Properties of Cardinality and intersection I 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 A?


What I did was this:
(11 - 5) + (12 - 5) = 13.
But I'm not sure if I have to substract the 5 twice or only once. And the fact that I don't know if B and C are disjoint gets me confused =/
 A: \begin{align}
\#(A\cap (B\cup C)) 
&= \#((A\cap B)\cup (A\cap C))\\ &=\#(A\cap B)+\#(A\cap C)-\#(A\cap B\cap C)\\ &=11+12-5=18.
\end{align} 
Note that $A$ cannot be fewer than $18$ since $A\cap(B\cup C)\subseteq A$, thus $\#(A)\geq18$.
Now, if we can find an example, where $\#(A)=18$, we'll have minimized the cardinality of $A$. For such an example, let $A=\{1,\ldots,18\}$, $B=\{1,\ldots,11\}$, and $C=\{7,\ldots,18\}$.
A: Note that
$$A = (A\cap B^c \cap C^c) \bigcup (A \cap B \cap C^c) \bigcup (A\cap B^c \cap C) \bigcup (A \cap B \cap C)$$
In the above, we have written $A$ as the union of four disjoint sets.
Hence, we get that
$$\#A = \#(A\cap B^c \cap C^c) + \#(A \cap B \cap C^c) + \#(A\cap B^c \cap C) + \#(A \cap B \cap C) \,\,\,\,\,\,\,\, (\star)$$
Now note that we can write $A \cap B$ and $A \cap C$ as a disjoint union as follows.
$$(A \cap B) = (A \cap B \cap C^c) \bigcup (A \cap B \cap C)$$ and 
$$(A \cap C) = (A \cap B \cap C) \bigcup (A \cap B^c \cap C)$$
Hence, we get that
$$\#(A \cap B) = \#(A \cap B \cap C^c) + \#(A \cap B \cap C)$$
and
$$\#(A \cap C) = \#(A \cap B \cap C) + \#(A \cap B^c \cap C)$$
Rearranging, we get that
$$\#(A \cap B \cap C^c) = \#(A \cap B) - \#(A \cap B \cap C)$$
$$\#(A \cap B^c \cap C) = \#(A \cap C) - \#(A \cap B \cap C)$$
Plugging the above two in $(\star)$, we get that
$$\#A = \#(A\cap B^c \cap C^c) + \#(A \cap B) + \#(A \cap C) - \#(A \cap B \cap C) \,\,\,\,\,\,\, (\dagger)$$
Plugging in the given values in $(\dagger)$, we get
$$\#A = \#(A\cap B^c \cap C^c) + 11 + 12 - 5 = \#(A\cap B^c \cap C^c) + 18$$
Note that $\#X \geq 0$ for any set $X$ and hence, the minimum value of $\#A$ is when $\#(A\cap B^c \cap C^c) = 0$, which gives us
$$\#A = 18$$
Note that the minimum is attained when $A \subseteq B \cup C$.
