A computer programming team has 15 members.

Question

(a) How many ways can a group of eight be chosen to work on a project?

What I got: C(15,8)

(b) Suppose nine team members are women and six are men.

- i. How many groups of eight can be chosen that contain four women and four men?

What I got: C(9,4)*C(6,4)

- ii. How many groups of eight can be chosen that contain at least one man?

What I got: C(15,8) - 1 (minus 1 for the case all are women)

- iii. How many groups of eight can be chosen that contain at most three women?

What I got: C(9,1)C(6,7) + C(9,2)P(6,6) + C(9,3)P(6,5)

(c) Suppose two team members refuse to work together on projects. How many groups of eight can be chosen to work on a project?

What I got: C(15,8) - C(13,6)

(d) Suppose two team members insist on either working together or not at all on projects. How many groups of eight can be chosen to work on a project?

What I got: C(13,6) + C(13,8)

I didn't know how to do the formatting of the combinations , sorry about that. I just wanted to see if I was approaching this question right. I feel that b)iii) is wrong though. Thanks for anyone looking at this question!

Answer 1

b ii) To form a team of all women, there are $\binom{9}{1}$ options, we just have to exclude one woman. Hence rather than subtract by $1$, you should subtract by $9$.

part $c$, why is it $\binom{12}{6}$, how do you obtain $12$?

For part $d$, same comment as part $c$.

Answer 2

To add to Siong Thye Goh, (b) (iii) is right in form, but wrong on the details. The number of groups of eight that have exactly two women is ${9\choose 2}{6\choose 6}$ (which is just ${9\choose 2}$) and the number that have exactly three is ${9\choose 3}{6\choose 5}.$ There are no groups of eight than contain exactly one, or zero women. You seem to have included the case of one (but not zero) and this is reflected in the fact that ${6\choose 7}$ does not make sense (or is zero, depending on how you think about it).