# Combinations and Different Ways for Statistical Measurement

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.

#### Solution Preview

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 ...

#### Solution Summary

The following posting helps with problems involving combinations and different methods for statistical measurement.