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Fundamental counting principle

Learning goals

Identifies successive choices in a counting situation.
Applies the fundamental counting principle.
Distinguishes when order matters from when it does not.
Checks for overcounting or omitted cases.

Difficulty: 1 – 5

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Fundamental counting principle 1

Jordan is choosing an outfit for school from \(5\) different T-shirts, \(3\) different pairs of pants, and \(2\) different pairs of shoes. a) How many outfits can Jordan make by choosing one item from each category? b) Jordan decides to wear one particular blue T-shirt. How many outfit choices remain?

Hints

- How many successive choices are being made? - Multiply the number of options at each stage. - What changes when one category has only one available choice?

Solution

1. Apply the fundamental counting principle. There are \(5\) choices for the T-shirt, \(3\) choices for the pants, and \(2\) choices for the shoes, so the total is \(5\cdot3\cdot2=30\). 2. When the T-shirt is fixed, there is \(1\) shirt choice, \(3\) pants choices, and \(2\) shoe choices. The number of outfits is \(1\cdot3\cdot2=6\).

Answer

a) \(30\) outfits b) \(6\) outfits

Fundamental counting principle 2

Four soccer teams, A, B, C, and D, play in a small league. Each team plays every other team twice: once at home and once away. How many games are played altogether?

Hints

- Choose the home team first and then choose one of its opponents. - Make sure each game is counted once, by its home team.

Solution

1. Each of the \(4\) teams hosts one game against each of the other \(3\) teams. 2. Apply the fundamental counting principle: \(4 \cdot 3 = 12\) home games. Every league game has exactly one home team, so this counts each game once.

Answer

\(12\) games

Fundamental counting principle 3

Lena builds an ice cream sundae by choosing exactly one ice cream flavor and exactly one topping. Flavors: chocolate, vanilla, and strawberry Toppings: whipped cream, sprinkles, and chocolate sauce How many different sundaes can Lena make?

Hints

- Choose a flavor and then a topping. - Each flavor can be paired with every topping. - Multiply the number of choices at the two stages.

Solution

1. There are \(3\) choices for the flavor. 2. For each flavor, there are \(3\) choices for the topping. 3. By the fundamental counting principle, \(3 \cdot 3 = 9\) sundaes are possible.

Answer

\(9\) different sundaes

Fundamental counting principle 4

A two-digit puzzle code uses only the digits \(5,6,7,\) and \(8\). Digits may repeat. a) Make an organized list of all possible codes. b) How many codes are possible?

Hints

- How many choices are available for the first digit? - For each first digit, how many choices are available for the second digit? - Organize the codes by their first digit. - A table can help ensure that no pair is missed.

Solution

1. Organize the list by the first digit: \(55,56,57,58\); \(65,66,67,68\); \(75,76,77,78\); and \(85,86,87,88\). 2. There are \(4\) choices for the first digit and \(4\) choices for the second digit. By the fundamental counting principle, \(4\cdot4=16\).

Answer

a) \(55,56,57,58,65,66,67,68,75,76,77,78,85,86,87,88\) b) \(16\) codes

Fundamental counting principle 5

In a coded language, every word has exactly five characters. Each character is either a dot “•” or a dash “—”. How many different words are possible?

Hints

- How many positions must be filled? - How many choices are available at each position? - Does choosing one character change the choices at the next position? - What repeated multiplication represents the total?

Solution

1. Each of the \(5\) positions has \(2\) choices: dot or dash. 2. Apply the fundamental counting principle: \(2\cdot2\cdot2\cdot2\cdot2=2^5=32\).

Answer

\(32\) words

Fundamental counting principle 6

How many different four-digit numbers can be formed using only the odd digits \(1,3,5,7,\) and \(9\) if digits may repeat? Use the fundamental counting principle.

Hints

- How many odd digits are available? - Think of four empty positions. How many choices can fill each one? - Does allowing repetition keep the same number of choices at every position? - Which operation combines the numbers of choices across the positions?

Solution

1. There are \(5\) available digits. 2. Each of the \(4\) positions can contain any of the \(5\) digits. 3. By the fundamental counting principle, the number of four-digit numbers is \(5\cdot5\cdot5\cdot5=5^4=625\).

Answer

\(625\) four-digit numbers

Fundamental counting principle 7

Leon is creating a character for a video game. He chooses one of \(3\) character classes, one of \(5\) hairstyles, and one of \(4\) armor colors. How many different characters can he create?

Hints

- How many hairstyle choices are available for each character class? - How many armor colors can be paired with each class-and-hairstyle choice? - Multiply the numbers of choices across the three categories.

Solution

1. There are \(3\) choices for the character class, \(5\) choices for the hairstyle, and \(4\) choices for the armor color. 2. By the fundamental counting principle, the total is \(3\cdot5\cdot4=60\).

Answer

\(60\) characters

Fundamental counting principle 8

A sports team is designing a logo. Each logo uses one symbol and one background color. The symbol choices are a star, circle, or square. The background choices are red, blue, green, or yellow. a) How many symbol-and-background logos are possible? b) The team adds a border that can be black or white. How many logos are now possible using one symbol, one background color, and one border color?

Hints

- First count the symbol-and-background combinations. - How many backgrounds can be paired with one symbol? - The border adds another independent choice. - Multiply the numbers of choices across the categories.

Solution

1. There are \(3\) symbol choices and \(4\) background choices, so \(3\cdot4=12\) logos are possible. 2. Each of those \(12\) logos can have either of \(2\) border colors. Therefore, \(12\cdot2=24\) logos are possible. 3. Equivalently, apply the fundamental counting principle directly: \(3\cdot4\cdot2=24\).

Answer

a) \(12\) logos b) \(24\) logos

Fundamental counting principle 9

Each row of dots represents one T-shirt color. Each dot in a row represents a different pants color that can be paired with that T-shirt color. How many outfits are possible? If two more T-shirt colors are added, how many outfits are possible then? How many additional outfits is that?

Figure for problem 537439

Hints

- Count the rows and the dots in each row. - Think about what one additional T-shirt color contributes. - Compare the number of outfits before and after the change.

Solution

1. The diagram has \(4\) rows with \(3\) dots in each, giving \(4\cdot3=12\) outfits. 2. Adding two T-shirt colors gives \(6\) rows of \(3\) outfits, or \(6\cdot3=18\) outfits. 3. The increase is \(18-12=6\) outfits.

Answer

Initially: \(12\) outfits; after the change: \(18\) outfits; increase: \(6\) outfits.

Fundamental counting principle 10

An ice cream order includes one flavor and one sauce. How many different orders can be made by following the choices in the tree?

Figure for problem 537502

Hints

- Follow a branch from the beginning all the way to a terminal choice. - Count the number of complete paths, not just the first set of branches.

Solution

1. The tree has \(3\) flavor branches, each ending in \(2\) sauce choices. 2. Counting its complete paths gives \(3\cdot2=6\) possible orders.

Answer

\(6\) different orders.

Fundamental counting principle 11

A theater group is putting together costumes from the choices in the tree. How many different complete outfits can they make? Explain how you counted.

Figure for problem 537512

Hints

- Follow one branch all the way from the root to a shoe choice. - How many endings are available under each choice of bottoms? - Check that every complete path represents a different costume.

Solution

1. The tree has \(2\) top choices. Each top leads to \(3\) choices of bottoms, giving \(2\cdot3=6\) partial costumes. 2. Each of these leads to \(2\) shoe choices, giving \(6\cdot2=12\) complete costumes.

Answer

\(12\) costumes, because the tree gives \(2\cdot3\cdot2\) complete paths.

