Probability/Counting Rules Question

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SUMMARY

The forum discussion centers on a probability problem involving the assignment of staff to projects, specifically focusing on the distribution of scientists and lab technicians. The project director has 6 scientists and 3 lab technicians, and the goal is to assign 4 staff to the first project with at least 3 being scientists. The correct answer to this problem is determined to be 750 ways, although one participant initially calculated 735 due to an oversight in counting combinations. The discussion highlights the importance of considering all possible combinations and conditions when solving probability problems.

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  • Understanding of combinatorial mathematics, specifically combinations
  • Familiarity with basic probability concepts
  • Knowledge of how to apply the counting principle in probability
  • Experience with binomial coefficients, denoted as C(n, k)
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  • Study the principles of combinatorial counting in probability theory
  • Learn how to calculate binomial coefficients and their applications
  • Explore advanced probability problems involving multiple conditions
  • Practice solving similar staff assignment problems using combinatorial methods
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Students studying probability and combinatorics, educators teaching mathematical concepts, and anyone interested in solving complex counting problems in project management scenarios.

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

So I tried this problem, and i don't get 750 which is pretty frustrating...
Heres what I did:

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

1st group 2nd group 3rd group
------------ ----------- ------------

3S 1LT 1S 2LT 2S 0LT
3S 1LT 2S 1LT 1S 1LT
3S 1LT 3S 0LT 0S 2LT

Let's say there are 4 scientists on 1st project:

1st group 2nd group 3rd group
---------- --------- ---------
4S 0LT 2S 1LT 0S 2LT
4S 0LT 1S 2LT 1S 1LT

Counting up all these (omitting combinations which equal 1):

C(6,3)C(3,1)C(3,1)+C(6,3)C(3,1)C(3,2)C(2,1)+
C(6,3)C(3,1)+C(6,4)C(3,1)+C(6,4)C(2,1)C(3,2)
=180+360+60+45+90=735

...help :confused:
 
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Looks to me like you omitted one case for when there are 4 scientists on the 1st project. There isn't any condition that says there has to be a scientist on the 2nd project, is there?

By the way, you said you had trouble with two questions?
 

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