Problem: Let $X$ be a set with $2023$ elements. Consider the set
\[
\mathcal{S} = \left\{(A,B,C): A,B,C\subseteq X,~ A\cap B = \emptyset, ~B\cap C = \emptyset,~ \text{ and }C\cap A = \emptyset\right\}.
\]
Then what is the cardinality of the set $\mathcal{S}$?
Solution: Let $X = \left\{ 1,2,\dots,2023 \right\} $. Note that for any $n \leq 2023$ we have the following choices:
-
$n \in A,~ n\notin B,\text{ and } n \notin C$;
-
$n \notin A,~ n\in B,\text{ and } n \notin C$;
-
$n \notin A,~ n\notin B,\text{ and } n \in C$ and
-
$n \notin A,~ n\notin B,\text{ and } n \notin C$.
Therefore, the total number of elements in the set $\mathcal{S} $ is $4^{2023}$.
Equivalently, let $D = X\setminus \left( A\cup B\cup C \right) $, then each of $2023$ elements will be in exactly one of the four of $A,B,C$ and $D$. This is also equivalent to the total number of functions from the set $X$ to ${0,1,2,3}$.