Venn diagram counting problem? 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!
 A: 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 $(B\cap C) \cup A^c = A^c$.
A: 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 $C$. 
A: 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?
