Math II · S-CP.9

Using Permutations and Combinations to Compute Probabilities of Compound Events

Counting methods let students compute probabilities when listing every outcome is too slow, which is essential for cards, codes, schedules, selections, and complex events.

Concept Statistics and Probability

Domain Conditional Probability and the Rules of Probability

Read time 8 minutes

What this learning objective is really asking you to learn

This objective asks students to use counting methods to compute probabilities of compound events. A compound event involves more than one outcome feature or more than one stage. For simple situations, students can list all outcomes. But for larger situations, listing becomes inefficient or impossible. Permutations and combinations provide efficient counting tools.

A permutation is an arrangement where order matters. For example, choosing a president, vice president, and treasurer from a group is a permutation because assigning Ana as president and Ben as vice president is different from assigning Ben as president and Ana as vice president. A lock code is also order-sensitive: 1234 is different from 4321.

A combination is a selection where order does not matter. Choosing a committee of 3 students from 10 is a combination because the group {Ana, Ben, Carlos} is the same committee regardless of the order in which the names are listed. Drawing a 5-card poker hand is usually a combination because the order of cards in the hand does not matter.

The objective asks students to use these counting tools in probability. The basic probability structure is still

\[probability = favorable outcomes / total outcomes\].

Permutations and combinations help count the numerator and denominator.

For example, how many ways can 3 winners be chosen in order from 10 contestants? That is \(10P3 = 10 \cdot 9 \cdot 8 = 720\). How many ways can 3 committee members be chosen from 10? That is \(10C3 = 120\).

The hard part is not pressing a calculator button. The hard part is deciding whether order matters, whether repetition is allowed, and what counts as favorable. This objective is about efficient sample-space counting.

Why students should learn this math

Students should learn permutations and combinations because many probability problems are too large to solve by listing outcomes. Card hands, lottery tickets, passwords, seating arrangements, team selections, schedules, genetic combinations, tournament outcomes, and random samples can involve hundreds, thousands, millions, or billions of possibilities. Counting methods make such problems manageable.

This is practical in daily life and technology. A password system's security depends on the number of possible passwords. A lottery's odds depend on combinations. A card game's probabilities depend on how many hands satisfy a condition. A school schedule depends on arrangements. A hiring panel or committee selection depends on combinations. A seating chart depends on permutations. A random sample in statistics depends on counting possible selections.

Students also learn an important modeling question: does order matter? This is a major thinking skill. If the order of selected people creates different roles, use permutations. If only the group matters, use combinations. If digits can repeat in a code, the count differs from a code with no repetition. If cards are drawn and kept, the sample space changes after each draw. Counting requires attention to the situation.

This objective also helps students see why probability can be unintuitive. Many compound events have surprisingly small probabilities because the total number of outcomes is huge. The chance of winning a large lottery is tiny not because the math is mysterious, but because the denominator is enormous. Counting methods reveal the denominator.

The “why” is that combinatorics is the engine behind many probabilities. Without counting, probability becomes guessing. With counting, students can reason about complex systems efficiently.

The historical machinery: combinatorics and games of chance

Permutations and combinations developed from counting problems in arrangements, selections, games, and algebra. Games of chance helped drive early probability theory because players wanted to know odds. Card games, dice games, lotteries, and wagers all require counting possible outcomes.

Combinatorics, the mathematics of counting arrangements and selections, became a major branch of mathematics. It connects to probability, algebra, computer science, cryptography, statistics, and optimization. The binomial coefficients used in combinations also appear in Pascal's Triangle and the Binomial Theorem. This connects directly to later algebraic expansion.

Historically, the development of probability and combinatorics went hand in hand. To compute probability, mathematicians needed to count equally likely cases. As problems became more complex, systematic counting methods became essential.

Today, combinatorics is central to computer science and data security. Password counts, encryption keys, search spaces, algorithm complexity, and network structures all involve counting. The classroom distinction between permutations and combinations is a small entry into a large field.

Where this fits in the big map of mathematics

This objective comes near the end of the Math II probability sequence. Students already understand sample spaces, unions, intersections, conditional probability, independence, and multiplication rules. Now they learn counting tools for large sample spaces.

It connects to the Fundamental Counting Principle. If one choice has \(m\) options and the next has \(n\) options, the combined process has \(mn\) options, when choices are independent in the counting sense. This principle underlies permutations.

