 12{p rSub { size 8{2} } } {}.

The following tree diagram results.

Figure 5
The tree diagrams shows the possible outfits.

We count the number of branches in the tree, and see that there are 12 different possibilities. This time the method involves three steps. First, the woman chooses a blouse. She has two choices: blouse one or blouse two. Now suppose she chooses blouse one. This takes us to step two of the process which consists of choosing a skirt. She has three choices for a skirt, and let us suppose she chooses skirt two. Now that she has chosen a blouse and a skirt, we have moved to the third step of choosing a pair of pumps. Since she has two pairs of pumps, she has two choices for the last step. Let us suppose she chooses pumps two. She has chosen the outfit consisting of blouse one, skirt two, and pumps two, or b1s2p2b1s2p2. By looking at the different branches on the tree, one can easily see the other possibilities.

The important thing to observe here, again, is that this is a three step process. There are two choices for the first step of choosing a blouse. For each choice of a blouse, there are three choices of choosing a skirt, and for each combination of a blouse and a skirt, there are two choices of selecting a pair of pumps. All in all, we have 232=12232=12 different possibilities.

The tree diagrams help us visualize the different possibilities, but they are not practical when the possibilities are numerous. Besides, we are mostly interested in finding the number of elements in the set and not the actual possibilities. But once the problem is envisioned, we can solve it without a tree diagram. The two examples we just solved may have given us a clue to do just that.

Let us now try to solve Example 10 without a tree diagram. Recall that the problem involved three steps: choosing a blouse, choosing a skirt, and choosing a pair of pumps. The number of ways of choosing each are listed below.

Table 1
The Number of ways of choosing a blouse The number of ways of choosing a skirt The number of ways of choosing pumps
2 3 2

By multiplying these three numbers we get 12, which is what we got when we did the problem using a tree diagram.

The procedure we just employed is called the multiplication axiom.

THE MULTIPLICATION AXIOM

If a task can be done in m ways, and a second task can be done in n ways, then the operation involving the first task followed by the second can be performed in m· nm·n ways.

The general multiplication axiom is not limited to just two tasks and can be used for any number of tasks.

Example 12

Problem 1

A truck license plate consists of a letter followed by four digits. How many such license plates are possible?

Solution

Since there are 26 letters and 10 digits, we have the following choices for each.

Table 2
Letter Digit Digit Digit Digit
26 10 10 10 10

Therefore, the number of possible license plates is 2610101010=260,0002610101010=260,000.

Example 13

Problem 1

In how many different ways can a 3-question true-false test be answered?

Solution

Since there are two choices for each question, we have

Table 3
Question 1 Question 2 Question 3
2 2 2

Applying the multiplication axiom, we get 222=8222=8 size 12{2 cdot 2 cdot 2=8} {} different ways.

We list all eight possibilities below.

TTT, TTF, TFT, TFF, FTT, FTF, FFT, FFF

The reader should note that the first letter in each possibility is the answer corresponding to the first question, the second letter corresponds to the answer to the second question and so on. For example, TFF, says that the answer to the first question is given as true, and the answers to the second and third questions false.

Example 14

Problem 1

In how many different ways can four people be seated in a row?

Solution

Suppose we put four chairs in a row, and proceed to put four people in these seats.

There are four choices for the first chair we choose. Once a person sits down in that chair, there are only three choices for the second chair, and so on. We list as shown below.

Table 4
4 3 2 1

So there are altogether 4321=244321=24 size 12{4 cdot 3 cdot 2 cdot 1="24"} {} different ways.

Example 15

Problem 1

How many three-letter word sequences can be formed using the letters A,B,CA,B,C size 12{ left lbrace A,B,C right rbrace } {} if no letter is to be repeated?

Solution

The problem is very similar to Example 14.

Imagine a child having three building blocks labeled AA size 12{A} {}, BB size 12{B} {}, and CC size 12{C} {}. Suppose he puts these blocks on top of each other to make word sequences. For the first letter he has three choices, namely AA size 12{A} {}, BB size 12{B} {}, or CC size 12{C} {}. Let us suppose he chooses the first letter to be a BB size 12{B} {}, then for the second block which must go on top of the first, he has only two choices: AA size 12{A} {} or CC size 12{C} {}. And for the last letter he has only one choice. We list the choices below.

Table 5
3 2 1

Therefore, 6 different word sequences can be formed.

Finally, we'd like to illustrate this with a tree diagram.

Figure 6
A Tree diagram showing all the possible outcomes.

All six possibilities are displayed in the tree diagram.

