We are tasked with determining how many ways the committee positions can be filled, based on the given table.

Step-by-step analysis:

Each position has a certain number of people interested in it. Let's break it down:

1. Faculty member: 3 candidates (Mr. Barnes, Dr. Ward, Ms. Robinson).
2. Male student: 5 candidates (Alonzo, Justin, Ravi, Miguel, Deandre).
3. Female student: 5 candidates (Keisha, Salma, Maria, Donna, Ivanna).
4. Parent: 3 candidates (Mrs. Chang, Dr. Jenkins, Ms. Peterson).
5. School board member: 3 candidates (Mrs. Price, Mr. Russell).

Since the selections are independent, we multiply the number of choices for each position:

$\text{Total ways} = 3 \times 5 \times 5 \times 3 \times 3$

Now, let's calculate it:

$3 \times 5 = 15$ $15 \times 5 = 75$ $75 \times 3 = 225$ $225 \times 3 = 675$

Final Answer:

There are 675 ways to fill the five committee positions.

Would you like more details or have any questions?

Here are 5 relative questions to expand this topic:

1. What if the committee required two faculty members instead of one?
2. How would the number of ways change if there were more candidates for each position?
3. What if the selection of male and female students had to be made from the same pool of students?
4. Can you determine the probability of selecting a specific individual for each position?
5. How would adding a sixth committee position affect the total number of combinations?

Tip: When dealing with independent events in probability or counting, you multiply the number of choices for each event to get the total number of combinations.