Math 2 Quiz: Fundamental Counting Principle
4 questions · exam conditions
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Fundamental Counting PrincipleQuestion 1 of 4

In a tournament, 8 teams are divided into two groups of 4 teams each. Within each group, every team plays every other team exactly once. How many games are played in total during the group stage?

12
24
28
56
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Math 2 Quiz

Math 2 Quiz: Fundamental Counting Principle

Practice Fundamental Counting Principle in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Fundamental Counting Principle, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In a tournament, 8 teams are divided into two groups of 4 teams each. Within each group, every team plays every other team exactly once. How many games are played in total during the group stage?

  1. 12 (correct answer)
  2. 24
  3. 28
  4. 56
Explanation: In each group of 4 teams, every team plays every other team exactly once. This is equivalent to finding the number of ways to choose 2 teams from 4 to form a game: (42)=4!2!2!=6\binom{4}{2} = \frac{4!}{2!2!} = 6 games per group. Since there are 2 groups, the total number of games is 2×6=122 \times 6 = 12. Choice B (24) incorrectly calculates as if each team plays every other team in both groups. Choice C (28) incorrectly calculates (82)−(42)\binom{8}{2} - \binom{4}{2} as if teams from different groups also play. Choice D (56) calculates as if every team plays every other team in the entire tournament.

Question 2

A restaurant offers a prix fixe menu where customers must select exactly one appetizer, one main course, and one dessert. There are 4 appetizers, 6 main courses, and 3 desserts available. However, if a customer chooses the fish main course, they cannot select the chocolate dessert due to dietary restrictions. How many different complete meals are possible?

  1. 71
  2. 72
  3. 69 (correct answer)
  4. 68
Explanation: Without restrictions, there would be 4×6×3=724 \times 6 \times 3 = 72 possible meals. However, the fish main course cannot be paired with the chocolate dessert, which eliminates certain combinations. When fish is chosen (4 ways to choose appetizer × 1 way to choose fish × 1 way to choose chocolate), we lose 4×1×1=44 \times 1 \times 1 = 4 combinations. Therefore, the total is 72−4=6972 - 4 = 69 possible meals. Choice A (71) incorrectly subtracts only 1 restriction. Choice B (72) ignores the restriction entirely. Choice D (68) incorrectly subtracts 4 from both the fish-chocolate and some other imagined restriction.

Question 3

A computer generates 6-character passwords where each character is either a digit (0-9) or a lowercase letter (a-z). The password must contain at least one digit and at least one letter. Additionally, consecutive characters cannot be the same. How many such passwords are possible?

  1. 35^6 - 2 \times 26^6 - 2 \times 10^6 + 35
  2. 35 \times 35^5 - 26 \times 25^5 - 10 \times 9^5 + 1
  3. 36 \times 35^5 - 26 \times 25^5 - 10 \times 9^5 (correct answer)
  4. 36^6 - 26^6 - 10^6 + 1
Explanation: We use inclusion-exclusion principle combined with the constraint that consecutive characters cannot be the same. Total characters available: 10+26=3610 + 26 = 36. For passwords with no consecutive repeats: First character: 36 choices. Each subsequent character: 35 choices (any except the previous one). Total with no consecutive repeats: 36×35536 \times 35^5. Now subtract invalid cases: All letters (no digits) with no consecutive repeats: 26×25526 \times 25^5. All digits (no letters) with no consecutive repeats: 10×9510 \times 9^5. Final answer: 36×355−26×255−10×9536 \times 35^5 - 26 \times 25^5 - 10 \times 9^5. Choice A uses incorrect base calculations. Choice B incorrectly starts with 35 choices for the first character. Choice D ignores the consecutive character restriction entirely.

Question 4

A student has 5 different mathematics books, 4 different science books, and 3 different literature books. She wants to arrange them on a shelf so that books of the same subject are grouped together. Within each subject group, the books can be arranged in any order. How many different arrangements are possible?

  1. 103,680 (correct answer)
  2. 17,280
  3. 34,560
  4. 51,840
Explanation: This is a two-step counting problem. First, we arrange the three subject groups, then arrange books within each group. Step 1 - Arrange the 3 subject groups: 3!=63! = 6 ways. Step 2 - Arrange books within each subject: Mathematics books: 5!=1205! = 120 ways. Science books: 4!=244! = 24 ways. Literature books: 3!=63! = 6 ways. Total arrangements: 6×120×24×6=6×17,280=103,6806 \times 120 \times 24 \times 6 = 6 \times 17,280 = 103,680. Choice B (17,280) omits the factor for arranging subject groups. Choice C (34,560) uses an incorrect factor of 2 instead of 6 for subject arrangements. Choice D (51,840) uses an incorrect factor of 3 for subject arrangements.