Fundamental counting principle 12

A school art club offers \(3\) poster backgrounds and \(2\) lettering styles. Every background works with either style. If you organize the choices in a grid with one row per background and one column per lettering style, how many rows and columns are there? Use multiplication to find the number of possible posters.

Hints

- What does each row stand for? - How many choices fit into each row?

Solution

1. The grid needs \(3\) rows for backgrounds and \(2\) columns for lettering styles. 2. Each cell is one different combination, so \(3\cdot2=6\) posters are possible.

Answer

\(3\) rows, \(2\) columns, and \(3\cdot2=6\) possible posters.

Fundamental counting principle 13

Maya and Ben are playing a board game. One player must go first and the other second. Are “Maya first, Ben second” and “Ben first, Maya second” one arrangement or two? How many arrangements are there?

Hints

- What changes when the players switch places? - Does the game say who goes first?

Solution

1. The first player matters, so reversing the order changes the arrangement. 2. Maya can go first or Ben can go first. 3. There are two arrangements.

Answer

Two different arrangements. There are \(2\) in all.

Fundamental counting principle 14

A bicycle combination lock has \(3\) number wheels. Each wheel can display any digit from \(0\) through \(9\). a) How many different codes are possible? b) Tim remembers that his code uses only the digits \(1\), \(2\), \(3\), \(4\), and \(5\). How many codes satisfy this condition? c) With no information about the code, what is the probability of guessing it correctly on the first try? Give the result as a fraction and a percent.

Hints

- How many choices are available for each wheel? - How do you combine the numbers of choices for the separate positions? - What changes when only certain digits may be used? - For equally likely codes, probability is favorable outcomes divided by total outcomes.

Solution

1. a) Each of the \(3\) positions has \(10\) choices, so the number of codes is \(10 \cdot 10 \cdot 10 = 10^3 = 1000\). 2. b) Each position now has \(5\) choices, so the number of codes is \(5 \cdot 5 \cdot 5 = 5^3 = 125\). 3. c) Exactly one of the \(1000\) equally likely codes is correct. Therefore, \(P = \frac{1}{1000} = 0.001 = 0.1\%\).

Answer

a) \(1000\) codes b) \(125\) codes c) \(\frac{1}{1000} = 0.1\%\)

Fundamental counting principle 15

A raffle sells tickets with three-digit codes from \(\text{000}\) through \(\text{999}\). Each code appears exactly once. a) How many different ticket codes are there? b) A participant says, “The chance of drawing \(\text{000}\) is much smaller than the chance of drawing \(\text{527}\) because \(\text{000}\) is a special code.” Evaluate this claim mathematically. c) Suppose every ticket was sold. What is the probability that the winning code ends in \(7\)?

Hints

- Count the choices for all three positions, including codes that begin with \(0\). - Does a visible pattern make one equally likely ticket more or less likely than another? - If the last digit must be \(7\), how many choices remain for the first two positions?

Solution

1. a) Each of the three positions has \(10\) choices, so there are \(10 \cdot 10 \cdot 10 = 1000\) ticket codes. 2. b) Every ticket is equally likely to be drawn. Therefore, both \(\text{000}\) and \(\text{527}\) have probability \(\frac{1}{1000}\). A visually unusual pattern does not change its probability. 3. c) The last position is fixed as \(7\), while each of the first two positions has \(10\) choices. Thus, there are \(10 \cdot 10 \cdot 1 = 100\) favorable codes. 4. The probability is \(\frac{100}{1000} = \frac{1}{10} = 10\%\).

Answer

a) \(1000\) ticket codes b) The claim is false. Each code has probability \(\frac{1}{1000}\). c) \(\frac{1}{10} = 10\%\)

Fundamental counting principle 16

A school cafeteria lets students build a lunch by choosing exactly one appetizer and one main dish. There are \(6\) appetizers and exactly \(42\) possible lunches. How many main dishes are offered? Explain your reasoning.

Hints

- How would you find the total number of lunches if both category counts were known? - Write an equation with the number of main dishes as the unknown. - Which operation undoes multiplication by \(6\)?

Solution

1. Let \(m\) be the number of main dishes. By the fundamental counting principle, \(6m=42\). 2. Divide both sides by \(6\): \(m=42\div6=7\). 3. Check: \(6\cdot7=42\), so the cafeteria offers \(7\) main dishes.

Answer

\(7\) main dishes

Fundamental counting principle 17

A three-character code for a classroom game is created using this pattern: - First character: one uppercase letter from A through Z (\(26\) choices) - Second character: one digit from \(0\) through \(9\) (\(10\) choices) - Third character: one of the symbols “!”, “?”, “#”, “*”, or “&” (\(5\) choices) a) How many different codes are possible? b) A fourth character is added, and it can be any digit from \(0\) through \(9\). By what factor does the number of possible codes increase? Explain without calculating the new total.

Hints

- Identify the number of choices for each character. - What does one additional stage with \(10\) choices do to the total? - For part b, focus on the multiplier contributed by the new character.

Solution

1. Apply the fundamental counting principle to the three characters: \(26\cdot10\cdot5=1300\) possible codes. 2. The added character has \(10\) possible digits. Each original code can be extended in \(10\) ways. 3. Therefore, the number of possible codes is multiplied by a factor of \(10\).

Answer

a) \(1300\) codes b) The number of codes increases by a factor of \(10\).

Fundamental counting principle 18

A combination lock has four dials, each labeled with the digits \(0\) through \(9\). A person has forgotten the combination and sets all four dials randomly. Find the probability that the lock opens on the first attempt.

Hints

- Count the possible digits on one dial. - Use the fundamental counting principle to count all four-digit combinations. - Exactly one of those combinations is correct.

Solution

1. Each dial has \(10\) possible digits, and exactly one is correct. 2. There are \(10^4=10{,}000\) equally likely four-digit combinations. 3. Only one combination opens the lock, so the probability is \(\frac{1}{10000}=0.0001\).

Answer

\(\frac{1}{10000}=0.0001=0.01\%\)

Fundamental counting principle 19

A smoothie bar offers four fruits: apple, banana, cherry, and orange. A two-fruit smoothie always contains exactly two different fruits. How many different two-fruit smoothies are possible? List all of them.

Hints

- First count the choices for the first and second fruits. - Decide whether reversing the order creates a new smoothie. - List the combinations systematically to check your count.

Solution

1. There are \(4\) choices for the first fruit and \(3\) choices for the second fruit, giving \(4 \cdot 3 = 12\) ordered choices. 2. Each smoothie is counted twice because apple-banana and banana-apple are the same combination. Divide by \(2\): \(12 \div 2 = 6\). 3. The combinations are apple-banana, apple-cherry, apple-orange, banana-cherry, banana-orange, and cherry-orange.

Answer

There are \(6\) smoothies: apple-banana, apple-cherry, apple-orange, banana-cherry, banana-orange, and cherry-orange.

Fundamental counting principle 20

For each condition, determine how many three-digit numbers satisfy it. a) The tens digit is \(4\). b) The tens digit is \(4\), and the ones digit is \(0\). c) The tens digit is \(4\), but the ones digit is not \(0\).

Hints

- How many choices are possible for the hundreds digit of a three-digit number? - Count the choices for the tens and ones digits separately. - Use one factor for each digit position. - For part c, you can also use the results of parts a and b.

