1. **Problem Statement:** (A) How many ways can a 4-person subcommittee be selected from a committee of 8 people? (B) How many ways can a president, vice-president, secretary, and treasurer be chosen from a committee of 8 people? 2. **Formulas and Rules:** - For (A), selecting a subcommittee means choosing 4 people out of 8 without regard to order. This is a combination problem. - The formula for combinations is $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ where $n$ is the total number of people and $r$ is the number chosen. - For (B), choosing distinct officers means order matters because each position is different. This is a permutation problem. - The formula for permutations is $$P(n,r) = \frac{n!}{(n-r)!}$$ 3. **Calculations:** (A) Given $n=8$ and $r=4$: $$\binom{8}{4} = \frac{8!}{4!\times(8-4)!} = \frac{8!}{4!\times4!}$$ Calculate factorials: $$8! = 40320, \quad 4! = 24$$ So: $$\binom{8}{4} = \frac{40320}{24 \times 24} = \frac{40320}{576} = 70$$ (B) Given $n=8$ and $r=4$: $$P(8,4) = \frac{8!}{(8-4)!} = \frac{8!}{4!}$$ Calculate factorials: $$8! = 40320, \quad 4! = 24$$ So: $$P(8,4) = \frac{40320}{24} = 1680$$ 4. **Explanation:** - For (A), since the order does not matter, we use combinations and find there are 70 ways. - For (B), since each position is unique, order matters, so we use permutations and find there are 1680 ways. **Final answers:** (A) 70 (B) 1680