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Math Help - Organized counting

  1. #1
    Newbie Jason2JZ's Avatar
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    Exclamation Organized counting

    So question2....

    There are 12 questions on an examination, and each student must answer 8 questions including at least 4 of the first 5 questions. how many different combinations of questions could a student choose to answer?
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  2. #2
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    Organized counting

    There are two ways to do what's required:

    (1) Answer all 5 of the first 5 questions, and 3 of the remaining 7.
    (2) Answer 4 of the first 5 five questions, and 4 of the remaining 7.

    For (1), there's only one way to answer 5 of 5, and 7 choose 3 = 35 ways to select 3 from 7. So there are 35 ways to accomplish (1).

    For (2), there are 5 ways to select 4 of the first 5, and 7 choose 4 = 35 ways to select 4 of the remaining 7. So there are 5 * 35 = 175 ways to accomplish (2).

    In total, then, I believe the answer would be 35+175=210 ways to do either (1) or (2).

    - Steve J
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  3. #3
    Newbie Jason2JZ's Avatar
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    ahh ic thank you!
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  4. #4
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    Hello, Jason2JZ!

    There are 12 questions on an examination.
    Each student must answer 8 questions including at least 4 of the first 5 questions.
    How many different combinations of questions could a student choose to answer?
    We must break this down into two cases . . .


    [1] The student answers exactly 4 of the first 5 questions.
    . . .He has {5\choose4} = 5 choices.
    Then he picks 4 more questions from the last 7 questions.
    . . .He has {7\choose4} = 35 choices.

    He has: 5 \times 35 \:=\:175 ways if he picks 4 of the first 5 questions.


    [2] The student answers all 5 of the first 5 questions.
    . . .He has 1 way to do this.
    The he picks 3 more question from the last 7 questions.
    . . .He has {7\choose3} = 35 choices.

    He has: 1 \times 35 \:=\:35 ways if he picks all 5 of the first 5 questions.


    Therefore, he has a total of: 175 + 35 \:=\:\boxed{210} ways to choose his 8 questons.

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