Permutations

In Example 15, we were asked to find the word sequences formed by using the letters A,B,CA,B,C size 12{ left lbrace A,B,C right rbrace } {} if no letter is to be repeated. The tree diagram gave us the following six arrangements.

ABCABC size 12{ ital "ABC"} {}, ACBACB size 12{ ital "ACB"} {}, BACBAC size 12{ ital "BAC"} {}, BCABCA size 12{ ital "BCA"} {}, CABCAB size 12{ ital "CAB"} {}, and CBACBA size 12{ ital "CBA"} {},

Arrangements like these, where order is important and no element is repeated, are called permutations.

Permutations

A permutation of a set of elements is an ordered arrangement where each element is used once.

Example 17

Problem 1

How many three-letter word sequences can be formed using the letters A,B,C,DA,B,C,D size 12{ left lbrace A,B,C,D right rbrace } {}?

Solution

There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third.

Table 6
4 3 2

Applying the multiplication axiom, we get 432=24432=24 size 12{4 cdot 3 cdot 2="24"} {} different arrangements.

Example 18

Problem 1

How many permutations of the letters of the word ARTICLE have consonants in the first and last positions?

Solution

In the word ARTICLE, there are 4 consonants.

Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. We show as follows:

Table 7
4           3

Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions.

So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on. We get

Table 8
4 5 4 3 2 1 3

So the total permutations are 4543213=14404543213=1440 size 12{4 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1 cdot 3="1440"} {}.

Example 19

Problem 1

Given five letters A,B,C,D,EA,B,C,D,E size 12{ left lbrace A,B,C,D,E right rbrace } {}. Find the following:

  1. The number of four-letter word sequences.

  2. The number of three-letter word sequences.

  3. The number of two-letter word sequences.

Solution

The problem is easily solved by the multiplication axiom, and answers are as follows:

  1. a. The number of four-letter word sequences is 5432=1205432=120 size 12{5 cdot 4 cdot 3 cdot 2="120"} {}.

  2. b. The number of three-letter word sequences is 543=60543=60 size 12{5 cdot 4 cdot 3="60"} {}.

  3. c. The number of two-letter word sequences is 54=2054=20 size 12{5 cdot 4="20"} {}.

We often encounter situations where we have a set of nn size 12{n} {} objects and we are selecting rr size 12{r} {} objects to form permutations. We refer to this as permutations of nn size 12{n} {} objects taken rr size 12{r} {} at a time, and we write it as nPrnPr size 12{n"Pr"} {}.

Therefore, Example 19 can also be answered as listed below.

  1. The number of four-letter word sequences is 5P4=1205P4=120 size 12{5P4="120"} {}.

  2. The number of three-letter word sequences is 5P3=605P3=60 size 12{5P3="60"} {}.

  3. The number of two-letter word sequences is 5P2=205P2=20 size 12{5P2="20"} {}.

Before we give a formula for nPrnPr size 12{n"Pr"} {}, we'd like to introduce a symbol that we will use a great deal in this as well as in (Reference).

Factorial

n!=nn1n2n3321n!=nn1n2n3321 size 12{n!=n left (n - 1 right ) left (n - 2 right ) left (n - 3 right ) dotsaxis 3 cdot 2 cdot 1} {}.

Where nn size 12{n} {} is a natural number.

0 ! = 1 0 ! = 1 size 12{0!=1} {}
(8)

Now we define nPrnPr size 12{n"Pr"} {}.

The Number of Permutations of n n size 12{n} {} Objects Taken r r size 12{r} {} at a Time

nPr=nn1n2n3nr+1nPr=nn1n2n3nr+1 size 12{n"Pr"=n left (n - 1 right ) left (n - 2 right ) left (n - 3 right ) dotsaxis left (n - r+1 right )} {}, or
nPr=n!nr!nPr=n!nr! size 12{n"Pr"= { {n!} over { left (n - r right )!} } } {}

Where nn size 12{n} {} and rr size 12{r} {} are natural numbers.

The reader should become familiar with both formulas and should feel comfortable in applying either.

Example 22

Problem 1

Compute the following using both formulas.

  1. 6P36P3 size 12{6P3} {}

  2. 7P27P2 size 12{7P2} {}

Solution

We will identify nn size 12{n} {} and rr size 12{r} {} in each case and solve using the formulas provided.

