# Counting Rules

• May 10th 2006, 06:30 AM
skhan
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
• May 10th 2006, 02:54 PM
ThePerfectHacker
Quote:

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.
• May 10th 2006, 05:52 PM
skhan
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?
• May 10th 2006, 07:03 PM
ThePerfectHacker
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.
• May 10th 2006, 09:24 PM
JakeD
Quote:

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.