Solution

1. For part a, the hundreds digit has \(9\) choices, the tens digit has \(1\) choice, and the ones digit has \(10\) choices. Thus, \(9\cdot1\cdot10=90\). 2. For part b, the hundreds digit has \(9\) choices, while the tens and ones digits are fixed. Thus, \(9\cdot1\cdot1=9\). 3. For part c, the hundreds digit has \(9\) choices, the tens digit is fixed, and the ones digit has \(9\) nonzero choices. Thus, \(9\cdot1\cdot9=81\). Equivalently, \(90-9=81\).

Answer

a) \(90\) numbers b) \(9\) numbers c) \(81\) numbers

Fundamental counting principle 21

a) List all four-digit numbers that use only the digits \(0\) and \(3\). Remember that a four-digit number cannot begin with \(0\). b) How many numbers are in your list?

Hints

- Which digit must be in the thousands place? - List the numbers in increasing order. - A tree diagram can help organize the choices for the last three positions.

Solution

1. The thousands digit must be \(3\). Each of the other three positions can be either \(0\) or \(3\). 2. The numbers are \(3000\), \(3003\), \(3030\), \(3033\), \(3300\), \(3303\), \(3330\), and \(3333\). 3. There are \(2\) choices for each of the last three positions, so \(2^3=8\) numbers.

Answer

a) \(3000\), \(3003\), \(3030\), \(3033\), \(3300\), \(3303\), \(3330\), \(3333\) b) \(8\) numbers

Fundamental counting principle 22

Use the digits \(4\), \(5\), and \(6\) exactly once each to form a three-digit number. a) Use the fundamental counting principle to determine how many different numbers are possible. b) List all possible numbers to verify your count.

Hints

- Count the choices available for each place as digits are used. - Multiply the stage counts for part a. - For the verification list, organize outcomes by the hundreds digit.

Solution

1. There are \(3\) choices for the hundreds digit, \(2\) choices for the tens digit, and \(1\) choice for the ones digit, so \(3\cdot2\cdot1=6\) numbers are possible. 2. The six numbers are \(456,465,546,564,645,654\), confirming the product-rule count.

Answer

a) \(6\) numbers b) \(456,465,546,564,645,654\)

Fundamental counting principle 23

A three-digit puzzle code uses only the digits \(2,4,6,\) and \(8\). a) How many codes are possible if digits may repeat? b) How many codes are possible if the first digit must be \(8\) and no digit may repeat?

Hints

- Count the choices for each position. - When repetition is allowed, does the number of choices change from one position to the next? - In part b, the first position is fixed and each chosen digit is removed from the available set. - You can list the six codes to check your count.

Solution

1. With repetition allowed, each of the \(3\) positions has \(4\) choices, so there are \(4^3=64\) codes. 2. In part b, the first digit is fixed as \(8\). There are \(3\) choices for the second digit and \(2\) remaining choices for the third digit. 3. Therefore, the number of codes is \(1\cdot3\cdot2=6\).

Answer

a) \(64\) codes b) \(6\) codes

Fundamental counting principle 24

A puzzle code uses the digits \(1,2,3,4,\) and \(5\), and digits may repeat. Lucas claims, “A three-digit code has exactly \(5\) times as many possibilities as a two-digit code.” Determine whether Lucas is correct by calculating both numbers of possibilities and explaining the comparison.

Hints

- Count the available digits for each position. - Use the fundamental counting principle for the two-digit code. - Repeat for the three-digit code. - Compare the two totals using multiplication.

Solution

1. A two-digit code has \(5\) choices for each position, so there are \(5^2=25\) codes. 2. A three-digit code has \(5\) choices for each position, so there are \(5^3=125\) codes. 3. Since \(25\cdot5=125\), the three-digit code has exactly \(5\) times as many possibilities. Lucas is correct.

Answer

Lucas is correct. There are \(25\) two-digit codes and \(125\) three-digit codes, and \(125=5\cdot25\).

Fundamental counting principle 25

A game code uses the letters A, B, C, and D, and letters may repeat. a) How many different two-letter codes are possible? b) How many different four-letter codes are possible? c) Without calculating a new total, explain how the number of possibilities changes when the code length increases from four letters to five letters.

Hints

- How many letters are available for each position? - Apply the fundamental counting principle for each code length. - What multiplier is contributed by one additional position?

Solution

1. A two-letter code has \(4\) choices for each position, so there are \(4^2=16\) codes. 2. A four-letter code has \(4\) choices for each position, so there are \(4^4=256\) codes. 3. Adding a fifth position gives \(4\) possible extensions for every four-letter code. Therefore, the number of codes is multiplied by \(4\).

Answer

a) \(16\) codes b) \(256\) codes c) The number of codes is multiplied by \(4\).

Fundamental counting principle 26

Devon is planning a visit to a waterfront festival. He adds the choices at each stage and gets \(7\) plans. Explain why that count is wrong, and find the number of complete plans.

Figure for problem 537518

Hints

- What counts as one complete result in this tree? - Count complete paths rather than individual branches. - Think about how the number of endings changes after an earlier choice.

Solution

1. Each complete outcome follows one choice from each stage of the tree. 2. The branches give \(2\cdot3\cdot2=12\) complete paths. 3. Devon added the counts for separate stages rather than counting the combinations across stages.

Answer

There are \(12\) complete plans. Devon added the choices at each stage rather than counting the combinations of choices across stages.

Fundamental counting principle 27

A workshop assigns each session to one room and one time block. Count the valid room-and-time combinations from the tree. Explain why multiplying the number of rooms by the number of time blocks would give the wrong answer.

Figure for problem 537552

Hints

- Count the endpoints under each room branch. - Does every room have the same number of available time choices? - Try grouping full room-and-time paths without including an option the tree leaves out.

Solution

1. Rooms A, B, and C each have \(3\) time choices, giving \(3\cdot3=9\) combinations. 2. Room D has only \(2\) time choices. 3. The total is \(9+2=11\). A single product would wrongly count the unavailable room-D and time-3 pair.

Answer

\(11\) combinations. Room D has fewer time choices than the other rooms, so not every room can be paired with every time block.

Fundamental counting principle 28

An inspector chooses a first station from \(7\) stations, then a different second station, and then chooses one of \(2\) report formats. How many ordered inspection plans are possible?

Hints

- Count the choices available at each stage. - Account for the fact that the second station must differ from the first. - Multiply the stage counts because each partial plan can be completed in every later way.

Solution

1. There are \(7\) choices for the first station. 2. After that choice, \(6\) different stations remain for the second station. 3. There are \(2\) report formats for each ordered station pair. 4. The total is \(7\cdot6\cdot2=84\).

Answer

There are \(84\) ordered inspection plans.

Fundamental counting principle 29

A data dashboard can be configured by choosing one of \(4\) regions, one of \(3\) chart types, and one time period. The dashboard has \(96\) possible configurations. How many time-period choices are available?

Hints

- Represent the unknown stage count with a variable. - Write the product of the choices at all stages. - Use the known total to solve for the missing factor.

Solution

1. Let \(t\) be the number of time-period choices. 2. The fundamental counting principle gives \(4\cdot3\cdot t=96\). 3. Since \(12t=96\), \(t=8\).

Answer

There are \(8\) time-period choices.

Fundamental counting principle 30

On a field trip, students choose one of \(2\) morning tours, one of \(3\) midday workshops, and one of \(2\) afternoon demonstrations. Every combination is available. How many different full-day schedules can a student make?

Hints

- How many choices does the student make to build one schedule? - Can any morning choice go with any midday and afternoon choice?

Solution

1. There are \(2\) choices for the morning, \(3\) for midday, and \(2\) for the afternoon. 2. Multiply the choices at the three stages: \(2\cdot3\cdot2=12\).

Answer

\(12\) full-day schedules.

