Problem $1.9$ Path to Combinatorics. 
Problem $1.9$ Find the number of ordered triples of sets $(A,B,C)$ such that $A\cup B\cup C=\{1,2,3....,2003\}$ and $A\cap B\cap C$ is empty.

My Attempt: Consider the following cases: 
1) Case $1$: $A,B$ and $C$ are all pairwise disjoint. Then each element $s\in S$ has $3$ choices; it can go to set $A,B$ or $C$. Thus there will $3^{2003}$ such triples.     
2) Case $2$: Either $(A\cup B)\cap C=\phi$, $(B\cup C)\cap A=\phi$ or $(C\cup A)\cap B=\phi.$ In this case each element in set $S$ has $2$ choices; it can belong to $(A\cup B)$ or $C$ and so on. Thus the number of such triples is $2^{2003}.$
Thus we have $3^{2003}+2^{2003}$ triples. Is this answer/approach right? I am asking this because the textbook, which I am following has no solutions.
 A: You are on the right track, but missed some details.
To have $A\cap B\cap C=\emptyset$, we allow each element to be in any combination of the three sets except in all at the same time (would lead to $A\cap B\cap C\ne \emptyset$) as well as except in none of them (would lead to wrong union). Hence each element is left with exactly six (instead of $2^3$) possibilities. We end up with $$6^{2003}.$$
A: Consider the set of functions $S=\{f~:~\{1,2,3,\dots,2003\}\to\{a,b,c,ab,ac,bc\}\}$.
Consider the set of triples $T=\{(A,B,C)~:~A\cup B\cup C=\{1,2,\dots,2013\},A\cap B\cap C=\emptyset\}$
Associate to each function an ordered triple $(A,B,C)$ in an "obvious" way forming a bijection between the set of ordered triples of sets with the property you desire and the set of functions described above.
Explicitly, the bijection $\sigma~:~T\to S$can be defined as $\sigma((A,B,C))=f$ such that $f(x)=\begin{cases} a&\text{if}~ x\in A\setminus (B\cup C)\\ b&\text{if}~x\in B\setminus (A\cup C)\\ c&\text{if}~x\in C\setminus(A\cup C)\\ ab&\text{if}~x\in (A\cap B)\setminus C\\ac&\text{if}~x\in (A\cap C)\setminus B\\ bc&\text{if}~x\in (B\cap C)\setminus A\end{cases}$
In the other direction, $\sigma^{-1}(f)=(A,B,C)$ where $A=\{x~:~f(x)=a~\text{or}~f(x)=ab~\text{or}~f(x)=ac\}$, similarly $B=\{x~:~f(x)=b~\text{or}~f(x)=ab~\text{or}~f(x)=bc\}$ and again for $C$.
It should be clear that $\sigma$ is indeed a bijection and is well defined, but feel free to go through the effort of proving that directly.

Now that we know there is a bijection between $S$ and $T$, we know $|S|=|T|$ and so to count the number of ordered triples satisfying your properties we can simply count the number of functions.  Counting the number of functions can be seen via multiplication principle.


*

*pick the value of $f(1)$ (six choices)

*pick the value of $f(2)$ (six choices)

*$\vdots$

*pick the value of $f(2003)$ (six choices)


By multiplication principle, the total number of outcomes is the product of number of choices at each step, yielding a final total of $\underbrace{6\cdot 6\cdot 6\cdots 6}_{2003~\text{times}}=6^{2003}$
