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Math Help - Counting Rules

  1. #1
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    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
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  2. #2
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    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.
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  3. #3
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    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?
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  4. #4
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    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.
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  5. #5
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    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.
    Last edited by JakeD; May 10th 2006 at 10:41 PM.
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