It connects to factorial notation. n! means the product \(n(n-1)(n-2)...1\). Factorials count arrangements.

It connects to combinations and Pascal's Triangle. \(nCk\) counts the number of ways to choose \(k\) items from \(n\) without regard to order. These numbers appear in binomial expansion.

It connects to probability by giving denominators and numerators. Counting favorable outcomes and total outcomes is often the main work.

It connects to statistics because random sampling without replacement uses combinations. It connects to computer science because arrangements, selections, and search spaces are fundamental.

The big-map role is efficient counting. Students learn to build sample-space sizes without listing every outcome.

How to execute the skill technically

Start with the question: does order matter?

If order matters, use permutations. The number of ways to arrange \(r\) objects chosen from \(n\) distinct objects without replacement is

\[nPr = n! / (n-r)!\].

Example: How many ways can gold, silver, and bronze medals be awarded among 8 runners?

Order matters because gold-silver-bronze roles differ. The count is

\[8P3 = 8 \cdot 7 \cdot 6 = 336\].

If order does not matter, use combinations. The number of ways to choose \(r\) objects from \(n\) distinct objects is

\[nCr = n! / (r!(n-r)!)\].

Example: How many 3-person committees can be chosen from 8 students?

Order does not matter. The count is

\[8C3 = 56\].

Now apply to probability. Suppose 5 cards are dealt from a 52-card deck. What is the probability all 5 are hearts?

Total 5-card hands:

52C5.

Favorable all-heart hands: choose 5 from the 13 hearts:

13C5.

So the probability is

\[13C5 / 52C5\].

Students do not always need to compute the decimal. The exact combination expression is often meaningful.

Another example: A 4-digit code uses digits 0 through 9 with no repetition. What is the probability a randomly generated code starts with 7?

Total codes: \(10 \cdot 9 \cdot 8 \cdot 7\).

Favorable codes starting with 7: fix the first digit as 7, then choose the next three without repetition from the remaining 9 digits: \(9 \cdot 8 \cdot 7\).

Probability:

\[(9 \cdot 8 \cdot 7) / (10 \cdot 9 \cdot 8 \cdot 7) = 1/10\].

This example shows that counting can simplify elegantly.

Permutations versus combinations in context

The most important habit is to decide what would count as a different outcome.

If selecting a password, order matters. \(ABCD\) and \(DCBA\) are different. If selecting a study group, order usually does not matter. Ana, Ben, and Carlos form the same group regardless of listing order.

If awarding first, second, and third place, order matters. If choosing three finalists with no ranking, order does not matter.

If drawing cards into a poker hand, order usually does not matter. If recording the sequence of cards drawn, order matters.

If selecting meal toppings for a pizza, order usually does not matter. If arranging books on a shelf, order matters.

Students should not choose permutation or combination based on key words alone. They should ask: would changing the order create a new outcome in this situation? That question is the core of the skill.

For an app, this objective should include a sorting interaction: show scenarios and ask students to drag them into “order matters” or “order does not matter.” Only after that should the app ask for calculations.

Worked probability example: committee with a condition

Suppose a class has 12 students: 7 juniors and 5 seniors. A committee of 4 students is chosen at random. What is the probability the committee has exactly 2 juniors and 2 seniors?

Total committees:

12C4.

Favorable committees: choose 2 juniors from 7 and 2 seniors from 5:

\[7C2 \cdot 5C2\].

So the probability is

\[(7C2 \cdot 5C2) / 12C4\].

Compute:

\(7C2 = 21\), \(5C2 = 10\), and \(12C4 = 495\).

So the probability is

\[210/495 = 14/33\].

This example shows the standard structure: count all possible selections, count favorable selections, divide. It also shows why combinations are correct: a committee is a group, and order does not matter.

Worked probability example: ordered awards

Now suppose the same 12 students are competing for first, second, and third prize. What is the probability all three winners are juniors?

Total ordered outcomes:

\[12P3 = 12 \cdot 11 \cdot 10\].

Favorable ordered outcomes: choose ordered winners from the 7 juniors:

\[7P3 = 7 \cdot 6 \cdot 5\].

So the probability is

\[(7P3)/(12P3) = (7 \cdot 6 \cdot 5)/(12 \cdot 11 \cdot 10) = 210/1320 = 7/44\].

This time order matters because first, second, and third are different roles. The same three students in a different order create a different outcome.