Fundamental counting principle 31

Four students volunteer to carry a banner. Two students are needed, and their order does not matter. Paolo counts \(4\cdot3=12\) pairs. Explain why this counts each pair more than once, and find the actual number of pairs.

Hints

- Does swapping two chosen students make a new team? - How many orders can the same two people appear in? - Adjust the original count so each pair is included once.

Solution

1. Choosing the same two students in opposite orders does not create a new pair. 2. Every unordered pair appears twice in the \(4\cdot3=12\) ordered choices. 3. Divide by \(2\) to get \(12\div2=6\) different pairs.

Answer

Each pair is counted in two orders. There are \(6\) different pairs.

Fundamental counting principle 32

A recreation center offers \(4\) courts and several possible starting times. Each court can be booked at every available starting time. Together, there are \(28\) different court-and-time choices. How many starting times are offered?

Hints

- How does the multiplication principle connect the number of courts and time slots? - Which factor is unknown?

Solution

1. The number of combinations is the number of courts multiplied by the number of starting times. 2. So \(4\cdot n=28\), where \(n\) is the number of starting times. 3. \(n=28\div4=7\).

Answer

\(7\) starting times.

Fundamental counting principle 33

During an elective week, each student chooses one course from each category: - Athletics: soccer, basketball, or volleyball - Arts: painting, ceramics, photography, or theater - Languages: Spanish or French a) How many different three-course schedules are possible? b) One student cannot choose ceramics. How many schedules are available to that student? c) The school adds a fifth arts course. Explain why the total number of schedules increases by more than \(1\).

Hints

- Decide how many choices are available at each stage of a schedule. - Which stage changes when ceramics is removed? - In part c, consider how a new arts course can be combined with the choices in the other categories.

Solution

1. There are \(3\) athletics choices, \(4\) arts choices, and \(2\) languages. The total is \(3\cdot4\cdot2=24\) schedules. 2. Without ceramics, there are \(3\) arts choices, giving \(3\cdot3\cdot2=18\) schedules. 3. With a fifth arts course, the total is \(3\cdot5\cdot2=30\), an increase of \(30-24=6\). Each new arts choice can be paired with every combination of athletics and language courses.

Answer

a) \(24\) schedules b) \(18\) schedules c) The number increases by \(6\), since the new course pairs with \(3\cdot2\) athletics-and-language choices.

Fundamental counting principle 34

A locker combination lock has \(4\) number wheels. Each wheel can display any digit from \(0\) through \(9\), including \(0\) in the first position. a) How many different four-digit codes are possible? b) How many codes use only the digits \(1\), \(2\), and \(3\)? c) A secret code is chosen at random, with every allowed code equally likely. A person guesses once. In which case is a correct first guess more likely? Compare the numbers of possible codes. - Case 1: The first digit is known to be \(5\). - Case 2: Every position is known to contain an even digit from \(0, 2, 4, 6, 8\).

Hints

- How many digits can go in each position? - For part b, apply the same counting process to the smaller set of digits. - Compare how many codes remain possible after each piece of information in part c.

Solution

1. Each of the \(4\) positions has \(10\) choices, so there are \(10\cdot10\cdot10\cdot10=10{,}000\) codes. 2. With digits \(1\), \(2\), and \(3\), each position has \(3\) choices, giving \(3\cdot3\cdot3\cdot3=81\) codes. 3. In Case 1, the first digit is fixed and the other three have \(10\) choices each, giving \(1000\) possible codes. 4. In Case 2, each position has \(5\) choices, giving \(5\cdot5\cdot5\cdot5=625\) possible codes. Each is equally likely, so a correct guess is more likely in Case 2.

Answer

a) \(10{,}000\) codes b) \(81\) codes c) Case 2. It has \(625\) possible codes, compared with \(1000\) in Case 1.

Fundamental counting principle 35

Three-digit numbers are formed using the digits \(3\), \(5\), \(7\), and \(9\). a) How many different numbers can be formed if digits may repeat? b) How many different numbers can be formed if no digit may be used more than once? c) For a number selected at random from part a and from part b, find the probability that it is divisible by \(5\).

Hints

- Count the choices available for each position. - What last digit makes a whole number divisible by \(5\)? - For equally likely outcomes, probability is favorable outcomes divided by total outcomes.

Solution

1. a) Each of the \(3\) positions has \(4\) choices, so there are \(4 \cdot 4 \cdot 4 = 4^3 = 64\) numbers. 2. b) There are \(4\) choices for the first digit, \(3\) for the second, and \(2\) for the third. Therefore, there are \(4 \cdot 3 \cdot 2 = 24\) numbers. 3. c) A number is divisible by \(5\) when its last digit is \(5\). In part a, the first two positions each have \(4\) choices, giving \(4 \cdot 4 = 16\) favorable numbers. Thus, \(P = \frac{16}{64} = \frac{1}{4}\). 4. In part b, after fixing \(5\) as the last digit, the first two positions have \(3\) and \(2\) choices. This gives \(3 \cdot 2 = 6\) favorable numbers, so \(P = \frac{6}{24} = \frac{1}{4}\).

Answer

a) \(64\) numbers b) \(24\) numbers c) \(\frac{1}{4} = 25\%\) in both cases

Fundamental counting principle 36

A fair six-sided die is rolled twice to form a two-digit number. The first roll is the tens digit, and the second roll is the ones digit. a) How many different numbers can be formed? b) How many different numbers can be formed if the two rolls must show different values? c) For the sample spaces in parts a and b, find the probability that the sum of the digits is exactly \(7\).

Hints

- Treat the tens digit and ones digit as separate choices. - List the ordered pairs of values from \(1\) through \(6\) that add to \(7\). - Use the correct total number of outcomes for each sample space.

Solution

1. a) Each roll has \(6\) possible results, so there are \(6 \cdot 6 = 36\) possible two-digit numbers. 2. b) The first roll has \(6\) choices. The second roll then has \(5\) choices because it must differ from the first. Therefore, there are \(6 \cdot 5 = 30\) numbers. 3. c) The ordered pairs with digit sum \(7\) are \((1, 6)\), \((2, 5)\), \((3, 4)\), \((4, 3)\), \((5, 2)\), and \((6, 1)\). There are \(6\) favorable outcomes, and all have different digits. 4. For part a, \(P = \frac{6}{36} = \frac{1}{6}\). For part b, \(P = \frac{6}{30} = \frac{1}{5}\).

Answer

a) \(36\) numbers b) \(30\) numbers c) Part a: \(\frac{1}{6}\); part b: \(\frac{1}{5}\)

Fundamental counting principle 37

A four-character code uses the letters A, B, C, and D. a) Find the number of possible codes when letters may repeat. b) Find the number of possible codes when each of the four letters is used exactly once. c) Find the probability that a randomly selected code from part a contains no A. d) Find the probability that a randomly selected code from part b begins with the letters AD.

Hints

- If one letter is excluded, how many choices remain for each position? - When every letter must be used once, the number of choices decreases after each position is filled. - In part d, the first two positions are fixed. Which letters remain?

Solution

1. When letters may repeat, each position has \(4\) choices, giving \(4\cdot4\cdot4\cdot4=256\) codes. 2. Without repetition, the choices decrease as letters are used: \(4\cdot3\cdot2\cdot1=24\) codes. 3. Excluding A leaves \(3\) choices at each position, giving \(3\cdot3\cdot3\cdot3=81\) favorable codes out of \(256\). The probability is \(\frac{81}{256}\). 4. Fix A and D as the first two letters. The remaining B and C have \(2\cdot1=2\) orders out of \(24\) total codes, so the probability is \(\frac{2}{24}=\frac{1}{12}\).

