Choosing $3$ out of $n$ objects from a circular arrangement How many ways are there to choose $3$ out of $n$ objects from a circular arrangement, such that no two adjacent objects are picked?
Here's an attempt of mine:
Let $S$ be the set of ways to choose $3$ out of $n$ objects, which I will call $a$, $b$, $c$.
Let $A$ be the set of ways to choose $3$ out of $n$ objects such that $a$ and $b$ are adjacent.
Let $B$ be the set of ways to choose $3$ out of $n$ objects such that $b$ and $c$ are adjacent.
Let $C$ be the set of ways to choose $3$ out of $n$ objects such that $a$ and $c$ are adjacent.
${\lvert A' \cap B' \cap C' \rvert}$ = ${\lvert S \rvert}$ - ${\lvert A \cup B \cup C \rvert}$ = ${\lvert S \rvert}$ - ${\lvert A \rvert}$ - ${\lvert B \rvert}$ - ${\lvert C \rvert}$ + ${\lvert A \cap B \rvert}$ + ${\lvert B \cap C \rvert}$ + ${\lvert A \cap C \rvert}$ - ${\lvert A \cap B \cap C \rvert}$
I know that ${\lvert S \rvert}$ = ${n\choose 3}$
I think that ${\lvert A \rvert}$ = ${\lvert B \rvert}$ = ${\lvert C \rvert}$ = $n$ since there are $n$ ways to pick a group of $2$ objects out of $n$ objects
I think that ${\lvert A \cap B \rvert}$ = ${\lvert B \cap C \rvert}$ = ${\lvert A \cap C \rvert}$ = $2n$ since there are $n$ ways to pick a group of $3$ objects out of $n$ objects, and there are $2$ permutations (e.g. $a$,$b$,$c$ and $c$,$b$,$a$ for ${\lvert A \cap B \rvert}$)
I think that ${\lvert A \cap B \cap C \rvert}$ = $0$ since that would require $a$,$b$,$c$ to all be adjacent to each other pairwise which is not possible.
This means that
${\lvert A' \cap B' \cap C' \rvert}$ = ${n\choose 3}$ - $3n$ + $6n$ + $0$ = $\frac{(n^3-3n^2+20n)}{6}$
But I'm pretty sure that the answer has a $9n^2$ instead of $3n^2$. Can anyone point out my mistake(s)?
 A: Your mistakes are in $|A|=n$, where you are not choosing third object at all; and in $|A\cap B|=2n$ which should be $|A\cap B|=n$ since there is only one way to choose a triplet of adjacent objects.
There is a direct approach avoiding principle of inclusion-exclusion. Let number of objects located between three objects to be chosen be $p,q,r$. Then
$$p+q+r=n-3$$
We can choose first object in $n$ ways. And we can choose remaining two in one-to-one correspondence with number of positive integral solutions of above equation, namely $\binom{n-3-1}{3-1}$. But any triplet will be counted three times since any object could be the leader. Taking care of overcounting, required answer is
$$n \times \binom{n-4}{2}\times \frac{1}{3}=\frac{n^3-9n^2+20n}{6}$$
as you know.
A: Your mistake lies in mixing calculations that deal with an unordered choice of 3 objects, e.g. $|S|={n \choose 3}$, and calculations that deal with an ordered choice.
Your definitions of $A,B$ and $C$ only make sense if there is an order to $a$, $b$ and $c$, because otherwise how do you determine which one of the three chosen objects is $a$, a.s.o?
If you use your inclusion exclusion principle approach and go with a strictly ordered version of the choices, you get
$$|S|=n(n-1)(n-2),$$
$$|A|=|B|=|C| = 2n(n-2),$$
$$|A \cap B| = |A \cap C|= |B \cap C| = 2n,$$
leading to
$$|A' \cap B' \cap C'| = n(n-1)(n-2) - 6n(n-2) + 6n = n^3-9n^2 + 20n.$$
Of course, we finally want unordered choices, so divide the above by 6.
Note that this formula is correct only for all natural $n \ge 4$. For small $n$ (like $n=3$) the reasoning is not correct because the chosen elements can wrap around, so in this case $|A \cap B \cap C| = 1$ (unordered), as $S=A=B=C$.