Counting with repetition

Some problems allow repetition. For example, a 4-digit PIN using digits 0 through 9 allows repeated digits unless stated otherwise. The total number of PINs is

\[10 \cdot 10 \cdot 10 \cdot 10 = 10,000\].

This is not 10P4, because 10P4 assumes no repetition. If the problem says no digit may repeat, then the count is

\[10 \cdot 9 \cdot 8 \cdot 7 = 5040\].

This distinction is crucial in password and code problems. Repetition rules can change the sample space dramatically.

Why counting is a modeling decision

Counting methods are not just formulas. They are decisions about what makes outcomes different. Does order matter? Can items repeat? Are selections made with or without replacement? Are there roles? Are there restrictions? Are all outcomes equally likely?

If those questions are answered incorrectly, the probability will be wrong even if the arithmetic is perfect. That is why students should not jump straight to \(nPr\) or \(nCr\). They should first describe the sample space in words.

A strong student can say: “This is a combination because the committee has no roles,” or “This is a permutation because president, vice president, and treasurer are different positions,” or “This uses repeated multiplication because digits can repeat.” That explanation is part of the answer.

Common misconceptions and how to avoid them

One common mistake is using permutations when combinations are needed. This overcounts groups by counting the same group in multiple orders.

Another mistake is using combinations when permutations are needed. This undercounts arrangements where roles or order matter.

A third mistake is ignoring whether repetition is allowed. Codes with repeated digits have different counts from codes without repeated digits.

A fourth mistake is counting total outcomes and favorable outcomes using inconsistent methods. Both numerator and denominator must describe the same kind of outcome.

A fifth mistake is relying on calculator buttons without understanding. Students should know what \(nPr\) and \(nCr\) count.

The big takeaway

Permutations and combinations are efficient counting tools for probability. Use permutations when order matters. Use combinations when order does not matter. Probability still means favorable outcomes divided by total outcomes, but counting methods make large sample spaces manageable. This objective prepares students for compound probability, lotteries, cards, sampling, password security, binomial reasoning, and advanced combinatorics.

Problem Library

Problems in the App From This Objective

201 problems across 15 archetypes in the app.

use factorial or multiplication principle.

12 problems Warmup Practice Mixed Review Assessment

Problem 1

Count permutations of distinct objects for arrange 5 distinct books.

Problem 2

Count permutations of distinct objects for choose president, vice president, and secretary from 8 students.

Problem 3

Count permutations of distinct objects for arrange 4 distinct letters.

Problem 4

Count permutations of distinct objects for order r objects from n distinct objects.

Problem 5

Count permutations of distinct objects for arrange 3 distinct paintings on a wall.

Problem 6

Count permutations of distinct objects for line up 6 distinct toys.

Problem 7

Count permutations of distinct objects for select a gold, silver, and bronze medalist from 10 runners.

Problem 8

Count permutations of distinct objects for form a 4-digit number using distinct digits from 1 to 7.

Problem 9

Count permutations of distinct objects for assign 3 different tasks to 5 employees, where each employee gets at most one task.

Problem 10

Count permutations of distinct objects for arrange 7 distinct musical instruments.

Open in simulator

Problem 11

Count permutations of distinct objects for create a 3-letter code using distinct letters from the English alphabet (26 letters).

Problem 12

Count permutations of distinct objects for choose a first, second, and third place winner from 12 contestants.

divide by repeated factorials.

12 problems Warmup Practice Mixed Review Assessment

Problem 13

Count permutations with repeated objects for arrangements of BALLOON.

Problem 14

Count permutations with repeated objects for arrangements of MISSISSIPPI.

Problem 15

Count permutations with repeated objects for arrangements of AABBC.

Problem 16

Count permutations with repeated objects for n objects with repeats a, b, c.

Open in simulator

Problem 17

Count permutations with repeated objects for arrangements of BANANA.

Problem 18

Count permutations with repeated objects for arrangements of BOOKKEEPER.

Problem 19

Count permutations with repeated objects for arrangements of MATHEMATICS.

Problem 20

Count permutations with repeated objects for arrangements of STATISTICS.

Problem 21

Count permutations with repeated objects for arrangements of SUCCESS.

Problem 22

Count permutations with repeated objects for arrangements of ENGINEERING.

Problem 23

Count permutations with repeated objects for arrangements of PARALLEL.

Problem 24

Count permutations with repeated objects for arrangements of ALABAMA.

use `n choose r` when order does not matter.