Answer

a) \(256\) codes b) \(24\) codes c) \(\frac{81}{256}\) d) \(\frac{1}{12}\)

Fundamental counting principle 38

A raffle randomly awards a grand prize and a second prize to \(12\) participants. Each participant has an equal chance to win each available prize, and no one may win twice. a) How many possible outcomes are there for awarding the two different prizes? b) Julia and Tim are participants. Find the probability that Julia wins the grand prize and Tim wins the second prize. c) Find the probability that Julia and Tim both win a prize, regardless of who wins which prize.

Hints

- Count choices for the grand prize, then the choices left for second prize. - Does it matter who receives which prize? - Think about both possible orders in which Julia and Tim could win.

Solution

1. There are \(12\) choices for the grand-prize winner and then \(11\) choices for the second-prize winner, giving \(12\cdot11=132\) equally likely outcomes. 2. Exactly one outcome has Julia winning the grand prize and Tim winning the second prize, so the probability is \(\frac{1}{132}\). 3. Julia and Tim can win in either order, giving \(2\) favorable outcomes. The probability is \(\frac{2}{132}=\frac{1}{66}\).

Answer

a) \(132\) outcomes b) \(\frac{1}{132}\) c) \(\frac{1}{66}\)

Fundamental counting principle 39

A combination lock has \(3\) wheels. Each wheel can display a digit from \(1\) through \(6\). a) How many different codes are possible? b) Find the probability that a randomly selected code contains the digits \(1\), \(2\), and \(3\), each exactly once. c) Find the probability that a randomly selected code has three identical digits.

Hints

- Each wheel has the same number of choices. - In part b, how many orders are possible for the fixed digits \(1\), \(2\), and \(3\)? - How many codes consist of one digit repeated three times?

Solution

1. a) Each wheel has \(6\) choices, so there are \(6 \cdot 6 \cdot 6 = 216\) codes. 2. b) The digits \(1\), \(2\), and \(3\) can be arranged in \(3 \cdot 2 \cdot 1 = 6\) orders. Therefore, \(P = \frac{6}{216} = \frac{1}{36}\). 3. c) The favorable codes are \(111, 222, 333, 444, 555, 666\), for a total of \(6\). Therefore, \(P = \frac{6}{216} = \frac{1}{36}\).

Answer

a) \(216\) codes b) \(\frac{1}{36}\) c) \(\frac{1}{36}\)

Fundamental counting principle 40

Eight runners compete in a \(100\)-meter race in lanes \(1\) through \(8\). Assume there are no ties and that all possible finish orders are equally likely. a) Find the probability that the runners in lanes \(1\), \(2\), and \(3\) win gold, silver, and bronze, respectively, in that exact order. b) Show that the probability that the runners in lanes \(1\), \(2\), and \(3\) take the first three places in any order is exactly six times the probability in part a. c) Find the probability that none of the runners in lanes \(1\), \(2\), or \(3\) finishes in the top three.

Hints

- Count the ordered ways to fill first, second, and third place from \(8\) runners. - In part b, how many orders are possible for the three specified runners? - In part c, how many runners remain eligible for the top three positions?

Solution

1. There are \(8\) choices for first place, \(7\) for second place, and \(6\) for third place, giving \(8 \cdot 7 \cdot 6 = 336\) equally likely ordered medal outcomes. 2. a) Exactly one outcome has lanes \(1\), \(2\), and \(3\) in that exact order, so \(P = \frac{1}{336}\). 3. b) The three specified runners can be arranged in the top three places in \(3 \cdot 2 \cdot 1 = 6\) orders. Thus, \(P = \frac{6}{336} = \frac{1}{56}\), which is six times \(\frac{1}{336}\). 4. c) The top three places must be filled by the other \(5\) runners. There are \(5 \cdot 4 \cdot 3 = 60\) favorable outcomes, so \(P = \frac{60}{336} = \frac{5}{28}\).

Answer

a) \(\frac{1}{336}\) b) \(\frac{6}{336} = \frac{1}{56} = 6 \cdot \frac{1}{336}\) c) \(\frac{5}{28}\)

Fundamental counting principle 41

A student club has \(12\) members: \(4\) girls and \(8\) boys. The club must fill three different offices: president, treasurer, and secretary. No member may hold more than one office. a) How many different ways can the three offices be filled? b) If the offices are assigned at random, what is the probability that all three offices are filled by girls?

Hints

- Does it matter which person holds each office? - How many choices are available for the first office, and how many remain for later offices? - For part b, which assignments qualify as wins, and which don't?

Solution

1. There are \(12\) choices for president, \(11\) for treasurer, and \(10\) for secretary, giving \(12\cdot11\cdot10=1320\) assignments. 2. If all three officeholders are girls, the counts are \(4\), \(3\), and \(2\), giving \(4\cdot3\cdot2=24\) favorable assignments. 3. The probability is \(\frac{24}{1320}=\frac{1}{55}\).

Answer

a) \(1320\) assignments b) \(\frac{1}{55}\)

Fundamental counting principle 42

A suitcase has a four-digit combination lock. Each wheel shows a digit from \(0\) through \(9\). The secret code is chosen at random so every code is equally likely. A person guesses at random from the codes still possible with the information they have. a) How many different codes are possible? b) What is the probability that a first guess is correct when the person knows nothing about the code? c) Suppose the first digit is known to be \(7\). How does this information change the probability of a correct first guess? Briefly explain. d) For a guess chosen without knowing any digits, what is the probability that at least one digit is in the wrong position compared with the secret code?

Hints

- Count the choices for each position. - How many codes open the lock? - Knowing one digit reduces the number of positions still to guess. - What is the complement of at least one digit being wrong?

Solution

1. Each of the \(4\) positions has \(10\) choices, giving \(10\cdot10\cdot10\cdot10=10{,}000\) codes. 2. One of those codes is correct, so the probability is \(\frac{1}{10000}=0.01\%\). 3. Knowing the first digit leaves \(10\cdot10\cdot10=1000\) possible codes. The probability is \(\frac{1}{1000}=0.1\%\), ten times the original probability. 4. At least one digit is wrong unless every digit is correct, so the probability is \(1-\frac{1}{10000}=\frac{9999}{10000}=99.99\%\).

Answer

a) \(10{,}000\) codes b) \(\frac{1}{10000}=0.01\%\) c) \(\frac{1}{1000}=0.1\%\); ten times as large d) \(\frac{9999}{10000}=99.99\%\)

Fundamental counting principle 43

A suitcase lock has three wheels: - The first wheel shows the digits \(0\) through \(9\). - The second wheel shows the letters A, B, C, D, and E. - The third wheel shows the four card suits \(\heartsuit\), \(\diamondsuit\), \(\clubsuit\), and \(\spadesuit\). A person has forgotten the correct combination and sets each wheel at random. a) How many different combinations are possible? b) What is the probability that all three positions are wrong?

Hints

- Count the choices available on each wheel. - If one setting on each wheel is correct, how many settings on that wheel are wrong? - Use the fundamental counting principle to count combinations where every wheel is wrong, then compare that count with the total number of combinations.

Solution

1. a) Multiply the number of choices for the three wheels: \(10 \cdot 5 \cdot 4 = 200\) combinations. 2. b) On the first wheel, \(9\) of the \(10\) digits are wrong. On the second, \(4\) of the \(5\) letters are wrong. On the third, \(3\) of the \(4\) suits are wrong. 3. By the fundamental counting principle, there are \(9 \cdot 4 \cdot 3 = 108\) combinations with all three positions wrong. Since all \(200\) combinations are equally likely, \(P(\text{all wrong}) = \frac{108}{200} = \frac{27}{50} = 0.54\).