  1. 6P3=654=1206P3=654=120 size 12{6P3=6 cdot 5 cdot 4="120"} {}, alternately 6P3=6!63!=6!3!=654321321=1206P3=6!63!=6!3!=654321321=120 size 12{6P3= { {6!} over { left (6 - 3 right )!} } = { {6!} over {3!} } = { {6 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1} over {3 cdot 2 cdot 1} } ="120"} {}

  2. 7P2=76=427P2=76=42 size 12{7P2=7 cdot 6="42"} {}, or 7P2=7!5!=765432154321=427P2=7!5!=765432154321=42 size 12{7P2= { {7!} over {5!} } = { {7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1} over {5 cdot 4 cdot 3 cdot 2 cdot 1} } ="42"} {}

Next we consider some more permutation problems to get further insight into these concepts.

Example 23

Problem 1

In how many different ways can 4 people be seated in a straight line if two of them insist on sitting next to each other?

Solution

Let us suppose we have four people AA size 12{A} {}, BB size 12{B} {}, CC size 12{C} {}, and DD size 12{D} {}. Further suppose that AA size 12{A} {} and BB size 12{B} {} want to sit together. For the sake of argument, we tie AA size 12{A} {} and BB size 12{B} {} together and treat them as one person.

The four people are ABAB size 12{ matrix { ital "AB" } } {}CDCD size 12{ matrix { ital "CD" } } {}. Since ABAB size 12{ matrix { ital "AB" } } {} is treated as one person, we have the following possible arrangements.

ABCD, ABDC, CABD, DABC, CDAB, DCABABCD size 12{ matrix { ital "AB" } matrix { ital "CD" } } {},ABDC size 12{ matrix { ital "AB" } matrix { ital "DC" } } {},{}CABD size 12{ matrix { C } matrix { ital "AB" } matrix { D } } {},DABC size 12{ matrix { D } matrix { ital "AB" } matrix { C } } {},CDAB size 12{ matrix { ital "CD" } matrix { ital "AB" } } {},DCAB size 12{ matrix { ital "DC" } matrix { ital "AB" } } {}
(9)

Note that there are six more such permutations because AA size 12{A} {} and BB size 12{B} {} could also be tied in the order BABA size 12{ ital "BA"} {}. And they are

BACD,BADC,CBAD,DBAC,CDBA, DCBABACD size 12{ matrix { ital "BA" } matrix { ital "CD" } } {},BADC size 12{ matrix { ital "BA" } matrix { ital "DC" } } {},CBAD size 12{ matrix { C } matrix { ital "BA" } matrix { D } } {},DBAC size 12{ matrix { D } matrix { ital "BA" } matrix { C } } {},CDBA size 12{ matrix { ital "CD" } matrix { ital "BA" } } {},DCBA size 12{ matrix { ital "DC" } matrix { ital "BA" } } {}
(10)

So altogether there are 12 different permutations.

Let us now do the problem using the multiplication axiom.

After we tie two of the people together and treat them as one person, we can say we have only three people. The multiplication axiom tells us that three people can be seated in 3!3! size 12{3!} {} ways. Since two people can be tied together 2!2! size 12{2!} {} ways, there are 3!2!=123!2!=12 size 12{3!2!="12"} {} different arrangements.

Example 24

Problem 1

You have 4 math books and 5 history books to put on a shelf that has 5 slots. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books?

Solution

We first do the problem using the multiplication axiom.

Since the math books go in the first three slots, there are 4 choices for the first slot, 3 for the second and 2 for the third. The fourth slot requires a history book, and has five choices. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The choices are shown below.

Table 9
4 3 2 5 4

Therefore, the number of permutations are 43254=48043254=480.

Alternately, we can see that 432432 is really same as 4P34P3, and 5454 is 5P25P2.

So the answer can be written as (P3)(5P2)=480(P3)(5P2)=480.

Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are 5P25P2 permutations of history books that can be placed in the last two slots. Hence the multiplication axiom applies, and we have the answer (4P3)(5P2)(4P3)(5P2).

We summarize.

  1. Permutations

    A permutation of a set of elements is an ordered arrangement where each element is used once.

  2. Factorial

    n!=nn1n2n3321n!=nn1n2n3321 size 12{n!=n left (n - 1 right ) left (n - 2 right ) left (n - 3 right ) dotsaxis 3 cdot 2 cdot 1} {}.

    Where nn size 12{n} {} is a natural number.

    0 ! = 1 0 ! = 1 size 12{0!=1} {}
    (11)
  3. Permutations of n n size 12{n} {} Objects Taken r r size 12{r} {} at a Time

    nPr=nn1n2n3nr+1nPr=nn1n2n3nr+1 size 12{n"Pr"=n left (n - 1 right ) left (n - 2 right ) left (n - 3 right ) dotsaxis left (n - r+1 right )} {}, or nPr=n!nr!nPr=n!nr! size 12{n"Pr"= { {n!} over { left (n - r right )!} } } {}

    Where nn size 12{n} {} and rr size 12{r} {} are natural numbers.

