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Combination question help

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    A history exam paper contains eight questions, four in part A and four in B. Candidates are required to attempt five questions. In how many ways can this be done if at least two questions from part and at least two questions from B must be attempted


    2. Relevant equations



    3. The attempt at a solution
    I tried with (4 2) * (4 2) * (4 1) where (n r) is the no. of combinations of r objects from n objects. But it is the wrong answer. Give me the right solution and also why my reasoning in wrong
     
  2. jcsd
  3. Sep 5, 2011 #2
    Re: combination

    I would try (4C3)(4C2)+(4C2)(4C3). You can just do it once and multiply by 2. Does that give the right answer?
     
  4. Sep 5, 2011 #3
    Re: combination

    no that gives the wrong answer. i need a solution with explanation not just a formula.
     
  5. Sep 5, 2011 #4

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    Re: combination

    Hi batballbat! :smile:

    The number of answers given is either 8, 7, 6, or 5.
    How many ways to give say 6 answers?
    And how many of those satisfy the condition of at least 2 in each part?
     
  6. Sep 5, 2011 #5
    Re: combination

    i didnt understand your question
     
  7. Sep 5, 2011 #6

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    Re: combination

    The problem requires a candidate to give at least 5 answers.
    So he could for instance give 6 answers.
    At this stage I'm not looking yet whether he chooses questions from part A or from part B.
    Just the total number of combinations of questions he could choose.
    So he could for instance choose 1,2,3,4,5,6.
    Or he could choose 1,2,5,6,7,8.
    How many combinations if he chooses exactly 6 questions?
     
  8. Sep 5, 2011 #7

    tiny-tim

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    hi batballbat! :smile:
    because your (4,1) doesn't come from anywhere …

    what is the group of 4 things from which 1 has to be chosen?? :confused:
     
  9. Sep 5, 2011 #8
    Re: combination

    8c6........................................................
     
  10. Sep 5, 2011 #9
    Re: combination

    i know i am wrong with that reasoning. but i got (4 1) because the remaining questions is 4 after we have chosen 2 from each. And for the fifth question we have to choose one from the remaining
    yes there is repetition so this does not give the answer
     
  11. Sep 5, 2011 #10

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    Re: combination

    Yep! :smile:
    (without the dots :wink:)

    Now, say questions 1,2,3,4 are in part A, and questions 5,6,7,8 are in part B.
    If we choose questions 1,2,3,4,5,6, then you will have 4 questions from part A and 2 questions from part B.
    Which combination of questions wouldn't have at least 2 questions in each part?
     
  12. Sep 5, 2011 #11
    Re: combination

    4............................................. without the dots
     
  13. Sep 5, 2011 #12

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    Re: combination

    :rolleyes:

    Can you name a specific combination of questions?
     
  14. Sep 5, 2011 #13

    tiny-tim

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    ah! i see the logic now! :smile:

    nooo that doesn't work, because the new 4 isn't independent of the previous 2 …

    that ()()() formula only works for separate sets

    sometimes there's isn't a one-step answer to these probability questions, and you have to consider different cases

    try splitting it up into two cases, the way ArcanaNoir :smile: suggested …

    what do you get? :wink:
     
  15. Sep 5, 2011 #14
    Re: combination

    sorry my brain was repairing when i answered that
     
  16. Sep 5, 2011 #15
    Re: combination

    i still dont have the solution
     
  17. Sep 5, 2011 #16
    Re: combination

    Can you post the answer? Sometimes it's easier to help someone when you are sure you have it right before suggesting something. Plus, I didn't understand the question to mean you could attempt six questions. "required to attempt five" is not the same to me as "required to attempt at least five".
     
  18. Sep 5, 2011 #17
    Re: combination

    the 4 combinations are 5678(1,2,3,4)
     
  19. Sep 5, 2011 #18

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    Re: combination

    @ARCANANOIR :smile:: Well, your answer is the right one if the candidate answers exactly 5 questions...

    But as yet, I'm at a loss how to explain that.
    TBH my current tactic is to let the OP think about it for a while, since he has to be able to solve this and other problems himself.

    Do you want to give it a try?
     
  20. Sep 5, 2011 #19
    Re: combination

    No, I'm satisfied knowing I got the right answer according to my interpretation of the question. I still have my own probability nightmares to work on.
     
  21. Sep 5, 2011 #20

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    Re: combination

    Hah! So you got one answer for free! :wink:


    Errr... those are not combinations?

    I meant combinations like:
    1,2,3,4, 5,6
    1,2,3,4, 5,7
    3,4, 5,6,7,8
    1,4, 5,6,7,8

    These are 4 combinations of possible answers.
    Do they satisfy the criterion at least 2 in each part?
     
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