Answer

a) \(200\) combinations b) \(\frac{27}{50} = 0.54 = 54\%\)

Fundamental counting principle 44

A building manager has three different keys for three different doors. The manager randomly assigns one key to each lock, using every key exactly once. Find the probability of each event: a) All three keys are assigned to the correct locks. b) No key is assigned to its correct lock. c) Exactly one key is assigned to its correct lock.

Hints

- List all assignments systematically. - How many ways can three objects be assigned to three positions? - For each event, count the listed assignments that satisfy the condition. - If two keys are correctly assigned, what must happen to the third key?

Solution

1. There are \(3 \cdot 2 \cdot 1 = 6\) equally likely assignments of the three keys to the three locks. 2. a) Exactly one assignment matches every key to its correct lock, so \(P = \frac{1}{6}\). 3. b) Label the correct assignment \((1, 2, 3)\). The two assignments with no correct matches are \((2, 3, 1)\) and \((3, 1, 2)\). Therefore, \(P = \frac{2}{6} = \frac{1}{3}\). 4. c) The assignments with exactly one correct match are \((1, 3, 2)\), \((3, 2, 1)\), and \((2, 1, 3)\). Therefore, \(P = \frac{3}{6} = \frac{1}{2}\).

Answer

a) \(\frac{1}{6}\) b) \(\frac{1}{3}\) c) \(\frac{1}{2}\)

Fundamental counting principle 45

Two explorers must choose paths through two different cave systems. At each fork, they choose randomly among the available paths, so all choices at that fork are equally likely. In each cave, exactly one complete path reaches the goal. - Cave A has \(2\) consecutive forks with \(5\) choices at each fork. - Cave B has \(3\) consecutive forks with \(3\) choices at each fork. Use calculations to determine which cave gives the greater probability of reaching the goal by guessing.

Hints

- Count the number of complete paths in each cave. - Use the product of the choices at successive forks. - How does the number of possible paths affect the chance when only one path succeeds?

Solution

1. Cave A has \(5\cdot5=25\) complete paths. Exactly one succeeds, so the probability is \(\frac{1}{25}\). 2. Cave B has \(3\cdot3\cdot3=27\) complete paths. Exactly one succeeds, so the probability is \(\frac{1}{27}\). 3. Since \(\frac{1}{25}>\frac{1}{27}\), Cave A gives the greater chance of reaching the goal.

Answer

Cave A: \(\frac{1}{25}\), compared with \(\frac{1}{27}\) for Cave B.

Fundamental counting principle 46

Anya wants to join one sports club and one music club. There are \(5\) sports clubs—soccer, basketball, swimming, tennis, and rock climbing—and \(4\) music clubs—choir, orchestra, band, and guitar. a) How many sports-and-music club combinations are possible? b) The swimming club is canceled. How many combinations remain? c) Rock climbing and band meet at the same time, so that one pair cannot be chosen together. How many of the combinations from part b are still possible?

Hints

- A table can show every sports choice paired with every music choice. - How does removing one entire sports option change the product? - How many individual pairs are removed by one specific scheduling conflict?

Solution

1. Apply the fundamental counting principle: \(5\cdot4=20\) combinations. 2. After swimming is canceled, \(4\) sports choices and \(4\) music choices remain, giving \(4\cdot4=16\) combinations. 3. Exactly one of those combinations—rock climbing with band—is not allowed. Therefore, \(16-1=15\) combinations remain.

Answer

a) \(20\) combinations b) \(16\) combinations c) \(15\) combinations

Fundamental counting principle 47

A four-digit code for a puzzle box uses only the digits \(1,2,3,4,\) and \(5\). a) How many codes are possible if digits may repeat? b) How many codes are possible if no digit may be used more than once? c) How many codes begin with an even digit and end with an odd digit? For this part, digits may repeat.

Hints

- Count the choices available for each position separately. - Without repetition, how does the number of choices change after each digit is used? - Which available digits are even, and which are odd?

Solution

1. With repetition allowed, each of the four positions has \(5\) choices, so there are \(5^4=625\) codes. 2. Without repetition, the successive numbers of choices are \(5,4,3,\) and \(2\). Thus, there are \(5\cdot4\cdot3\cdot2=120\) codes. 3. For part c, the first digit has \(2\) choices, \(2\) or \(4\). Each middle position has \(5\) choices, and the last digit has \(3\) choices, \(1,3,\) or \(5\). Therefore, there are \(2\cdot5\cdot5\cdot3=150\) codes.

Answer

a) \(625\) codes b) \(120\) codes c) \(150\) codes

Fundamental counting principle 48

A bag contains six balls labeled A, B, C, D, E, and F. Two balls are drawn in order without replacement. Order matters, so \((\mathrm{A}, \mathrm{B})\) and \((\mathrm{B}, \mathrm{A})\) are different outcomes. a) How many possible outcomes are there? b) How many outcomes contain the letter A? c) How many outcomes consist only of vowels?

Hints

- How does drawing without replacement change the number of choices on the second draw? - Count separately the outcomes with A first and the outcomes with A second. - Which letters from A through F are vowels?

Solution

1. a) There are \(6\) choices for the first draw and \(5\) remaining choices for the second draw. Therefore, there are \(6 \cdot 5 = 30\) outcomes. 2. b) A can appear first, followed by any of the other \(5\) letters, or second, preceded by any of the other \(5\) letters. Thus, there are \(5 + 5 = 10\) outcomes containing A. 3. c) The only vowels are A and E. Without replacement, the vowel-only outcomes are \((\mathrm{A}, \mathrm{E})\) and \((\mathrm{E}, \mathrm{A})\), for a total of \(2\).

Answer

a) \(30\) outcomes b) \(10\) outcomes c) \(2\) outcomes

Fundamental counting principle 49

A clothing store sells T-shirts in \(4\) colors (red, blue, green, and yellow), \(3\) sizes (S, M, and L), and \(2\) neck styles (V-neck and crew neck). a) How many different T-shirt options are available? b) A box contains exactly one T-shirt of each option. If one T-shirt is selected at random, what is the probability that it is red and size L? c) What is the probability that the selected T-shirt has a crew neck?

Hints

- Count the choices for color, size, and neck style. - How many complete options satisfy each condition? - For equally likely options, divide favorable outcomes by the total number of options.

Solution

1. There are \(4\cdot3\cdot2=24\) different T-shirt options. 2. A red size-L shirt can have either neck style, so there are \(2\) favorable options. The probability is \(\frac{2}{24}=\frac{1}{12}\). 3. A crew-neck shirt can have any of the \(4\) colors and \(3\) sizes, giving \(4\cdot3=12\) favorable options. The probability is \(\frac{12}{24}=\frac{1}{2}\).

Answer

a) \(24\) options b) \(\frac{1}{12}\) c) \(\frac{1}{2}\)

Fundamental counting principle 50

A raffle ticket has a color (blue or white) and a three-digit code. The first digit is \(1\), \(2\), or \(3\). The second and third digits can each be any digit from \(0\) through \(9\). Exactly one ticket exists for every possible combination, and all tickets are equally likely to be drawn. A “super ticket” has identical second and third digits, such as Blue-122 or White-300. How many tickets are there altogether? How many are super tickets? Find the probability of drawing a super ticket.

Hints

- Separate the ticket into color, first digit, second digit, and third digit. - Count the choices for each part. - For a super ticket, how many matching pairs are possible for the final two digits?

