The question is to count how many 3-element antichains there are in $\bf{2}^{n}$, where $\bf{2}^{n}$ is the Boolean algebra on $n$ elements.
One example is in $\bf{2}^{3}$, there are exactly two antichains of three elements, $\{1,2,3\}$ and $\{12,13,23\}$.
In general, I'm looking for a simple formula to count the 3-element antichains in $\bf{2}^{n}$. My approach is to count the number of ways to assign each element in $[n]$ to one of the following 8 sets:
$A\setminus ((A\cap B)\cup (A\cap C))$, $B\setminus((A\cap B)\cup (B\cap C))$, $C\setminus((A\cap C)\cup (B\cap C))$,
$(A\cap B)\setminus(A\cap B\cap C)$, $(A\cap C)\setminus(A\cap B\cap C)$, $(B\cap C)\setminus(A\cap B\cap C)$,
$A\cap B\cap C$, $[n]\setminus(A\cup B\cup C)$.
The extra conditions are:
$A\setminus ((A\cap B)\cup (A\cap C))\neq\emptyset$,
$B\setminus((A\cap B)\cup (B\cap C))\neq\emptyset$,
$C\setminus((A\cap C)\cup (B\cap C))\neq\emptyset$.
Then the total number of 3-element antichains should be $8^{n}-3\cdot 7^{n}+3\cdot 6^{n}-5^{n}$, where $8^{n}$ is the number of choices where there are no extra conditions, $7^{n}$ is the number of choices when one of the extra conditions is not followed, $6^{n}$ is the number of choices when two of the extra conditions are not followed and $5^{n}$ is the number of choices when all of the extra conditions are not followed.
Back to the case $\bf{2}^{3}$, $8^{3}-3\cdot 7^{3}+3*6^{3}-5^{3}=6\neq 2$. I couldn't see where I did wrong. Any help and suggestion is welcome.
There are three scenarios when we have a three-element anti chain.
(1) There are some integers in each of $A$, $B$, and $C$.
(2) There are some integers in each of $D$, $E$ and $F$.
(3) a. There are some integers in each of $A$, $B$, $D$ and $E$, but not in $C$ and $F$.
b. There are some integers in each of $A$, $C$, $D$ and $F$, but not in $B$ and $E$.
c. There are some integers in each of $B$, $C$, $E$ and $F$, but not in $A$ and $D$.
The first two scenarios each contributes $8^{n}-3\cdot 7^{n}+3\cdot 6^{n}-5^{n}$ anti chains. But the case when $A$, $B$, $C$, $D$, $E$, $F$ are all non empty which contributes $$8^{n}-6\cdot 7^{n}+15\cdot 6^{n}-20\cdot 5^{n}+15\cdot 4^{n}-6\cdot 3^{n}+2^{n}$$ is counted twice. Thus, in total, the first two scenarios contribute $$ 2\cdot(8^{n}-3\cdot 7^{n}+3\cdot 6^{n}-5^{n})-(8^{n}-6\cdot 7^{n}+15\cdot 6^{n}-20\cdot 5^{n}+15\cdot 4^{n}-6\cdot 3^{n}+2^{n})$$
number of anti chains.
For the third scenario, since each sub-scenario has two subsets empty, in total there are $3\cdot(6^{n}-4\cdot 5^{n}+6\cdot 4^{n}-4\cdot 3^{n}+2^{n})$ anti chains.
Since anti chains are not order sensitive, we have to divide the total number of anti chains by $6$. Therefore,
$$\frac{1}{6}(2(8^{n}-3\cdot 7^{n}+3\cdot 6^{n}-5^{n})-(8^{n}-6\cdot 7^{n}+15\cdot 6^{n}-20\cdot 5^{n}+15\cdot 4^{n}-6\cdot 3^{n}+2^{n})+3(6^{n}-4\cdot 5^{n}+6\cdot 4^{n}-4\cdot 3^{n}+2^{n}))$$
is the total number of three-element anti chains in $\bf{2}^{n}$.
This formula is verify in $\bf{2}^{3}$.