Final answer:
To determine the number of ways to select a five-member committee with at least one woman from a department of 7 women and 9 men, subtract the number of all-men committee combinations from the total combinations. The total number is the combination of 16 select 5 minus the combination of 9 select 5, which equals 4242 committees.
Step-by-step explanation:
The student has asked: 'How many ways are there to select a committee of five members of the department if at least one woman must be on the committee?' when considering a department with 7 women and 9 men.
To solve the problem, we will use the concept of combinations. There's a total of 16 faculty members, and the question specifies that the committee must include at least one woman.
The total number of ways to select a committee of five without any restrictions is the combination of 16 faculty members taken 5 at a time, which is:
C(16,5) = 16! / [(16-5)! * 5!]
However, we need at least one woman on the committee, so we must subtract the number of committees that have no women (and therefore, all men)
C(9,5) = 9! / [(9-5)! * 5!]
Therefore, the number of committees with at least one woman is the total number of committees minus the number of all-men committees:
C(16,5) - C(9,5)
When calculating these values, you find:
C(16,5) = 4368
C(9,5) = 126
So, the number of ways to select a committee with at least one woman is:
4368 - 126 = 4242
This represents the total number of possible committees with the desired condition.