I have attempted part of this practice exercise in the text, but I need help in understanding the rest of it. I thought maybe another perspective would enlighten me. Thanks!
There are 7 women and 9 men on a faculty.
a.) How many ways are there to select a committee of 5 members if at least 1 member must be a woman?
b.) How many ways are there to select a committee of 5 members if at least 1 man and 1 woman must be on the committee?
For the first part, I so far have that there could be 4 men and 1 woman for the committee. There could also be 3 men and 2 women. My possibilities are: 4 men and 1 woman, or 3 men and 2 women. For part b, my possibilities are: 4 women and 1 man, or 3 women and 2 men.
After that, I am not sure how to proceed.
For part a), where you are asked to determine how many ways there are to select a committee of 5 people from a group of faculty members made up of 7 women and 9 men, consider the following. If AT LEAST 1 member must be a woman, primarily consider the gender composition of ALL POSSIBLE 5 person committees that you could select in this situation, regardless of whether the "at least 1 is a woman" requirement is satisfied or not:
(1) You choose 5 of the 9 men available AND 0 of the 7 women available, OR
(2) You choose 4 of the 9 men available AND 1 of the 7 women available, OR
(3) You choose 3 of the 9 men available AND 2 of the 7 women available, OR
(4) You choose 2 of the 9 men available AND 3 of the 7 women available, OR
(5) You choose 1 of the 9 men available AND 4 of the 7 women available, OR
(6) You choose 0 of the 9 men available AND 5 of the 7 women available.
Of the 6 possible outcomes in this problem, you have described situation #2 and #3, but failed to consider the other situations as possibilities. Clearly, situation #1 is the ONLY possible outcome that does NOT satisfy the requirement that AT LEAST one of the 5 committee members chosen is a woman. In mathematics, the word "and" generally means multiplication and the word "or" will generally lead to addition. So the answer to part a will be found if we can figure out how many ways there are for situation #2 to occur, and how many ways there are for situations #3, #4, #5 and #6 to occur. Adding up all of these numbers (because of the word "or" separating them) will yield the final answer.
For each possible ...
The following posting helps with problems involving combinations and different methods for statistical measurement.