Probability and Counting Rules - Question

[skhan]

Hello:
I was having trouble answering these two probability questions, so assistance from anyone would be much appreciated.

A project director runs a staff consisting of 6 scientists and 3 lab technicians. Three new projects have to be worked on and the director decides to assign 4 of her staff to the first project, 3 to the second project and 2 to the third project. In how many ways can this be accomplished if:

a) Each project requires one lab technician? ANS: 360
b) Of the 4 people assigned to the first project, at least 3 are scientists? ANS: 750

The way I go about answering part B gives my an answer of 24 which is way off. (2!/1!2!) x (3!/1!2!) x (4!/1!3!) = 2 x 3 x 4 = 24

..help :S

[OfficeShredder]

Since there are three lab technicians, there are 3*2*1=6 ways to arrange them. To sort out the scientists, there are 6 ways to pick the scientist for the third project. There are 5C2 ways to pick the scientists for the second project, and 1 way to pick the three remaining scientists for the first project.

6*6*5!/3!/2! = 6*5*4*3=360

For the second part, there are either 3 scientists and a technician or 4 scientists working on the first project. Looking at each case individually:

There are 6C4 = 15 ways to pick the four scientists for the first group. There are then 5C3 = 10 ways to pick the remaining three people for the second group, giving you a total of 150 ways to sort the people out of four scientists are in the first group.

If there are three scientists and a technician, then you have 3*6C3 = 60 ways to pick the first group. Additionally, there are again 5C3 ways to pick three people to fill the second group, and one way to pick the remaining two people for the third group, giving a total of 600 ways to sort the people if there is a technician in the first group.

600+150 = 750