Probability/Counting Rules Question

  • Thread starter Thread starter skhan
  • Start date Start date
  • Tags Tags
    Rules
AI Thread Summary
The discussion centers on a probability problem involving the assignment of staff to three projects, specifically focusing on how many ways to assign at least three scientists to the first project. The original poster calculates a total of 735 combinations but believes the correct answer should be 750. They detail their counting method and express frustration over the discrepancy, suggesting a potential omission in their calculations. Another participant points out that there is no requirement for a scientist to be assigned to the second project, which could affect the total count. The conversation highlights the complexities of combinatorial counting in probability scenarios.
skhan
Messages
2
Reaction score
0
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:
 
Physics news on Phys.org
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?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

Similar threads

Back
Top