1. Problem: How many 5-member committees can be formed from 9 sophomores and 12 seniors under different conditions? 2. Formula: The number of ways to choose $k$ members from $n$ is given by the combination formula: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ 3. a. All sophomores: Choose all 5 from 9 sophomores: $$\binom{9}{5} = \frac{9!}{5!4!} = 126$$ 4. b. All seniors: Choose all 5 from 12 seniors: $$\binom{12}{5} = \frac{12!}{5!7!} = 792$$ 5. c. 1 sophomore, 4 seniors: Choose 1 from 9 sophomores and 4 from 12 seniors: $$\binom{9}{1} \times \binom{12}{4} = 9 \times 495 = 4455$$ 6. d. 3 sophomores, 2 seniors: Choose 3 from 9 sophomores and 2 from 12 seniors: $$\binom{9}{3} \times \binom{12}{2} = 84 \times 66 = 5544$$ Final answers for problem 10: - a: 126 - b: 792 - c: 4455 - d: 5544