12 problems Warmup Practice Mixed Review Assessment

Problem 25

Count combinations for choose 3 students from 10.

Problem 26

Count combinations for choose 5 cards from 52.

Problem 27

Count combinations for choose a 4-person committee from 12.

Problem 28

Count combinations for choose r objects from n when order does not matter.

Problem 29

Count combinations for choose 2 toppings from 8 available.

Problem 30

Count combinations for select 3 books from 15 on a shelf.

Problem 31

Count combinations for form a 5-person subcommittee from a 20-person board.

Problem 32

Count combinations for pick 6 numbers from 49 for a lottery.

Problem 33

Count combinations for choose 2 flavors of ice cream from 10 available.

Problem 34

Count combinations for select 4 fruits from a basket of 10 different fruits.

Problem 35

Count combinations for select 3 students from a class of 25 for a competition.

Problem 36

Count combinations for form a committee of 3 men and 2 women from a group of 7 men and 5 women.

Open in simulator

determine whether order matters.

12 problems Warmup Practice Mixed Review Assessment

Problem 37

Decide whether the counting problem uses permutation or combination: choose 3 people for a committee.

Problem 38

Decide whether the counting problem uses permutation or combination: award gold, silver, and bronze among 10 runners.

Problem 39

Decide whether the counting problem uses permutation or combination: select 5 cards for a hand.

Problem 40

Decide whether the counting problem uses permutation or combination: create a 4-digit code with no repeated digits.

Problem 41

Decide whether the counting problem uses permutation or combination: arrange 7 different books on a shelf.

Problem 42

Decide whether the counting problem uses permutation or combination: choose 2 representatives from a class of 30 students.

Open in simulator

Problem 43

Decide whether the counting problem uses permutation or combination: elect a president, vice-president, and treasurer from 15 club members.

Problem 44

Decide whether the counting problem uses permutation or combination: select 4 toppings for a pizza from a list of 12 available toppings.

Problem 45

Decide whether the counting problem uses permutation or combination: form a 5-letter password using distinct letters from the alphabet.

Problem 46

Decide whether the counting problem uses permutation or combination: choose 3 flavors of ice cream from 20 available flavors for a triple scoop cone.

Problem 47

Decide whether the counting problem uses permutation or combination: determine the finishing order of the top 3 horses in a race of 12 horses.

Problem 48

Decide whether the counting problem uses permutation or combination: select 6 players for a basketball team from a roster of 10 players.

multiply number of options at each stage.

12 problems Warmup Practice Mixed Review Assessment

Problem 49

Use the multiplication principle for multi-stage choices: 3 shirts, 4 pants, 2 shoes.

Problem 50

Use the multiplication principle for multi-stage choices: license code with 2 letters then 3 digits, repeats allowed.

Problem 51

Use the multiplication principle for multi-stage choices: menu with 5 entrees, 3 sides, 4 drinks.

Problem 52

Use the multiplication principle for multi-stage choices: stages with a, b, and c options.

Problem 53

Use the multiplication principle for multi-stage choices: password with 4 digits followed by 2 uppercase letters, repeats allowed.

Problem 54

Use the multiplication principle for multi-stage choices: 4 hats, 5 shirts, 3 pants, 2 pairs of shoes.

Problem 55

Use the multiplication principle for multi-stage choices: travel from city A to B with 3 options, then B to C with 4 options, then C to D with 2 options.

Problem 56

Use the multiplication principle for multi-stage choices: car model with 6 exterior colors, 3 interior options, 2 engine types.

Problem 57

Use the multiplication principle for multi-stage choices: combo meal with 4 main dishes, 3 sides, 5 drinks, 2 desserts.

Problem 58

Use the multiplication principle for multi-stage choices: rolling a standard six-sided die three times.

Problem 59

Use the multiplication principle for multi-stage choices: answering a 5-question true/false test.

Open in simulator

Problem 60

Use the multiplication principle for multi-stage choices: 4-digit PIN code, digits can repeat.

account for required, forbidden, or fixed positions.

12 problems Warmup Practice Mixed Review Assessment

Problem 61

Count outcomes with restrictions for 4-digit code from 0-9 with no repeats and first digit cannot be 0.

Problem 62

Count outcomes with restrictions for arrange 5 people with Ana fixed first.

Problem 63

Count outcomes with restrictions for choose 3 students from 12 including one required student.

