If you understand this question, please limit your post to an explanation—don't post a solution. I'd like to prove it myself. :)
This is taken from Page 12 of Dan Saracino's "Abstract Algebra, A First Course". $A\triangle B$ is the symmetric difference of the sets $A$ and $B$.
If $X$ is a set and $A_1, ..., A_n$ are elements of $(P(X), \triangle$) then by Exercise 2.13 $A_1 \triangle A_2 \triangle ... \triangle A_n$ has an unambiguous meaning [that is, we can anywhere place parenthesis and get the same result]. Prove that for every $n \geq 1$, the elements of $X$ that are in $A_1 \triangle A_2 \triangle ... \triangle A_n$ are exactly those elements that are in $A_j$ for an odd number of $j$'s in $\{1, 2, ..., n\}$.
The part that's confusing me is "... are exactly those elements that are in $A_j$ for an odd number of $j$'s in $\{1, 2, ..., n\}$." It doesn't say "an odd numbered $j$," which wouldn't make sense anyway unless an ordering is specified. I thought it could mean "exactly those elements that appear in an odd number of elements of $P(X)$," but shouldn't every element of $X$ appears $2^{n-1}$ times?
I've gone through maybe 3 or 4 different possible interpretations, but have been able to find a counterexample to each one. Is my confusion unwarranted?
For example, $A\triangle B\triangle C$ is the set of all $x\in X$ which are either in exactly one of the three sets $A,B,$ and $C$, or which are in all three of them. Similarly, $A\triangle B\triangle C\triangle D$ is the set of all $x\in X$ which are either in exactly one of the sets $A,B,C,D$ or in exactly three of them. Note that here $A_1,\dots A_n$ is not a list of all the subsets of $X$; it's just some list of subsets of $X$ (which may not include all of them, or may include some of them more than once).