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Thread: refreshment on sets

  1. #1
    Senior Member
    Jan 2009

    refreshment on sets

    In an examination , part A was attempted by 60 students , part B by 52 students and part C by 45 students . 22 students attempted both parts A and B , 12 attempted both parts B and C , 20 attempted parts A and C and 4 attempted all three parts .

    (a) How many students attempted part A but not parts B and C ?

    (b) How many students attempted part Bbut not parts A and C ?

    (c) How many students attempted at least 2 parts ?

    For (A) , n(A-(B n C))=60-12=48 but it is wrong .

    i tried the same method on B and i hv no idea at all for c .

    THanks .
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  2. #2
    MHF Contributor Swlabr's Avatar
    May 2009
    I presume you know the answers (as you know that one is wrong) so I'll give you what I got:
    1) 22
    2) 22
    3) 46

    How I got them? Well, I started by looking at the unions of the sets:
    $\displaystyle |A \cup B| = |A|+|B|-|A \cap B| $, etc. Then, I couldn't think up some clever way so I decided to do the obvious thing and...draw a diagram!

    Draw three circles which all intersect one another. Label them A, B and C. There should be a triangle in the middle where they all intersect one another - this represents the set $\displaystyle A \cap B \cap C $. Similarly, where the circles A and B intersect this represents the intersection of the sets A and B, etc. We shall now put numbers in the spaces to represent the size of the sets.

    So you know what to put in $\displaystyle (A \cap B) \cap C $ - that's a 4. What about in $\displaystyle (A \cap B) \setminus C $ - that's simply 22-4=18. Then just continue putting numbers into this diagram until every space has a number in it. Questions 1 and 2 are easily read from this diagram, and question 3 isn't really hard to see on the diagram either...

    I hope that makes sense, and also that it helps!

    EDIT: essentially, the answer to part 1 is the cardinality of the set $\displaystyle (A \setminus B) \setminus C$, which is $\displaystyle |A| - |A \cap C| - |A \cap B| + |A \cap B \cap C|$ as you have "removed" $\displaystyle A \cap B \cap C$ twice (as it is in both $\displaystyle A \cap C $ and $\displaystyle A \cap B$) so you need to put one of them back.

    EDIT 2: Your diagram should look something like
    just with different numbers, and less stolen from someone else's web site...(it's too big to put in my post properly)
    Last edited by Swlabr; May 18th 2009 at 03:09 AM.
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