Open in simulator

Problem 64

Count outcomes with restrictions for arrange 6 books with two specific books together.

Problem 65

Count outcomes with restrictions for arrange 7 letters (A,B,C,D,E,F,G) such that A is not in the first position.

Problem 66

Count outcomes with restrictions for choose 4 items from 10, but item X cannot be chosen.

Problem 67

Count outcomes with restrictions for arrange 4 boys and 3 girls such that no two girls are together.

Problem 68

Count outcomes with restrictions for form a 3-digit number using digits 1,2,3,4,5,6 with no repeats, and the number must be even.

Problem 69

Count outcomes with restrictions for arrange 8 distinct books on a shelf where the first book is A and the last book is B.

Problem 70

Count outcomes with restrictions for choose 2 men and 2 women from 5 men and 4 women.

Problem 71

Count outcomes with restrictions for arrange 5 people (A,B,C,D,E) such that A, B, C are always together in that specific order (ABC).

Problem 72

Count outcomes with restrictions for arrange 7 different colored beads in a row, where red and blue beads must be next to each other.

favorable ordered outcomes over total ordered outcomes.

15 problems Warmup Practice Mixed Review Assessment

Problem 73

Compute probability using permutation counts for guess a 3-letter code from 26 letters with no repeats in exact order.

Problem 74

Compute probability using permutation counts for top 3 finishers in exact order from 8 runners.

Open in simulator

Problem 75

Compute probability using permutation counts for draw 2 cards in order and get ace then king.

Problem 76

Compute probability using permutation counts for f favorable ordered outcomes out of nPr total ordered outcomes.

Problem 77

Compute probability using permutation counts for predict the exact order of 5 horses in a race of 12.

Problem 78

Compute probability using permutation counts for guess a 4-digit PIN with no repeating digits in exact order from 10 digits.

Problem 79

Compute probability using permutation counts for draw 3 cards in order and get 3 hearts.

Problem 80

Compute probability using permutation counts for draw 3 cards in order and get a 7, then an 8, then a 9.

Problem 81

Compute probability using permutation counts for arrange 3 specific books out of 10 on a shelf in a particular order.

Problem 82

Compute probability using permutation counts for select a specific president, vice-president, and secretary from 15 candidates.

Problem 83

Compute probability using permutation counts for draw 2 red balls then 1 blue ball in order from a bag with 5 red and 3 blue balls without replacement.

Problem 84

Compute probability using permutation counts for form a 5-letter word using distinct letters from 'MATHS' in a specific order.

Problem 85

Compute probability using permutation counts for match the exact order of 6 distinct numbers drawn from 49.

Problem 86

Compute probability using permutation counts for arrange 2 specific people out of 7 in the first 2 seats of a row in a particular order.

Problem 87

Compute probability using permutation counts for draw 2 even numbers then 1 odd number in order from 1 to 10 without replacement.

favorable groups over total groups.

15 problems Warmup Practice Mixed Review Assessment

Problem 88

Compute probability using combination counts for 5-card hand has exactly 2 hearts.

Problem 89

Compute probability using combination counts for committee of 4 from 10 includes exactly 2 seniors among 3 seniors.

Problem 90

Compute probability using combination counts for lottery choose 6 from 49 all correct.

Problem 91

Compute probability using combination counts for f favorable groups from nCr total groups.

Problem 92

Compute probability using combination counts for 5-card hand has exactly 3 spades.

Problem 93

Compute probability using combination counts for 5-card hand has all red cards.

Problem 94

Compute probability using combination counts for 5-card hand has exactly 2 aces.

Problem 95

Compute probability using combination counts for committee of 3 from 7 men and 5 women includes exactly 2 men.

Problem 96

Compute probability using combination counts for committee of 5 from 12 people includes a specific person.

Problem 97

Compute probability using combination counts for selecting 3 red and 2 blue marbles from a bag of 5 red and 4 blue marbles.

Problem 98

Compute probability using combination counts for selecting 4 students from a class of 15 where 6 are honor students, and exactly 2 are honor students.

Problem 99

Compute probability using combination counts for lottery choose 5 from 50 has exactly 3 correct numbers.

Problem 100

Compute probability using combination counts for 13-card hand has exactly 4 spades.

Problem 101

Compute probability using combination counts for team of 6 from 8 boys and 7 girls has exactly 3 boys.

