# Number of ordered 3-tuples of subsets such that intersection of any two is empty

Let $$X$$ be a set with $$2021$$ 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 $$\mathcal{S}$$?

I can only think of the following cases:

• $$\{(\emptyset,\emptyset,\emptyset)\}$$
• $$\left|\{(\emptyset,\emptyset,C): \emptyset \neq C\subseteq X\}\right| = 2^{2021}-1$$. Since we can permute these sets, the total number will be $$\left(2^{2021}-1\right) \times 3$$.

These cases become very tedious and look complicated. Any help/hint will be appreciated.

• Perhaps try counting the functions $f : \{1,...,2021\} \to \{0,...,3\}$? Commented May 2, 2023 at 5:22
• The total number of functions will be $4^{2021}$, but I am not getting how it will help. Commented May 2, 2023 at 5:29
• Think about which set, if any, the element $1$ is in. What about $2$? Commented May 2, 2023 at 5:58

Let $$X = \{1,2,\ldots,2021\}$$. Let us count the ordered $$3$$-tuples $$(A,B,C)$$ such that intersection of any two is empty.
Following the hints provided in the comment, we note that for every $$n\le 2021$$, we have the following choices for $$n$$.
• $$n \in A,~ n\notin B,\text{ and } n \notin C$$
• $$n \notin A,~ n\in B,\text{ and } n \notin C$$
• $$n \notin A,~ n\notin B,\text{ and } n \in C$$ and
• $$n \notin A,~ n\notin B,\text{ and } n \notin C$$.
Therefore, the number of elements in the set $$\mathcal{S}$$ is $$4^{2021}$$.
• Equivalently let $D=X \backslash (A \cup B\cup C)$ so each of the $2021$ elements of $X$ will be in exactly one of the $4$ of $A,B,C,D$ Commented May 2, 2023 at 9:55