Find the number of triples $(A, B, C)$ of subsets of $[n]$ such that at least one of $A \cap B$, $A \cap C$, or $B \cap C$ is empty Find the number of triples $(A, B, C)$ of subsets of $[n]$ such that at least one of $A \cap B$, $A \cap C$, or $B \cap C$ is empty
I got $6^n\cdot 3-3\cdot 5^n+4^n$, not sure if this is correct.
 Please help me on this.
 A: I assume $[n]=\{1,2,\ldots,n\}$.  For an arbitrary triples $(A,B,C)$ (not requiring anything), each $k\in [n]$ can stay in one of the following list of sets:


*

*$[n]-(A\cup B\cup C)$

*$A-(B\cup C)$

*$B-(C\cup A)$

*$C-(A\cup B)$

*$B\cap C-A\cap B\cap C$

*$C\cap A -A\cap B\cap C$

*$A\cap B-A\cap B\cap C$

*$A\cap B\cap C$.


To meet your requirement, $k$ cannot stay in $A\cap B\cap C$ and must omit one of the sets $B\cap C$, $C\cap A$, and $A\cap B$.  
Let $n_A$, $n_B$, and $n_C$ denote the number of triples $(A,B,C)$ s.t. each $k\in [n]$ does not stay in $B\cap C$, $C\cap A$, and $A\cap B$, respectively.  To count $n_A$, each $k$ has only six choices from the list of sets above.  That is, we have $n_A=6^n$, and likewise, $n_B=n_C=6^n$.
Now, write $n_{A,B}$ for the number of triples $(A,B,C)$ s.t. each $k\in[n]$ does not stay in $B\cap C$ and $C\cap A$.  Define $n_{B,C}$ and $n_{C,A}$ similarly.  To count $n_{A,B}$, each $k$ has only $5$ choices from the list of sets above.  That is, $n_{A,B}=5^n$, and likewise, $n_{B,C}=n_{C,A}=5^n$.
Finally, let $n_{A,B,C}$ be the number of triples $(A,B,C)$ s.t. each $[k]$ stays in none of $B\cap C$, $C\cap A$, and $A\cap B$.  Then each $k$ has only $4$ possible places to be in the list above.  So $n_{A,B,C}=4^n$.  By PIE the answer is
$$(n_A+n_B+n_C)-(n_{A,B}+n_{B,C}+n_{C,A})+n_{A,B,C}=3\cdot 6^n-3\cdot 5^n+4^n.$$
Your answer is correct.
