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Suppose that

  • Set A has 5 elements
  • Set B has 6 elements
  • Set C has 7 elements
  • $\Omega$ has 10 elements

Determine the maximum and minimum number of elements the following set can have:

$ A^{c} \cup (B \cap C)$

I can see that there can be a maximum of 6 elements in $B \cap C$ but from there I'm at a loss. Would appreciate some pointers.

Thanks in advance!

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You're correct that the most number of elements $B \cap C$ can have is $6$. If $A$ has $5$ elements, then $A^c$ has $10 - 5 = 5$ elements. So the maximum number of elements in $A^c \cup (B\cap C)$ would seem to be $6 + 5 = 11$ elements. But since there are only $\,10\,$ elements in total (since we're given $\Omega$ has 10 elements), $A^C \cup (B \cap C) \leq 10$.

The minimum number of elements would be $5$, since $A^c$ has five elements, so we must have $5$ elements minimum in $A^c \cap (B\cap C)$. We can arrange $B, C$ such that the set $B\cap C$ consists of only those elements that are also in $A^c$. I.e., this minimum will be achieved when we have that $(B\cap C) \cup A^c = A^c$.

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    $\begingroup$ Shouldn't $(B\cap C) \cap A^c = \varnothing$ be $(B\cap C) \cup A^c = A^c$ or something equivalent? $\endgroup$ – Alraxite Jul 10 '13 at 21:16
  • $\begingroup$ Actually $(B \cap C) \cap A^{c}$ is the same as saying $B\cap C$ $\endgroup$ – zv.diego Jul 10 '13 at 21:39
  • $\begingroup$ Yes, Alraxite: that's exactly what I was trying to say. $\endgroup$ – amWhy Jul 10 '13 at 22:03
  • $\begingroup$ @amWhy: Needs another TU +1 $\endgroup$ – Amzoti Jul 11 '13 at 1:57
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Maximum: $10$, if only one element of $A$ is in $B\cap C$
Minimum: $5$, if none of the $5$ elements not in $A$ are in both $B$ and $C$.

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HINT: Either by drawing the Venn diagram or by working directly with the set algebra, you should be able to convince yourself that

$$\complement (A) \cup (B \cap C)$$

is the complement of

$$A\setminus(A\cap B\cap C)\;.$$

The question is then how large and how small you can make the intersection $A\cap B\cap C$. It clearly can’t have fewer than $0$ or more than $|A|=5$ elements. Both of these extremes can be achieved; can you see how?

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    $\begingroup$ But since there are only $10$ elements in the universal set, and considering the number of elements in $A,\, B$ and $C$, shouldn't there be at least one element in $A\cap B\cap C$? $\endgroup$ – Alraxite Jul 10 '13 at 20:59
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    $\begingroup$ @Alraxite: There need not be. It’s possible to arrange matters so that $A\cap B\cap C=\varnothing$, though of course the sets $A\cap B$, $A\cap C$, and $B\cap C$ cannot be empty. $\endgroup$ – Brian M. Scott Jul 10 '13 at 21:01
  • $\begingroup$ I was having the same questions as @Alraxite but i can see your point now. Appreciate your help, if only there could be more than one selected answer.. Thanks Brian! $\endgroup$ – zv.diego Jul 10 '13 at 21:12
  • $\begingroup$ @Diego: You’re welcome! $\endgroup$ – Brian M. Scott Jul 10 '13 at 21:24

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