1. ## Counting Rules

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) Of the 4 people assigned to the first project, at least 3 are scientists? ANS: 750

..one person tried to help me with this problem and got 735 but the answer is suppose to be 750

2. Originally Posted by skhan
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) Of the 4 people assigned to the first project, at least 3 are scientists? ANS: 750

..one person tried to help me with this problem and got 735 but the answer is suppose to be 750
That means,
Combinitions that there are exactly 3 scientists.
+Combinations that there are exactly 4 scientists.

Combinations there are three scientists exactly, multiply number of ways choosing 3 scientists $_6C_3=20$ by number of ways of choosing engineers $=_3C_1=3$ Thus, 60 different ways.

Combinations there are four scientists exactly is,
$_6C_4=15$

If you add them up you get 75 different combinations.

3. the answer is actually 750 different ways.

Another way the problem could be approached would be like this (below) - however this gave an asnwer of 735 which is close to the right answer

Let's say there are 3 scientists on the 1st project.

1st group
------------
3S 1LT
3S 1LT
3S 1LT

2nd group
-----------
1S 2LT
2S 1LT
3S 0LT

3rd group
-----------
2S 0LT
1S 1LT
0S 2LT

Do the same for 4 scientists on 1st project as well and the computate.

Does anyone get an asnwer of 750 exactly?

4. Maybe you did not phrase your question correctly and both of us are doing different problems.

Let me explain why it cannot be 750 by the way I understand. How can it be 750 is ALL the possibilities are less than 750!!!

Let me explain you are taking 4 people out of 6 scientist and 3 engineers. How many different groups can you form?

4S 0E=6C4*3C0=15
3S 1E=6C3*3C1=20*3=60
2S 2E=6C2*3C2=15*1=15
1S 3E=6C1*3C3=6*1=6
0S 4E=0
In total we have, 96 possibilites.
Something is wrong with what you are saying.

5. Originally Posted by ThePerfectHacker
That means,
Combinitions that there are exactly 3 scientists.
+Combinations that there are exactly 4 scientists.

Combinations there are three scientists exactly, multiply number of ways choosing 3 scientists $_6C_3=20$ by number of ways of choosing engineers $=_3C_1=3$ Thus, 60 different ways.

Combinations there are four scientists exactly is,
$_6C_4=15$

If you add them up you get 75 different combinations.
Now multiply the 75 by the number of ways to assign the remaining 5 people to the other two projects, which is $_5C_3 = 10.$ That makes 750.