Circular Permutations and Permutations with Similar Elements

Section Overview

In this section we will address the following two problems.

  1. In how many different ways can five people be seated in a circle?
  2. In how many different ways can the letters of the word MISSISSIPPI be arranged?

The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements.

Circular Permutations

Suppose we have three people named AA size 12{A} {}, BB size 12{B} {}, and CC size 12{C} {}. We have already determined that they can be seated in a straight line in 3!3! size 12{3!} {} or 6 ways. Our next problem is to see how many ways these people can be seated in a circle. We draw a diagram.

Figure 7
This figure shows how the people named A, B, and C can sit in a circle in different ways.

It happens that there are only two ways we can seat three people in a circle. This kind of permutation is called a circular permutation. In such cases, no matter where the first person sits, the permutation is not affected. Each person can shift as many places as they like, and the permutation will not be changed. Imagine the people on a merry-go-round; the rotation of the permutation does not generate a new permutation. So in circular permutations, the first person is considered a place holder, and where he sits does not matter.

Circular Permutations

The number of permutations of nn size 12{n} {} elements in a circle is n1!n1! size 12{ left (n - 1 right )!} {}

Example 26

Problem 1

In how many different ways can five people be seated at a circular table?

Solution

We have already determined that the first person is just a place holder. Therefore, there is only one choice for the first spot. We have

Table 10
1 4 3 2 1

So the answer is 24.

Example 27

Problem 1

In how many ways can four couples be seated at a round table if the men and women want to sit alternately?

Solution

We again emphasize that the first person can sit anywhere without affecting the permutation.

So there is only one choice for the first spot. Suppose a man sat down first. The chair next to it must belong to a woman, and there are 4 choices. The next chair belongs to a man, so there are three choices and so on. We list the choices below.

Table 11
1 4 3 3 2 2 1 1

So the answer is 144.

Now we address the second problem.

Permutations with Similar Elements

Let us determine the number of distinguishable permutations of the letters ELEMENT.

Suppose we make all the letters different by labeling the letters as follows.

E 1 LE 2 ME 3 NT E 1 LE 2 ME 3 NT size 12{E rSub { size 8{1} } ital "LE" rSub { size 8{2} } ital "ME" rSub { size 8{3} } ital "NT"} {}
(12)

Since all the letters are now different, there are 7!7! size 12{7!} {} different permutations.

Let us now look at one such permutation, say

LE 1 ME 2 NE 3 T LE 1 ME 2 NE 3 T size 12{ ital "LE" rSub { size 8{1} } ital "ME" rSub { size 8{2} } ital "NE" rSub { size 8{3} } T} {}
(13)

Suppose we form new permutations from this arrangement by only moving the E's. Clearly, there are 3!3! size 12{3!} {} or 6 such arrangements. We list them below.

LE 1 ME 2 NE 3 T LE 1 ME 2 NE 3 T size 12{ ital "LE" rSub { size 8{1} } ital "ME" rSub { size 8{2} } ital "NE" rSub { size 8{3} } T} {}
(14)
LE 1 ME 3 NE 2 T LE 1 ME 3 NE 2 T size 12{ ital "LE" rSub { size 8{1} } ital "ME" rSub { size 8{3} } ital "NE" rSub { size 8{2} } T} {}
(15)
LE 2 ME 1 NE 3 T LE 2 ME 1 NE 3 T size 12{ ital "LE" rSub { size 8{2} } ital "ME" rSub { size 8{1} } ital "NE" rSub { size 8{3} } T} {}
(16)
LE 3 ME 3 NE 1 T LE 3 ME 3 NE 1 T size 12{ ital "LE" rSub { size 8{3} } ital "ME" rSub { size 8{3} } ital "NE" rSub { size 8{1} } T} {}
(17)
LE 3 ME 2 NE 1 T LE 3 ME 2 NE 1 T size 12{ ital "LE" rSub { size 8{3} } ital "ME" rSub { size 8{2} } ital "NE" rSub { size 8{1} } T} {}
(18)
LE 3 ME 1 NE 2 T LE 3 ME 1 NE 2 T size 12{ ital "LE" rSub { size 8{3} } ital "ME" rSub { size 8{1} } ital "NE" rSub { size 8{2} } T} {}
(19)

Because the EE size 12{E} {}'s are not different, there is only one arrangement LEMENETLEMENET size 12{ ital "LEMENET"} {} and not six. This is true for every permutation.

Let us suppose there are nn size 12{n} {} different permutations of the letters ELEMENTELEMENT size 12{ ital "ELEMENT"} {}.