Solution

1. There are \(2\) color choices, \(3\) choices for the first digit, and \(10\) choices for each of the last two digits. Therefore, the total number of tickets is \(2 \cdot 3 \cdot 10 \cdot 10 = 600\). 2. For a super ticket, the color and first digit are still unrestricted. The matching pair of final digits can be \(00, 11, \ldots, 99\), giving \(10\) choices. Thus, there are \(2 \cdot 3 \cdot 10 = 60\) super tickets. 3. The probability is \(P = \frac{60}{600} = \frac{1}{10} = 10\%\).

Answer

There are \(600\) tickets, including \(60\) super tickets. The probability of drawing one is \(\frac{1}{10} = 10\%\).

Fundamental counting principle 51

Five students play a round-robin tennis tournament. Each student plays every other student exactly once. a) How many matches are played altogether? b) Just before the tournament begins, two more students join. How many additional matches must be scheduled so that every pair of students plays once?

Hints

- Count ordered choices for the two players in a match. - Each pair is counted twice, once in each order. - Find the totals for \(5\) and \(7\) students, and then subtract.

Solution

1. For \(5\) students, choose a first player in \(5\) ways and a second player in \(4\) ways. This counts every pair twice, so the number of matches is \(\frac{5 \cdot 4}{2} = 10\). 2. For \(7\) students, the total number of matches is \(\frac{7 \cdot 6}{2} = 21\). 3. The number of additional matches is \(21 - 10 = 11\).

Answer

a) \(10\) matches b) \(11\) additional matches

Fundamental counting principle 52

Use the digit cards \(1\), \(2\), \(3\), and \(4\) to form two two-digit numbers. Use each card exactly once. Reversing the order of the two factors does not create a new multiplication problem. a) How many different multiplication problems are possible if the digit \(1\) must be in the tens place of one factor? Explain using a counting principle. b) If the digit \(2\) must also be in the tens place of the other factor, how many of those problems remain? Explain.

Hints

- Fix the factor with tens digit \(1\) in the first position so factor reversal is not counted twice. - For part a), count the choices for its ones digit and then the arrangements of the remaining digits. - For part b), identify which positions are already fixed before counting the remaining assignments.

Solution

1. Put the factor whose tens digit is \(1\) first so reversing the factor order is not double-counted. 2. For part a), there are \(3\) choices for the ones digit paired with \(1\). After that choice, the remaining two digits can be arranged in \(2\) orders in the other factor. 3. By the fundamental counting principle, there are \(3\cdot2=6\) problems. 4. For part b), fixing \(2\) as the tens digit of the other factor leaves \(3\) and \(4\) for the two ones places. 5. The remaining digits can be assigned to those two ones places in \(2\) ways, so \(2\) problems remain.

Answer

a) \(6\) problems. b) \(2\) problems.

Fundamental counting principle 53

Digits may repeat. a) How many three-digit numbers can be formed using only the digits \(1,2,\) and \(3\)? b) How many three-digit numbers contain no digit \(0\)? c) How many three-digit numbers contain at least one digit \(0\)?

Hints

- Count the choices available for each digit position. - When zero is forbidden, how many digits remain available? - How many three-digit numbers are there from \(100\) through \(999\)? - For part c, count the complement: numbers with no zero.

Solution

1. For part a, each of the three positions has \(3\) choices, so \(3^3=27\). 2. For part b, each position can use any digit from \(1\) through \(9\), giving \(9^3=729\). 3. There are \(900\) three-digit numbers in all, from \(100\) through \(999\). The complement of “at least one zero” is “no zeros.” Therefore, \(900-729=171\).

Answer

a) \(27\) numbers b) \(729\) numbers c) \(171\) numbers

Fundamental counting principle 54

How many three-digit numbers contain the digit \(8\) exactly twice?

Hints

- Which pairs of positions can contain the two \(8\)s? - The remaining digit cannot also be \(8\). - Remember that the hundreds digit cannot be \(0\). - Count the choices for the free digit in each case.

Solution

1. If the two \(8\)s are in the hundreds and tens places, the ones digit can be any digit except \(8\), giving \(9\) choices. 2. If the two \(8\)s are in the hundreds and ones places, the tens digit can be any digit except \(8\), again giving \(9\) choices. 3. If the two \(8\)s are in the tens and ones places, the hundreds digit can be any nonzero digit except \(8\), giving \(8\) choices. 4. The cases do not overlap, so the total is \(9+9+8=26\).

Answer

\(26\) three-digit numbers

Fundamental counting principle 55

At an ice cream shop, you choose exactly one cup size, one flavor, and one sauce. <table> <tr> <th>Cup size</th> <th>Flavor</th> <th>Sauce</th> </tr> <tr> <td>Small</td> <td>Vanilla</td> <td>Chocolate</td> </tr> <tr> <td>Medium</td> <td>Chocolate</td> <td>Strawberry</td> </tr> <tr> <td>Large</td> <td>Strawberry</td> <td></td> </tr> <tr> <td></td> <td>Banana</td> <td></td> </tr> <tr> <td></td> <td>Mint</td> <td></td> </tr> </table> a) How many size-flavor-sauce combinations are possible? b) The mint and strawberry flavors are sold out. How many combinations remain? c) The shop adds a topping category with \(3\) choices. How many combinations are possible with the full original menu plus the toppings?

Hints

- Count the options in each category. - How does removing two flavors change one factor in the product? - Apply the fundamental counting principle by multiplying the choices across all categories.

Solution

1. The table shows \(3\) sizes, \(5\) flavors, and \(2\) sauces. By the fundamental counting principle, \(3\cdot5\cdot2=30\). 2. With two flavors sold out, \(3\) flavors remain. The number of combinations is \(3\cdot3\cdot2=18\). 3. Adding a topping category with \(3\) choices multiplies the original total by \(3\): \(30\cdot3=90\).

Answer

a) \(30\) combinations b) \(18\) combinations c) \(90\) combinations

Fundamental counting principle 56

A bicycle company offers one model that customers can customize by choosing one frame color, one seat style, and one handlebar style. There are \(36\) possible bicycles. a) There are \(3\) handlebar styles and \(4\) seat styles. How many frame colors are available? b) The company adds one frame color. How many possible bicycles are there now? c) A customer chooses one specific racing handlebar. Using the original number of frame colors, how many choices remain for the rest of the bicycle?

Hints

- Divide the total by the known category sizes to find the missing factor. - How does adding one frame color change the first factor? - When one category is fixed, multiply only the remaining category sizes.

Solution

1. Let \(c\) be the number of frame colors. The fundamental counting principle gives \(c\cdot4\cdot3=36\), so \(c=36\div12=3\). 2. After one color is added, there are \(4\) frame colors. The new total is \(4\cdot4\cdot3=48\). 3. When the handlebar is fixed, there are \(3\) original frame colors and \(4\) seat styles, giving \(3\cdot4=12\) choices.

Answer

a) \(3\) frame colors b) \(48\) bicycles c) \(12\) choices

Fundamental counting principle 57

An ice cream cone has three stacked scoops: bottom, middle, and top. Each scoop can be chocolate or vanilla, and flavors may repeat. The order of the scoops matters. a) How many different three-scoop cones are possible? b) How many flavors would need to be available for exactly \(27\) ordered three-scoop cones to be possible?

Hints

- Think of the bottom, middle, and top scoops as three successive choices. - How many choices are available at each position? - For part b, what number multiplied by itself three times equals \(27\)?

