Fundamental Counting Principle - Math 2

1. 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?
   Answer: 36 \times $35^5$ - 26 \times $25^5$ - 10 \times $9^5$ We use inclusion-exclusion principle combined with the constraint that consecutive characters cannot be the same. Total characters available: $10 + 26 = 36$. 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: $36 \times 35^5$. Now subtract invalid cases: All letters (no digits) with no consecutive repeats: $26 \times 25^5$. All digits (no letters) with no consecutive repeats: $10 \times 9^5$. Final answer: $36 \times 35^5 - 26 \times 25^5 - 10 \times 9^5$. 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.
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?
   Answer: 69 Without restrictions, there would be $4 \times 6 \times 3 = 72$ 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 $4 \times 1 \times 1 = 4$ combinations. Therefore, the total is $72 - 4 = 69$ 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.
3. 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?
   Answer: 103,680 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: $3! = 6$ ways. Step 2 - Arrange books within each subject: Mathematics books: $5! = 120$ ways. Science books: $4! = 24$ ways. Literature books: $3! = 6$ ways. Total arrangements: $6 \times 120 \times 24 \times 6 = 6 \times 17,280 = 103,680$. 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.
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?
   Answer: 12 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: $\binom{4}{2} = \frac{4!}{2!2!} = 6$ games per group. Since there are 2 groups, the total number of games is $2 \times 6 = 12$. Choice B (24) incorrectly calculates as if each team plays every other team in both groups. Choice C (28) incorrectly calculates $\binom{8}{2} - \binom{4}{2}$ as if teams from different groups also play. Choice D (56) calculates as if every team plays every other team in the entire tournament.

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