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Math Help - hard combination/permutation problem. please help!

  1. #1
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    hard combination/permutation problem. please help!

    In a committee, there is only one chairman, 2 secretaries, and 3 members. If there are only 6 people, how many ways are there to assign them these positions? If there are 2 candidates for chairman, 4 candidates for secretary positions, and 5 candidates for the members position, how many ways are there to get such a committee?

    please help
    thank you
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  2. #2
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    Re: hard combination/permutation problem. please help!

    i did 6 pick 1 x 6 pick 2 x 6 pick 3= 3600


    does anyone know if this is correct!?
    any help is greatly appreciated
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    Re: hard combination/permutation problem. please help!

    Hello, rsalort!

    In a committee, there is only one Chairman, 2 Secretaries, and 3 Members.
    If there are only 6 people, how many ways are there to assign them these positions?

    From the six people, choose one to be Chairman: . {6\choose1} = 6 choices.

    From the remaining five people, choose two to be Secretaries: . {5\choose2} = 10 choices.

    From the remaining three people. choose three to be Members: . {3\choose3} = 1 choice.

    Therefore, there are: . 6\cdot10\cdot1 \:=\:60 ways to form the committee.




    If there are 2 candidates for Chairman, 4 candidates for Secretary, and 5 candidates
    for the Members positions, how many ways are there to get such a committee?

    From the 2 Chairman candidates, choose one Chairman: . {2\choose1} = 2 choices.

    From the 4 Secretary candidates, choose two Secretaries: . {4\choose2} = 6 choices.

    From the 5 Member candidates, choose 3 Members: . {5\choose3} = 10 choices.

    Therefore, there are: . 2\cdot6\cdot10 \:=\:120 ways to form the committee.

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    Re: hard combination/permutation problem. please help!

    thank you soroban, i really appreciate it
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    Re: hard combination/permutation problem. please help!

    for the 2nd part of the question, are there 11 people or still only 6?
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    Re: hard combination/permutation problem. please help!

    Quote Originally Posted by agentmulder View Post
    for the 2nd part of the question, are there 11 people or still only 6?
    there are 11 people for 6 positions, but not any person can take any position, only two can be chairman, only 3 can be secretary, etc
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    Member agentmulder's Avatar
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    Re: hard combination/permutation problem. please help!

    Quote Originally Posted by rsalort View Post
    there are 11 people for 6 positions, but not any person can take any position, only two can be chairman, only 3 can be secretary, etc
    Oh, I see now, a person is not allowed to be a candidate for multiple positions in this problem.
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