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Math Help - finite math- venn diagrams and such.

  1. #1
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    finite math- venn diagrams and such.

    the problem is as follows:

    Partners in an accounting firm.

    Assume that each partner has at least one specialization.

    Specialization-------------Number

    Auditing---------------------13
    Consulting------------------14
    Tax--------------------------11
    Auditing and consulting-----8
    Auditing and tax-------------6
    Consulting and tax----------7
    All three---------------------3


    How many partners are there?

    i know that the problem would be set up like the picture I drew below, and that i'm trying to find ((A ∪ C) ∪ T), but i just can't figure it out.

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  2. #2
    MHF Contributor
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    Quote Originally Posted by sdsdsd View Post
    the problem is as follows:

    Partners in an accounting firm.

    Assume that each partner has at least one specialization.

    Specialization-------------Number

    Auditing---------------------13
    Consulting------------------14
    Tax--------------------------11
    Auditing and consulting-----8
    Auditing and tax-------------6
    Consulting and tax----------7
    All three---------------------3


    How many partners are there?

    i know that the problem would be set up like the picture I drew below, and that i'm trying to find ((A ∪ C) ∪ T), but i just can't figure it out.

    I am not an expert on Venn diagrams as these were not discussed very well in my school years long time ago. But I find them now interesting. I just use "common sense" now in figuring them out.

    I do not agree with your distribution as shown on the posted diagram.

    Take the case of the partners engaged in Auditing.
    Auditing ..........................13 partners
    Auditing and Consulting ......8
    All 3...(A +C +T) ...............3

    You forgot that the 3 partners in the (A +c +T) are engaged in A and C also that is why you put 8 in the in the "compartment" for (A +C). The number there should be 5 only.

    Then you put 13 on the compartment for A only. You forgot again that the 3 in the (A +C +T) and the 5 in the (A +C) are engaged in Auditing also.
    And then there are partners in the (A +T) that are egaged in Auditing also...these are 3 partners.
    So instead of 13, you should have put (13 -3 -5 -3) = 2 only in there.

    Following the reasonings above, you should come up with 2 only (instead of 14) in the C, and 1 only (instead of 11) in the T.

    In the end, the total partners are the sum of all the numbers in the diagram.
    A = 2
    C = 2
    T = 1
    A+C = 5
    A+T = 3
    C+T = 4
    A+C+T = 3
    -----------------
    Total = 20 partners -----------answer.
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  3. #3
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    ahhhhh thanks. also i should i have explained how i was doing my distribution... i knew that the 13 in A was a mixture of those who solely did accounting, those who did accounting and consulting, and those who did all 3, etc. i didn't mean that 13 solely did auditing. they were more like tentative numbers waiting to be subtracted from.

    anyway i failed to realize that in order to figure out the correct number of elements in each compartment, it is necessary to first subtract the "tri-intersection" from each of the "dual intersections". I went straight into subtracting the "dual intersections" from A, C, and T without first subtracting 3. Leaps of logic ftw.

    it's all clear now. i appreciate the help.
    Last edited by sdsdsd; September 11th 2008 at 05:11 PM.
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