Then there are n3!n3! size 12{n cdot 3!} {} permutations of the letters E1LE2ME3NTE1LE2ME3NT size 12{E rSub { size 8{1} } ital "LE" rSub { size 8{2} } ital "ME" rSub { size 8{3} } ital "NT"} {}.

But we know there are 7!7! size 12{7!} {} permutations of the letters E1LE2ME3NTE1LE2ME3NT size 12{E rSub { size 8{1} } ital "LE" rSub { size 8{2} } ital "ME" rSub { size 8{3} } ital "NT"} {}.

Therefore, n3!=7!n3!=7! size 12{n cdot 3!=7!} {}

Or n=7!3!n=7!3! size 12{n= { {7!} over {3!} } } {}.

This gives us the method we are looking for.

Permutations with Similar Elements

The number of permutations of nn size 12{n} {} elements taken nn size 12{n} {} at a time, with r1r1 size 12{r rSub { size 8{1} } } {} elements of one kind, r2r2 size 12{r rSub { size 8{2} } } {} elements of another kind, and so on, is

n ! r 1 ! r 2 ! . . . r k ! n ! r 1 ! r 2 ! . . . r k ! size 12{ { {n!} over {r rSub { size 8{1} } !r rSub { size 8{2} } ! "." "." "." r rSub { size 8{k} } !} } } {}
(20)

Example 29

Problem 1

Find the number of different permutations of the letters of the word MISSISSIPPI.

Solution

The word MISSISSIPPI has 11 letters. If the letters were all different there would have been 11!11! size 12{"11"!} {} different permutations. But MISSISSIPPI has 4 S's, 4 I's, and 2 P's that are alike.

So the answer is 11!4!4!2!11!4!4!2! size 12{ { {"11"!} over {4!4!2!} } } {}

Which equals 34,650.

Example 30

Problem 1

If a coin is tossed six times, how many different outcomes consisting of 4 heads and 2 tails are there?

Solution

Again, we have permutations with similar elements.

We are looking for permutations for the letters HHHHTT.

The answer is 6!4!2!=156!4!2!=15 size 12{ { {6!} over {4!2!} } ="15"} {}.

Example 31

Problem 1

In how many different ways can 4 nickels, 3 dimes, and 2 quarters be arranged in a row?

Solution

Assuming that all nickels are similar, all dimes are similar, and all quarters are similar, we have permutations with similar elements. Therefore, the answer is

9!4!3!2!=12609!4!3!2!=1260 size 12{ { {9!} over {4!3!2!} } ="1260"} {}
(21)

Example 32

Problem 1

A stock broker wants to assign 20 new clients equally to 4 of its salespeople. In how many different ways can this be done?

Solution

This means that each sales person gets 5 clients. The problem can be thought of as an ordered partitions problem. In that case, using the formula we get

20!5!5!5!5!=11,732,745,02420!5!5!5!5!=11,732,745,024 size 12{ { {"20"!} over {5!5!5!5!} } ="11","732","745","024"} {}
(22)

We summarize.

  1. Circular Permutations

    The number of permutations of nn size 12{n} {} elements in a circle is

    n 1 ! n 1 ! size 12{ left (n - 1 right )!} {}
    (23)
  2. Permutations with Similar Elements

    The number of permutations of nn size 12{n} {} elements taken nn size 12{n} {} at a time, with r1r1 size 12{r rSub { size 8{1} } } {} elements of one kind, r2r2 size 12{r rSub { size 8{2} } } {} elements of another kind, and so on, such that n=r1+r2++rkn=r1+r2++rk size 12{n=r rSub { size 8{1} } +r rSub { size 8{2} } + dotsaxis +r rSub { size 8{k} } } {} is

    n ! r 1 ! r 2 ! r k ! n ! r 1 ! r 2 ! r k ! size 12{ { {n!} over {r rSub { size 8{1} } !r rSub { size 8{2} } ! dotsaxis r rSub { size 8{k} } !} } } {}

    This is also referred to as ordered partitions.

Combinations

Suppose we have a set of three letters A,B,CA,B,C size 12{ left lbrace A,B,C right rbrace } {}, and we are asked to make two-letter word sequences. We have the following six permutations.

AB AB size 12{ ital "AB"} {} BA BA size 12{ ital "BA"} {} BC BC size 12{ ital "BC"} {} CB CB size 12{ ital "CB"} {} AC AC size 12{ ital "AC"} {} CA CA size 12{ ital "CA"} {}

Now suppose we have a group of three people A,B,CA,B,C size 12{ left lbrace A,B,C right rbrace } {} as Al, Bob, and Chris, respectively, and we are asked to form committees of two people each. This time we have only three committees, namely,

AB AB size 12{ ital "AB"} {} BC BC size 12{ ital "BC"} {} AC AC size 12{ ital "AC"} {}

When forming committees, the order is not important, because the committee that has Al and Bob is no different than the committee that has Bob and Al. As a result, we have only three committees and not six.