Open in simulator

Problem 102

Compute probability using combination counts for drawing 2 black balls from a bag containing 5 black and 3 white balls.

count complement and subtract from 1.

15 problems Warmup Practice Mixed Review Assessment

Problem 103

Compute probability of at least one event using complement counting: at least one head in 4 coin flips.

Problem 104

Compute probability of at least one event using complement counting: at least one ace in a 5-card hand.

Problem 105

Compute probability of at least one event using complement counting: at least one defective in 3 draws from 2 defective and 8 good.

Problem 106

Compute probability of at least one event using complement counting: at least one success in n independent trials with success probability p.

Problem 107

Compute probability of at least one event using complement counting: at least one 6 in 3 rolls of a fair die.

Problem 108

Compute probability of at least one event using complement counting: at least one tail in 5 flips of a fair coin.

Problem 109

Compute probability of at least one event using complement counting: at least one red card in 2 draws from a standard deck (without replacement).

Problem 110

Compute probability of at least one event using complement counting: at least one defective item in 4 draws from a batch of 5 defective and 15 good items (without replacement).

Problem 111

Compute probability of at least one event using complement counting: at least one even number in 2 rolls of a fair six-sided die.

Problem 112

Compute probability of at least one event using complement counting: at least one non-face card in 3 draws from a standard deck (with replacement).

Problem 113

Compute probability of at least one event using complement counting: at least one success in 2 independent trials, where success probability is 0.3.

Problem 114

Compute probability of at least one event using complement counting: at least one blue ball in 3 draws from an urn with 4 blue and 6 red balls (without replacement).

Problem 115

Compute probability of at least one event using complement counting: at least one vowel in a 3-letter sequence chosen randomly from the alphabet (with replacement).

Open in simulator

Problem 116

Compute probability of at least one event using complement counting: at least one even digit in a 2-digit number formed randomly (digits 0-9, with replacement).

Problem 117

Compute probability of at least one event using complement counting: at least one person out of 3 has a birthday in January.

multiply combinations across categories.

15 problems Warmup Practice Mixed Review Assessment

Problem 118

Compute probability with multiple categories selected: 5-card hand with 2 hearts and 3 clubs.

Problem 119

Compute probability with multiple categories selected: committee of 5 with 2 juniors from 6 and 3 seniors from 8.

Problem 120

Compute probability with multiple categories selected: choose 4 items with 1 red from 5, 1 blue from 4, and 2 green from 6.

Problem 121

Compute probability with multiple categories selected: select category counts r and s from groups R and S.

Problem 122

Compute probability with multiple categories selected: select 3 marbles from a bag with 4 red, 5 blue, and 6 green, such that 1 is red, 1 is blue, and 1 is green.

Problem 123

Compute probability with multiple categories selected: committee of 4 with 2 sophomores from 7 and 2 juniors from 9.

Problem 124

Compute probability with multiple categories selected: 7-card hand with 3 spades and 4 diamonds.

Problem 125

Compute probability with multiple categories selected: choose 6 books with 2 fiction from 10, 3 non-fiction from 8, and 1 poetry from 5.

Problem 126

Compute probability with multiple categories selected: task force of 5 with 2 engineers from 12 and 3 marketers from 10.

Problem 127

Compute probability with multiple categories selected: select 4 candies with 2 chocolates from 9 and 2 caramels from 7.

Problem 128

Compute probability with multiple categories selected: draw 5 balls from an urn containing 6 red, 7 blue, and 8 green balls, with 2 red, 2 blue, and 1 green.

Problem 129

Compute probability with multiple categories selected: debate team of 3 with 1 freshman from 5 and 2 sophomores from 6.

Open in simulator

Problem 130

Compute probability with multiple categories selected: choose 4 fruits from a basket with 7 apples, 6 oranges, and 5 bananas, such that 2 are apples, 1 is an orange, and 1 is a banana.

Problem 131

Compute probability with multiple categories selected: select 3 tools with 1 wrench from 4 and 2 screwdrivers from 5.

Problem 132

Compute probability with multiple categories selected: project team of 6 with 3 developers from 15, 2 designers from 8, and 1 tester from 7.

choose counting or sequential probability strategy.

15 problems Warmup Practice Mixed Review Assessment

Problem 133

Use permutations or combinations in a without-replacement probability problem: draw 3 cards in exact order ace, king, queen.

