# Number of subsets $Z$ of $X$ such that $(Y - Z)\cup(Z-Y)=\{3\}$

$X = \{1,2,3,4,...10\}$ and $Y = \{1,2,3,4,5\}$. The number of subsets $Z$ of $X$ such that $(Y - Z)\cup(Z-Y)=\{3\}$ is ?

What is most generalized approach for these kind of questions?

Number of subsets $Z$ of $X$ =1
Explanation: Given, $(Y - Z)\cup(Z-Y)=\{3\}$ and $Y = \{1,2,3,4,5\}$
To satisfy this condition $Z$ should be subset $Y$ and $Z$ should not . The only solution is $Y = \{1,2,3,4,5\}$ and $Z = \{1,2,4,5\}$
Note that $3\notin Z\setminus Y$ for any subset $Z$ of $X$. Hence, you need only find the number of subsets $Z$ of $X$ such that $Y\setminus Z=\{3\}$ and $Z\setminus Y=\emptyset$.
Let's think about this in somewhat more general terms: Suppose we have some set $X,$ with given subsets $Y,A\subseteq X,$ and that we're tasked to find the (number of) subsets $Z\subseteq X$ such that $$(Y\setminus Z)\cup(Z\setminus Y)=A.$$
Note that for any set $Z$ we have $(Y\setminus Z)\cap(Z\setminus Y)=\emptyset.$ Thus, in order to have $(Y\setminus Z)\cup(Z\setminus Y)=A,$ we must have $Y\setminus Z=Y\cap A$ and $Z\setminus Y=A\setminus Y.$ From the first of these conditions, it follows that $Z\cap Y=Y\setminus (Y\cap A),$ and so $Z\cap Y=Y\setminus A.$ Hence recalling the second of these conditions and the fact that $$Z=(Z\cap Y)\cup(Z\setminus Y),$$ we have $$Z=(Y\setminus A)\cup(A\setminus Y),$$ and so there is exactly one solution.