Assume a class has 19 members. a. In how many ways can a president, a vice president, and a secretary be selected? b. How many committees of 4 people can be chosen?

Let's solve each part of the problem step by step. --- ### **a. Selecting a President, Vice President, and Secretary** Since the positions of president, vice president, and secretary are distinct, the order in which we select the members matters. This is a permutation problem. **Number of ways to select the officers:** 1. **President:** There are 19 possible choices. 2. **Vice President:** After selecting the president, there are 18 remaining choices. 3. **Secretary:** After selecting the president and vice president, there are 17 remaining choices. **Total number of ways:** \[ 19 \times 18 \times 17 = 5,\!814 \text{ ways} \] --- ### **b. Choosing a Committee of 4 People** When forming a committee, the order in which the members are selected does not matter. This is a combination problem. **Number of ways to choose the committee:** We use the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n - k)!} \] where \( n = 19 \) and \( k = 4 \). Plugging in the values: \[ \binom{19}{4} = \frac{19!}{4! \times 15!} = \frac{19 \times 18 \times 17 \times 16}{4 \times 3 \times 2 \times 1} = \frac{93,\!024}{24} = 3,\!876 \text{ ways} \] --- **Summary:** - **a.** There are **5,814** ways to select a president, vice president, and secretary. - **b.** There are **3,876** ways to choose a committee of 4 people.