Problem 134

Use permutations or combinations in a without-replacement probability problem: draw 3 cards containing one ace and two non-aces, order irrelevant.

Problem 135

Use permutations or combinations in a without-replacement probability problem: select president and vice president from 10 with both seniors among 4 seniors.

Problem 136

Use permutations or combinations in a without-replacement probability problem: choose 2 winners from 20 including one named person.

Open in simulator

Problem 137

Use permutations or combinations in a without-replacement probability problem: draw 2 red cards then 1 black card in that exact order.

Problem 138

Use permutations or combinations in a without-replacement probability problem: draw 4 cards containing exactly two spades and two hearts, order irrelevant.

Problem 139

Use permutations or combinations in a without-replacement probability problem: select a chairman, secretary, and treasurer from 12 people, all three must be women from 7 women.

Problem 140

Use permutations or combinations in a without-replacement probability problem: form a committee of 3 from 15 people with exactly one engineer from 5 engineers.

Problem 141

Use permutations or combinations in a without-replacement probability problem: select 3 books from 10 and arrange them, such that the first is a novel (4 novels), the second a textbook (3 textbooks), and the third a biography (3 biographies).

Problem 142

Use permutations or combinations in a without-replacement probability problem: choose 2 red marbles and 1 blue marble from a bag of 5 red and 4 blue marbles, order irrelevant.

Problem 143

Use permutations or combinations in a without-replacement probability problem: draw 5 cards containing at least 3 hearts, order irrelevant.

Problem 144

Use permutations or combinations in a without-replacement probability problem: arrange 5 people in a line, with a specific person in the first position.

Problem 145

Use permutations or combinations in a without-replacement probability problem: select a team of 4 from 10 people, excluding two specific individuals.

Problem 146

Use permutations or combinations in a without-replacement probability problem: draw 3 cards in exact order: red, black, red.

Problem 147

Use permutations or combinations in a without-replacement probability problem: select 2 items from 6 (3 type A, 2 type B, 1 type C) and place them in order, such that the first is type A and the second is type B.

verify equivalent counts.

12 problems Warmup Practice Mixed Review Assessment

Problem 148

Compare two counting strategies for the same problem: at least one ace in 5 cards.

Problem 149

Compare two counting strategies for the same problem: one head and one tail in two flips.

Problem 150

Compare two counting strategies for the same problem: choose and arrange 3 from 8.

Problem 151

Compare two counting strategies for the same problem: committee with no seniors from 6 seniors and 9 juniors.

Problem 152

Compare two counting strategies for the same problem: selecting a president, vice-president, and secretary from 10 people.

Problem 153

Compare two counting strategies for the same problem: drawing 3 balls from a bag of 5 red and 5 blue, with at least one red.

Problem 154

Compare two counting strategies for the same problem: number of ways to form a 3-digit number using digits 1, 2, 3, 4, 5 without repetition.

Problem 155

Compare two counting strategies for the same problem: number of paths from (0,0) to (2,2) on a grid, only moving right or up.

Problem 156

Compare two counting strategies for the same problem: number of 5-card hands with at least one King or at least one Queen.

Problem 157

Compare two counting strategies for the same problem: choosing a committee of 3 from 10 people, where one is designated as chair.

Problem 158

Compare two counting strategies for the same problem: number of ways to seat 5 people around a circular table.

Problem 159

Compare two counting strategies for the same problem: number of subsets of a set with 4 elements.

Open in simulator

detect duplicated arrangements or missed restrictions.

12 problems Warmup Practice Mixed Review Assessment

Problem 160

Identify overcounting or undercounting in using 10P3 for a 3-person committee.

Problem 161

Identify overcounting or undercounting in using 10C3 for president, vice president, secretary.

Problem 162

Identify overcounting or undercounting in counting arrangements of AAB as 3!.

Problem 163

Identify overcounting or undercounting in counting at least one success by counting exactly one success.

Problem 164

Identify overcounting or undercounting in calculating the number of people who own a cat or a dog by adding the number of cat owners and dog owners.

Problem 165

Identify overcounting or undercounting in counting the number of distinct arrangements of the letters in 'MISSISSIPPI' as 11!.

Problem 166

Identify overcounting or undercounting in calculating the number of ways to choose 2 items from 5 with replacement as 5C2.

Problem 167

Identify overcounting or undercounting in calculating the probability of rolling a 4 on a die and flipping a head on a coin by adding P(4) + P(Head).

