Answer to Question #264118 in Statistics and Probability for Nathan

Question

The director of a small management consulting company is to form a team of 5 management consultants from the 12 available to run a workshop. Exactly 8 of the 12 have sound presentation skills. Two of the 8 with sound presentation skills are feuding and cannot both be on the same team. In how many ways can a team of 5 management consultants that includes 3 with sound presentation skills be selected?

Answer 1

In how many ways can 3 management consultants having  sound presentation skills be selected?

$$\dbinom{8}{3}=\dfrac{8!}{3!(8-3)!}=\dfrac{8(7)(6)}{1(2)(3)}=56$$

Two of the 8 with sound presentation skills are feuding and cannot both be on the same team.  Then In how many ways can 3 management consultants having  sound presentation skills be selected now?

$$\dbinom{8}{3}-(8-2)=50$$

In how many ways can 2 management consultants without  sound presentation skills be selected?

$$\dbinom{12-8}{5-3}=\dfrac{4!}{2!(4-2)!}=\dfrac{4(3)}{1(2)}=6$$

In how many ways can a team of 5 management consultants that includes 3 with sound presentation skills be selected now?

$$50(6)=300$$