Forming word sequences is an example of permutations, while forming committees is an example of combinations – the topic of this section.

Permutations are those arrangements where order is important, while combinations are those arrangements where order is not significant. From now on, this is how we will tell permutations and combinations apart.

In Example 32, there were six permutations, but only three combinations.

Just as the symbol nPrnPr size 12{n"Pr"} {} represents the number of permutations of nn size 12{n} {} objects taken rr size 12{r} {} at a time, nCrnCr size 12{ ital "nCr"} {} represents the number of combinations of nn size 12{n} {} objects taken rr size 12{r} {} at a time.

So in Example 32, 3P2=63P2=6 size 12{3P2=6} {}, and 3C2=33C2=3 size 12{3C2=3} {}.

Our next goal is to determine the relationship between the number of combinations and the number of permutations in a given situation.

In Example 32, if we knew that there were three combinations, we could have found the number of permutations by multiplying this number by 2!2! size 12{2!} {}. That is because each combination consists of two letters, and that makes 2!2! size 12{2!} {} permutations.

Example 34

Problem 1

Given the set of letters A,B,C,DA,B,C,D size 12{ left lbrace A,B,C,D right rbrace } {}. Write the number of combinations of three letters, and then from these combinations determine the number of permutations.

Solution

We have the following four combinations.

ABC BCD CDA BDA

Since every combination has three letters, there are 3!3! size 12{3!} {} permutations for every combination. We list them below.

ABC BCD CDA BDA

ACB BDC CAD BAD

BAC CDB DAC DAB

BCA CBD DCA DBA

CAB DCB ACD ADB

CBA DBC ADC ABD

The number of permutations are 3!3! size 12{3!} {} times the number of combinations. That is

4p3=3!4C34p3=3!4C3 size 12{4p3=3! cdot 4C3} {}
(24)

or 4C3=4P33!4C3=4P33! size 12{4C3= { {4P3} over {3!} } } {}

In general, nCr=nPrr!nCr=nPrr! size 12{ ital "nCr"= { {n"Pr"} over {r!} } } {}

Since nPr=n!nr!nPr=n!nr! size 12{n"Pr"= { {n!} over { left (n - r right )!} } } {}

We have, nCr=n!nr!r!nCr=n!nr!r! size 12{ ital "nCr"= { {n!} over { left (n - r right )!r!} } } {}

Summarizing,

  1. Combinations

    A combination of a set of elements is an arrangement where each element is used once, and order is not important.

  2. The Number of Combinations of n n size 12{n} {} Objects Taken r r size 12{r} {} at a Time

    nCr = n ! n r ! r ! nCr = n ! n r ! r ! size 12{ ital "nCr"= { {n!} over { left (n - r right )!r!} } } {}

    Where nn size 12{n} {} and rr size 12{r} {} are natural numbers.

Example 36

Problem 1

Compute:

  1. 5C35C3 size 12{5C3} {}
  2. 7C37C3 size 12{7C3} {}.
Solution

We use the above formula.

5C3=5!53!3!=5!2!3!=105C3=5!53!3!=5!2!3!=10 size 12{5C3= { {5!} over { left (5 - 3 right )!3!} } = { {5!} over {2!3!} } ="10"} {}
(25)
7C3=7!73!3!=7!4!3!=35.7C3=7!73!3!=7!4!3!=35 size 12{7C3= { {7!} over { left (7 - 3 right )!3!} } = { {7!} over {4!3!} } ="35"} {}.
(26)

Example 37

Problem 1

In how many different ways can a student select to answer five questions from a test that has seven questions, if the order of the selection is not important?

Solution

Since the order is not important, it is a combination problem, and the answer is

7C5=217C5=21 size 12{7C5="21"} {}.

Example 38

Problem 1

How many line segments can be drawn by connecting any two of the six points that lie on the circumference of a circle?

Solution

Since the line that goes from point AA size 12{A} {} to point BB size 12{B} {} is same as the one that goes from BB size 12{B} {} to AA size 12{A} {}, this is a combination problem.

It is a combination of 6 objects taken 2 at a time. Therefore, the answer is

6C2=6!4!2!=156C2=6!4!2!=15 size 12{6C2= { {6!} over {4!2!} } ="15"} {}
(27)

Example 39

Problem 1

There are ten people at a party. If they all shake hands, how many hand-shakes are possible?

