Theory set problem, determining the min and max number of elements from $(B \cup A) \bigtriangleup (C \cap A)$ Given that $$ |A| = 5 \\ |B| = 6 \\ |C| = 7 \\ |\Omega| = 10 \\ A \subseteq (B \cup C) $$
Determine the min and max numbers of elements from $$(B \cup A) \bigtriangleup (C \cap A)$$
I tried to solve this by first simplifying the above statement doing
$$[(B \cup A) \cap (C \cap A)^{c}] \cup [(B \cup A)^{c} \cap (C \cap A)] \\ [(B \cup A) \cap (C \cap A)^{c}] \cup [(B^{c} \cap A^{c}) \cap (C \cap A)] \\ [(B \cup A) \cap (C \cap A)^{c}] \cup \emptyset \\ (B \cup A) \cap (C \cap A)^{c} $$
Then i drew the venn diagrams that would represent the max and min case for $(B \cup A) \cap (C \cap A)^{c}$:


And according to this: 


*

*the minimum number of elements is 1 

*the maximum number of elements is 4


Question: Would this be correct? If so, is there any other faster way that can let me solve the exercise without drawing venn diagrams and trying to configure the elements for both cases?
Thanks so much in advance.
 A: Use the fact that $X\bigtriangleup Y=(X\cup Y)\setminus(X\cap Y)$, with $X=B\cup A$ and $Y=C\cap A$:
$$\begin{align*}
(B\cup A)\cup(C\cap A)&=(B\cup A\cup C)\cap(B\cup A\cup A)&&\text{distributivity}\\
&=(B\cup C)\cap(B\cup A)&&A\subseteq B\cup C\\
&=B\cup(C\cap A)&&\text{distributivity}
\end{align*}$$
and
$$\begin{align*}
(B\cup A)\cap(C\cap A)&=(B\cap C\cap A)\cup(A\cap C\cap A)&&\text{distributivity}\\
&=(B\cap C\cap A)\cup(C\cap A)\\
&=C\cap A\;,
\end{align*}$$
so
$$\begin{align*}
(B \cup A) \bigtriangleup (C \cap A)&=\Big(B\cup(C\cap A)\Big)\setminus(C\cap A)\\
&=B\setminus(C\cap A)\;.
\end{align*}$$
Corrected:
To make this large, you want to make $A\cap B\cap C$ as small as possible, and to make it small, you want to make $A\cap B\cap C$ as large as possible. If you put the $4$ elements of $\Omega\setminus B$ into both $A$ and $C$, you can arrange the rest so as to make $A\cap B\cap C=\varnothing$, and you certainly can’t make it any smaller! At the other extreme, if $A\subseteq B\cap C$, then $A\cap B\cap C=A$ and has cardinality $5$. Thus, the range of cardinalities for the symmetric difference is from $6-5=1$ to $6-0=6$.
A: A more systematic approach would be to partition $\Omega$ into $S_0, \ldots S_7$ as all possible combinations of $A$, $B$ and $C$, and translate
\begin{align}
A &= S_1 \uplus S_3 \uplus S_5 \uplus S_7, \\
B &= S_2 \uplus S_3 \uplus S_6 \uplus S_7, \\
C &= S_4 \uplus S_5 \uplus S_6 \uplus S_7, \\
(B \cup A) \bigtriangleup (C \cap A) &= (S_{1\uplus 2 \uplus 3 \uplus 5 \uplus 6 \uplus 7}) \bigtriangleup (S_{5 \uplus 7}) \\
&= S_1 \uplus S_2 \uplus S_3 \uplus S_6.
\end{align}
From $A \subseteq B \cup C$ we know that $S_1 = \varnothing$ and that gives us the following:
\begin{align}
s_1 &= 0\\
s_1+s_3+s_5+s_7 &= 5 \\
s_2+s_3+s_6+s_7 &= 6 \\
s_4+s_5+s_6+s_7 &= 7 \\
s_0+s_1+s_2+s_3+s_4+s_5+s_6+s_7 &= 10
\end{align}
where $s_i = |S_i|$ (don't forget about the $0 \leq s_i$ constraints). Now your question can be stated as: what is the maximum and minimum of $s_1+s_2+s_3+s_6$? Note that $s_1 = 0$ and $S_2 \uplus S_3 \uplus S_6$ is exactly $B \setminus (C \cap A)$ (as in Brian's answer).
I hope this helps ;-)
Edit: Cleared issue with $(B \cup A) \triangle (C\cap A)$ versus $(B \cup A) \triangle (C \cap A)^c$.
