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Venn Diagram tyoe question

  1. Nov 3, 2012 #1
    These kind of questions always get to me and I don't know how to solve them.

    Lets say that there are X many people that are in sports. Y of them are in soccer, Z of them are in cross country, and A of them are in basketball. And Y+Z+A>X

    How would i find out how many people do two sports or all three?
     
  2. jcsd
  3. Nov 3, 2012 #2

    arildno

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    Let S(2,3) be those who practice two or three sports, (Y,Z),(Y,A),(Z,A) those practicing two sports, and (Y,Z,A) those practicing 3.
    Then, S(2,3) equals the sum of those 4 disjoint groups.
    Agreed?
    Furthermore, let Y(0) be those ONLY playing soccer, and similarly for the 2 others.

    Then, we have the equation:
    S(2,3)+Y(0)+Z(0)+A(0)=X (*!*)

    Now, we have, of course, Y(0)=Y-(Y,Z)-(Y,Z,A) and so on.

    Now, inserting these into (*!*), we may simplify this to:

    Y+Z+A-S(2,3)-(Y,Z,A)=X (!!!)

    Therefore, in order to solve (!!!) for S(2,3) uniquely (knowing Y+Z+A and X), you need to know how many play 3 sports.

    Obviously, (Y,Z,A) must be less than or equal to S(2,3)

    This CAN help you in a specific case:
    If you know Y+Z+A-X=1, it follows immediately that (Y,Z,A)=0
     
    Last edited: Nov 3, 2012
  4. Nov 3, 2012 #3
    First, what do you mean by *!* is that some kind of factorial? And same thing with !!!? Other than those, I followed that pretty well.
     
  5. Nov 3, 2012 #4

    arildno

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    Those were NAMES I gave to my favourite equations. If you prefer to call them "Peter" and "Polly", by all means do so.
    :smile:
     
  6. Nov 3, 2012 #5
    Firstly find n(a^y)
    then n(y^z) , n(a^z) and n(a^y^z)
    The answer will be = n(a^Y)+n(y^z) + n(a^z) - 2*n(a^y^z)
    where n(a^y) denotes no. of players who play both soccer and basketball,
    n(a^y^z) denotes no. of players who play all the three games
     
    Last edited: Nov 3, 2012
  7. Nov 3, 2012 #6
    I'm sorry, you completly lost me there. I haven't learned about U or ^ yet.
     
  8. Nov 4, 2012 #7

    arildno

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    He is using ^ as the logical operator AND.
    There are many ways to split up a Venn problem, hopefully, the approach I gave you made sense (even though I gave my equations names, but didn't inform you on that)
     
  9. Nov 4, 2012 #8
    Ok. So now I'm trying to understand this more.
    I just made up these numbers
    There are 24 people.12 play soccer, 9 run cross, and 10 are in basketball. How many play two sports? How many play three?
    I tried putting in Y(0)=12-(y,z)-(y,z,a) but Im not.sure how to simplfy that.
     
  10. Nov 4, 2012 #9

    arildno

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    Well, X=24, Y+Z+A=31
    Thus, you have, by inserting in (!!!), and rearranging:
    S(2,3)+(Y,Z,A)=7 (agreed?)

    Now, this can refer to the following situations:
    a) There are 7 players who play two sports, none playing all
    b) There are 5 players who play two sports, and 1 playing all
    c) There are 3 players who play two sports, and 2 playing all
    d) There is 1 player who plays two sports, and 3 playing all
     
  11. Nov 4, 2012 #10

    arildno

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    In total, you have 70 unique arrangements satisfying the conditions you gave, with
    36 unique arrangements of the a)-solution
    21 unique arrangements of the b)-solution
    10 unique arrangements of the c)-solution
    3 unique versions of the d)-solution.
     
  12. Nov 4, 2012 #11
    So this question has multiple answers?
    Thank you for your guys help!
     
  13. Nov 4, 2012 #12

    arildno

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    With no further information given, yes.
    In exercises, there will usually be additional information to specify down to unique solution.
    You're welcome.
     
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