Problem 168

Identify overcounting or undercounting in calculating the probability of drawing two kings consecutively from a deck without replacement as (4/52) * (4/52).

Problem 169

Identify overcounting or undercounting in determining the number of ways to assign 3 different tasks to 3 different people from a group of 10 using 10C3.

Problem 170

Identify overcounting or undercounting in calculating the number of paths from (0,0) to (2,2) by moving only right or up as 4!.

Open in simulator

Problem 171

Identify overcounting or undercounting in calculating the number of possible passwords of length 5 using only lowercase letters, by considering only passwords with no repeated letters.

explain numerator, denominator, and event.

15 problems Warmup Practice Mixed Review Assessment

Problem 172

Interpret the combinatorial probability result in context: (13C2*39C3)/(52C5).

Problem 173

Interpret the combinatorial probability result in context: 1/(10P3).

Problem 174

Interpret the combinatorial probability result in context: (6C2*8C3)/(14C5).

Problem 175

Interpret the combinatorial probability result in context: 1-(48C5)/(52C5).

Problem 176

Interpret the combinatorial probability result in context: (5C2 * 4C1) / (12C3).

Problem 177

Interpret the combinatorial probability result in context: 1 / 7!.

Problem 178

Interpret the combinatorial probability result in context: 1 / (49C6).

Problem 179

Interpret the combinatorial probability result in context: (4C4 * 48C1) / (52C5).

Problem 180

Interpret the combinatorial probability result in context: (5C2 * 7C2) / (12C4).

Problem 181

Interpret the combinatorial probability result in context: 1 - (7C4) / (10C4).

Problem 182

Interpret the combinatorial probability result in context: 3! / (10P3).

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Problem 183

Interpret the combinatorial probability result in context: (1C1 * 51C4) / (52C5).

Problem 184

Interpret the combinatorial probability result in context: ((7C3) + (5C1 * 7C2)) / (12C3).

Problem 185

Interpret the combinatorial probability result in context: 3! / 5!.

Problem 186

Interpret the combinatorial probability result in context: (3P1 * 2P1) / (5P2).

catch order-matters confusion, restriction mistakes, and denominator errors.

15 problems Warmup Practice Mixed Review Assessment

Problem 187

Correct the permutations/combinations probability error: A student uses combinations for ranked prizes.

Problem 188

Correct the permutations/combinations probability error: A student uses permutations for a committee.

Problem 189

Correct the permutations/combinations probability error: A student uses 52P5 as the denominator for a 5-card hand.

Problem 190

Correct the permutations/combinations probability error: A student counts at least one ace as exactly one ace.

Problem 191

Correct the permutations/combinations probability error: A student calculates the number of ways to choose 3 pizza toppings from 10 using 10P3.

Problem 192

Correct the permutations/combinations probability error: A student uses combinations to find the number of ways to arrange 5 distinct books on a shelf.

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Problem 193

Correct the permutations/combinations probability error: A student uses 12C3 to select a president, vice-president, and secretary from 12 candidates.

Problem 194

Correct the permutations/combinations probability error: A student uses combinations to find the probability of drawing a King as the first card and a Queen as the second card.

Problem 195

Correct the permutations/combinations probability error: A student calculates the probability of drawing 2 red balls from a bag of 5 red and 5 blue balls using 10P2 as the total possible outcomes.

Problem 196

Correct the permutations/combinations probability error: A student calculates the probability of a specific arrangement of 6 unique items using 6C6 in the denominator.

Problem 197

Correct the permutations/combinations probability error: A student calculates the number of distinct permutations of the letters in 'STATISTICS' as 10!.

Problem 198

Correct the permutations/combinations probability error: A student finds the number of unique 6-digit numbers that can be formed using the digits 1, 1, 2, 3, 4, 5 as 6!.

Problem 199

Correct the permutations/combinations probability error: A student calculates the probability of getting at least one tail in 4 coin flips by only considering the case of exactly one tail.

Problem 200

Correct the permutations/combinations probability error: A student calculates the probability of drawing at least one blue marble from a bag of 5 red and 3 blue marbles (drawing 2 marbles) by only counting P(exactly 1 blue).

Problem 201

Correct the permutations/combinations probability error: A student calculates the number of ways to select a group of 3 people from 5 men and 4 women such that at least one woman is selected, by only counting groups with exactly one woman.