## 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

1. No-data Group Title

I thought that it would be (11 - 5) + (12 - 5) = 13

2. No-data Group Title

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.

3. No-data Group Title

And also I don't know if I can figure out the maximun cardinality with this information. Or even the cardinality of set A.

4. No-data Group Title

Can you see the math symbols? I think because I can't =/

5. aacehm Group Title

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.

6. aacehm Group Title

If they aren't disjoint, your solution of 13 sounds right.

7. No-data Group Title

@aacehm You mean $$#(A \cap B) + #(A \cap C)$$

8. aacehm Group Title

Yeah, but the number signs aren't working.

9. aacehm Group Title

sorry

10. No-data Group Title

Is this equality right? $$\#A\cap(B\cap C) = \#A + \# (B\cap C)$$

11. No-data Group Title

This is so confusing =/

12. aacehm Group Title

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

13. aacehm Group Title

is there information given about which sets are disjoint?

14. aacehm Group Title

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

15. mathmate Group Title

The minimum cardinality of A is $$\#(A\cap B)\cup (A \cap C)=\#(A\cap B) +\#(A \cap C) - \#(A\cap B \cap C)=11+12-5=18$$

16. mathmate Group Title

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17. No-data Group Title

Thank you, this is what i got: $$A\cap(B\cup C)\subseteq A$$ Then $$\#A\geq \#[ A\cap(B\cup C)]$$ $$\#[ A\cap(B\cup C)]=\#[(A\cap B)\cup(A\cap C)]$$ $$\#[ A\cap(B\cup C)]=\#(A\cap B)+\#(A\cap C)-\#(A\cap\ B\cap C)$$ $$\#[ A\cap(B\cup C)]=11+12-5=18$$

18. No-data Group Title

Then as you said @mathmate the miminum cardinality of A is 18

19. mathmate Group Title

Yep, that's very convincing! Good job!