Solution

Note that between any two people there is only one hand shake. Therefore, we have

10C2=45   hand-shakes.10C2=45   hand-shakes. size 12{"10"C2="45"" hand-shakes" "." } {}
(28)

Example 40

Problem 1

The shopping area of a town is in the shape of square that is 5 blocks by 5 blocks. How many different routes can a taxi driver take to go from one corner of the shopping area to the opposite cater-corner?

Solution

Let us suppose the taxi driver drives from the point AA size 12{A} {}, the lower left hand corner, to the point BB size 12{B} {}, the upper right hand corner as shown in the figure below.

Figure 8
 This grid shows how a taxi driver must get from point A to point B.

To reach his destination, he has to travel ten blocks; five horizontal, and five vertical. So if out of the ten blocks he chooses any five horizontal, the other five will have to be the vertical blocks, and vice versa. Therefore, all he has to do is to choose 5 out of ten.

The answer is 10C5, or 252.

Alternately, the problem can be solved by permutations with similar elements.

The taxi driver's route consists of five horizontal and five vertical blocks. If we call a horizontal block HH size 12{H} {}, and a vertical block a VV size 12{V} {}, then one possible route may be as follows.

HHHHHVVVVVHHHHHVVVVV size 12{ ital "HHHHHVVVVV"} {}
(29)

Clearly there are 10!5!5!=252  permutations10!5!5!=252  permutations size 12{ { {"10"!} over {5!5!} } ="252"" permutations"} {}.

Further note that by definition 10C5=10!5!5!10C5=10!5!5! size 12{"10"C5= { {"10"!} over {5!5!} } } {}.

Example 41

Problem 1

If a coin is tossed six times, in how many ways can it fall four heads and two tails?

Solution

First we solve this problem using Section 7 technique–permutations with similar elements.

We need 4 heads and 2 tails, that is

HHHHTTHHHHTT size 12{ ital "HHHHTT"} {}
(30)

There are 6!4!2!=156!4!2!=15 size 12{ { {6!} over {4!2!} } ="15"} {} permutations.

Now we solve this problem using combinations.

Suppose we have six spots to put the coins on. If we choose any four spots for heads, the other two will automatically be tails. So the problem is simply

6C4=156C4=15 size 12{6C4="15"} {}.

Incidentally, we could have easily chosen the two tails, instead. In that case, we would have gotten

6C2=156C2=15 size 12{6C2="15"} {}.

Further observe that by definition

6C4=6!2!4!6C4=6!2!4! size 12{6C4= { {6!} over {2!4!} } } {}
(31)

and 6C2=6!4!2!6C2=6!4!2! size 12{6C2= { {6!} over {4!2!} } } {}

Which implies

6C4=6C2.6C4=6C2 size 12{6C4=6C2} {}.
(32)

Combinations: Involving Several Sets

So far we have solved the basic combination problem of rr size 12{r} {} objects chosen from nn size 12{n} {} different objects. Now we will consider certain variations of this problem.

Example 42

Problem 1

How many five-people committees consisting of 2 men and 3 women can be chosen from a total of 4 men and 4 women?

Solution

We list 4 men and 4 women as follows:

M1M2M3M4W1W2W3W4M1M2M3M4W1W2W3W4 size 12{M rSub { size 8{1} } M rSub { size 8{2} } M rSub { size 8{3} } M rSub { size 8{4} } W rSub { size 8{1} } W rSub { size 8{2} } W rSub { size 8{3} } W rSub { size 8{4} } } {}
(33)

Since we want 5-people committees consisting of 2 men and 3 women, we'll first form all possible two-man committees and all possible three-woman committees. Clearly there are 4C2 = 6 two-man committees, and 4C3 = 4 three-woman committees, we list them as follows:

Table 12
2-Man Committees 3-Woman Committees
M 1 M 2 M 1 M 2 size 12{M rSub { size 8{1} } M rSub { size 8{2} } } {} W 1 W 2 W 3 W 1 W 2 W 3 size 12{W rSub { size 8{1} } W rSub { size 8{2} } W rSub { size 8{3} } } {}
M 1 M 3 M 1 M 3 size 12{M rSub { size 8{1} } M rSub { size 8{3} } } {} W 1 W 2 W 4 W 1 W 2 W 4 size 12{W rSub { size 8{1} } W rSub { size 8{2} } W rSub { size 8{4} } } {}
M 1 M 4 M 1 M 4 size 12{M rSub { size 8{1} } M rSub { size 8{4} } } {} W 1 W 3 W 4 W 1 W 3 W 4 size 12{W rSub { size 8{1} } W rSub { size 8{3} } W rSub { size 8{4} } } {}
M 2 M 3 M 2 M 3 size 12{M rSub { size 8{2} } M rSub { size 8{3} } } {} W 2 W 3 W 4 W 2 W 3 W 4 size 12{W rSub { size 8{2} } W rSub { size 8{3} } W rSub { size 8{4} } } {}
M 2 M 4 M 2 M 4 size 12{M rSub { size 8{2} } M rSub { size 8{4} } } {}  
M 3 M 4 M 3 M 4 size 12{M rSub { size 8{3} } M rSub { size 8{4} } } {}  

