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No-data Group Title

Cardinality properties: Suppose you have three sets A, B, and C, satisfying the following conditions: \( \# (A\cap B)=11 \) \( \# (A\cap C)=12\) \( \# (A\cap B \cap C)=5\) What is the minimun cardinality of set A?

  • one year ago
  • one year ago

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  1. No-data Group Title
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    I thought that it would be (11 - 5) + (12 - 5) = 13

    • one year ago
  2. No-data Group Title
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    I'm not sure if I'm right and I don't know if the fact that B a C are disjoint set could afect the cardinality.

    • one year ago
  3. No-data Group Title
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    And also I don't know if I can figure out the maximun cardinality with this information. Or even the cardinality of set A.

    • one year ago
  4. No-data Group Title
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    Can you see the math symbols? I think because I can't =/

    • one year ago
  5. aacehm Group Title
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    well, if B and C are disjoint, wouldn't it make sense that the cardinality of A is at least \[(A \cap B) + (A \cap C)\]? Because the minimum cardinality of B has to be 11, which means that A is at least 11, but there are also 12 elements that are not in B but have to be in A as well.

    • one year ago
  6. aacehm Group Title
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    If they aren't disjoint, your solution of 13 sounds right.

    • one year ago
  7. No-data Group Title
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    @aacehm You mean \( #(A \cap B) + #(A \cap C) \)

    • one year ago
  8. aacehm Group Title
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    Yeah, but the number signs aren't working.

    • one year ago
  9. aacehm Group Title
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    sorry

    • one year ago
  10. No-data Group Title
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    Is this equality right? \( \#A\cap(B\cap C) = \#A + \# (B\cap C) \)

    • one year ago
  11. No-data Group Title
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    This is so confusing =/

    • one year ago
  12. aacehm Group Title
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    That equality doesn't sound right.\[A = \left\{5, 6, 7..100\right\}\] \[B = \left\{1, 2, 3\right\}\] \[C = \left\{3,4,5\right\}\] Then the left side of your equation would 1, and the right side would be 97

    • one year ago
  13. aacehm Group Title
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    is there information given about which sets are disjoint?

    • one year ago
  14. aacehm Group Title
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    Actually, from statement 3, you can infer that B and C aren't disjoint, because if they were, then it would equal 23, not 5

    • one year ago
  15. mathmate Group Title
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    The minimum cardinality of A is \( \#(A\cap B)\cup (A \cap C)=\#(A\cap B) +\#(A \cap C) - \#(A\cap B \cap C)=11+12-5=18 \)

    • one year ago
  16. mathmate Group Title
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    |dw:1356902619299:dw|

    • one year ago
  17. No-data Group Title
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    Thank you, this is what i got: \( A\cap(B\cup C)\subseteq A\) Then \(\#A\geq \#[ A\cap(B\cup C)]\) \(\#[ A\cap(B\cup C)]=\#[(A\cap B)\cup(A\cap C)]\) \(\#[ A\cap(B\cup C)]=\#(A\cap B)+\#(A\cap C)-\#(A\cap\ B\cap C)\) \(\#[ A\cap(B\cup C)]=11+12-5=18\)

    • one year ago
  18. No-data Group Title
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    Then as you said @mathmate the miminum cardinality of A is 18

    • one year ago
  19. mathmate Group Title
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    Yep, that's very convincing! Good job!

    • one year ago
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