Solution

1. Each of the \(3\) scoop positions has \(2\) choices, so the number of cones is \(2^3=8\). 2. Let \(n\) be the number of available flavors. The number of ordered cones is \(n^3\). 3. Solve \(n^3=27\). Since \(3^3=27\), the shop would need \(3\) flavors.

Answer

a) \(8\) cones b) \(3\) flavors

Fundamental counting principle 58

An ice cream shop has four flavors: strawberry, chocolate, vanilla, and mint. A cone has three stacked scoops, and the order from bottom to top matters. a) How many cones are possible if flavors may repeat? b) How many cones are possible if all three flavors must be different and one scoop must be chocolate?

Hints

- Fill the bottom, middle, and top positions one at a time. - For part b, first choose the position of the chocolate scoop. - The other two scoops must be different from chocolate and from each other. - Multiply the numbers of choices at each stage.

Solution

1. With repetition allowed, each of the \(3\) positions has \(4\) choices, so there are \(4^3=64\) cones. 2. For part b, choose the position of the chocolate scoop in \(3\) ways. 3. Choose the flavor for the first remaining position from the other \(3\) flavors, then choose the flavor for the last position from the remaining \(2\) flavors. 4. The total is \(3\cdot3\cdot2=18\).

Answer

a) \(64\) cones b) \(18\) cones

Fundamental counting principle 59

The school art club prints flyers. Members choose one of \(3\) papers (plain, card, or glossy), \(2\) sizes (small or large), \(2\) layouts (portrait or landscape), and \(2\) ink colors. The printer cannot make large, landscape flyers on glossy paper, no matter which ink color is chosen. How many possible flyer settings work?

Hints

- Count all combinations before applying the restriction. - Which choices would make the printer refuse a flyer? - Does the ink color change the number of settings that are not allowed?

Solution

1. Without restrictions, the fundamental counting principle gives \(3\cdot2\cdot2\cdot2=24\) options. 2. The unavailable glossy, large, landscape combination has \(2\) ink colors. 3. Subtract those \(2\) combinations: \(24-2=22\) options remain.

Answer

\(22\) possible flyer settings.

Fundamental counting principle 60

A four-digit lock has digits \(0\) through \(9\) at each position. One code is selected at random. Find the probability of each event. a) All four digits are identical. b) The code begins with \(19\). c) The sum of the four digits is exactly \(2\).

Hints

- Use the fundamental counting principle for all possible codes. - In part a, choose the one digit that repeats. - In part b, the fixed first two digits leave choices only for the remaining positions. - In part c, separate the possible digit patterns that sum to \(2\).

Solution

1. There are \(10^4 = 10{,}000\) possible codes. 2. There are \(10\) codes with four identical digits: \(0000, 1111, \ldots, 9999\). Thus, \(P = \frac{10}{10000} = \frac{1}{1000}\). 3. The first two digits are fixed, and each of the final two positions has \(10\) choices. There are \(10 \cdot 10 = 100\) favorable codes, so \(P = \frac{100}{10000} = \frac{1}{100}\). 4. A digit sum of \(2\) occurs either when one position contains \(2\) and the others contain \(0\), giving \(4\) codes, or when two positions contain \(1\) and the others contain \(0\), giving \(6\) codes. Thus, there are \(10\) favorable codes, so \(P = \frac{10}{10000} = \frac{1}{1000}\).

Answer

a) \(\frac{1}{1000} = 0.1\%\) b) \(\frac{1}{100} = 1\%\) c) \(\frac{1}{1000} = 0.1\%\)

Fundamental counting principle 61

A safe manufacturer offers two keypad-lock systems: System A: A four-digit code using only the digits \(1, 2, 3, 4, 5, 6\). System B: A three-digit code using any digit from \(0\) through \(9\). a) Find the number of possible codes for each system. b) Which system is more secure against a random guess? Justify your answer using the numbers of possible codes. c) What is the minimum number of positions System A would need to allow more than \(50{,}000\) different codes?

Hints

- Use the fundamental counting principle to count the codes for each system. - How does a larger number of possible codes affect the chance of a correct random guess? - For part c, compare successive powers of \(6\) with \(50{,}000\).

Solution

1. a) System A has \(6\) choices for each of \(4\) positions, so it has \(6 \cdot 6 \cdot 6 \cdot 6 = 6^4 = 1296\) codes. 2. System B has \(10\) choices for each of \(3\) positions, so it has \(10 \cdot 10 \cdot 10 = 10^3 = 1000\) codes. 3. b) System A is more secure against one random guess because \(1296 > 1000\), so \(\frac{1}{1296} < \frac{1}{1000}\). 4. c) Check successive powers of \(6\): \(6^5 = 7776\), \(6^6 = 46{,}656\), and \(6^7 = 279{,}936\). Because \(6^6 < 50{,}000 < 6^7\), System A needs at least \(7\) positions.

Answer

a) System A: \(1296\) codes; System B: \(1000\) codes b) System A, because it has more possible codes and therefore a smaller probability of a correct random guess. c) \(7\) positions

Fundamental counting principle 62

Four friends—Anna, Ben, Clara, and David—sit at random in a row on a bench with four seats. Find the probability of each event: a) Anna sits in one of the two end seats. b) Anna and Ben sit next to each other. c) The four friends sit in exact alphabetical order from left to right: Anna, Ben, Clara, David.

Hints

- How many total orders are possible for four people in four seats? - If Anna is fixed in an end seat, how many ways can the other three be arranged? - What possible places could Anna and Ben occupy if they must sit together? - How many seating orders are exactly alphabetical?

Solution

1. There are \(4\cdot3\cdot2\cdot1=24\) equally likely seating orders. 2. Anna can sit in either end seat. The other three friends have \(3\cdot2\cdot1=6\) orders, giving \(2\cdot6=12\) favorable arrangements. The probability is \(\frac{12}{24}=\frac{1}{2}\). 3. For Anna and Ben to sit together, their block has \(3\) possible positions. They can switch places, and Clara and David can switch places, giving \(3\cdot2\cdot2=12\) arrangements. The probability is \(\frac{12}{24}=\frac{1}{2}\). 4. Only one order is alphabetical, so the probability is \(\frac{1}{24}\).

Answer

a) \(\frac{1}{2}\) b) \(\frac{1}{2}\) c) \(\frac{1}{24}\)

Fundamental counting principle 63

A costume team builds an outfit by choosing one top, one pair of pants, and one hat. The team has \(4\) tops, \(3\) pairs of pants, and \(2\) hats. a) How many outfits are possible? b) One particular yellow top cannot be worn with one particular green pair of pants. All other combinations are allowed. How many outfits remain possible? c) Create two different sets of clothing categories and category sizes that would each produce exactly \(24\) outfits.

Hints

- How many complete outfits contain the one forbidden top-and-pants pair? - For part c, find different factorizations of \(24\). - Assign each factor to the number of choices in a clothing category.

Solution

1. The total number of outfits is \(4\cdot3\cdot2=24\). 2. The forbidden yellow-top-and-green-pants pair can be combined with either of the \(2\) hats, so it eliminates \(2\) outfits. Therefore, \(24-2=22\) outfits remain. 3. For part c, choose category sizes whose product is \(24\). For example, \(2\) shoe choices, \(3\) scarf choices, and \(4\) jacket choices give \(2\cdot3\cdot4=24\). Another example is \(2\) belt choices, \(2\) glove choices, and \(6\) cape choices, giving \(2\cdot2\cdot6=24\).

Answer

a) \(24\) outfits b) \(22\) outfits c) Example 1: \(2\) shoe choices, \(3\) scarf choices, and \(4\) jacket choices Example 2: \(2\) belt choices, \(2\) glove choices, and \(6\) cape choices

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