For every 2-man committee there are four 3-woman committees that can be chosen to make a 5-person committee. If we choose M1M2M1M2 as our 2-man committee, then we can choose any of W1W2W3W1W2W3, W1W2W4W1W2W4, W1W3W4W1W3W4, or W2W3W4W2W3W4 as our 3-woman committees. As a result, we get

M1M2,W1W2W3M1M2 ,W1W2W4M1M2 , W1W3W4M1M2 ,W2W3W4M1M2,W1W2W3M1M2 ,W1W2W4M1M2 ,W1W3W4M1M2 ,W2W3W4
(34)

Similarly, if we choose M1M3M1M3 as our 2-man committee, then, again, we can choose any of W1W2W3W1W2W3, W1W2W4W1W2W4, W1W3W4W1W3W4, or W2W3W4W2W3W4 as our 3-woman committees.

M1M3 ,W1W2W3M1M3 ,W1W2W4 M1M3 , W1W3W4 M1M3 , W2W3W4M1M3 ,W1W2W3M1M3 ,W1W2W4M1M3 ,W1W3W4 M1M3 ,W2W3W4
(35)

And so on.

Since there are six 2-man committees, and for every 2-man committee there are four 3- woman committees, there are altogether 64=2464=24 five-people committees.

In essence, we are applying the multiplication axiom to the different combinations.

Example 43

Problem 1

A high school club consists of 4 freshmen, 5 sophomores, 5 juniors, and 6 seniors. How many ways can a committee of 4 people be chosen that includes

  1. One student from each class?
  2. All juniors?
  3. Two freshmen and 2 seniors?
  4. No freshmen?
  5. At least three seniors?
Solution
  1. Applying the multiplication axiom to the combinations involved, we get

    4C15C15C16C1=6004C15C15C16C1=600 size 12{4C1 cdot 5C1 cdot 5C1 cdot 6C1="600"} {}
    (36)
  2. We are choosing all 4 members from the 5 juniors, and none from the others.

    5C4=55C4=5 size 12{5C4=5} {}
    (37)
  3. 4C26C2=904C26C2=90 size 12{4C2 cdot 6C2="90"} {}

  4. Since we don't want any freshmen on the committee, we need to choose all members from the remaining 16. That is

    16C4=182016C4=1820 size 12{"16"C4="1820"} {}
    (38)
  5. Of the 4 people on the committee, we want at least three seniors. This can be done in two ways. We could have three seniors, and one non-senior, or all four seniors.

    6C314C1+6C4=2956C314C1+6C4=295 size 12{6C3 cdot "14"C1+6C4="295"} {}
    (39)

Example 44

Problem 1

How many five-letter word sequences consisting of 2 vowels and 3 consonants can be formed from the letters of the word INTRODUCE?

Solution

First we select a group of five letters consisting of 2 vowels and 3 consonants. Since there are 4 vowels and 5 consonants, we have

4C25C34C25C3 size 12{4C2 cdot 5C3} {}
(40)

Since our next task is to make word sequences out of these letters, we multiply these by 5!5! size 12{5!} {}.

4C25C35!=72004C25C35!=7200 size 12{4C2 cdot 5C3 cdot 5!="7200"} {}.

Example 45

Problem 1

A standard deck of playing cards has 52 cards consisting of 4 suits each with 13 cards. In how many different ways can a 5-card hand consisting of four cards of one suit and one of another be drawn?

Solution

We will do the problem using the following steps. Step 1. Select a suit. Step 2. Select four cards from this suit. Step 3. Select another suit. Step 4. Select a card from that suit.

Applying the multiplication axiom, we have

Table 13
Ways of selecting a suit Ways if selecting 4 cards from this suit Ways if selecting the next suit Ways of selecting a card from that suit
4C1 4C1 size 12{4C1} {} 13 C4 13 C4 size 12{"13"C4} {} 3C1 3C1 size 12{3C1} {} 13 C1 13 C1 size 12{"13"C1} {}
4C113C43C113C1=111,540.4C113C43C113C1=111,540.
(41)

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