TITLE: Combinatorics: Selecting objects arranged in a circle
QUESTION [15 upvotes]: If $n$ distinct objects are arranged in a circle, I need to prove that the number of ways of selecting three of these $n$ things so that no two of them are next to each other is $\frac{1}{6}n(n-4)(n-5)$.
Initially I can select $1$ object in $n$ ways. Then its neighbours cannot be selected. So I will have $n-3$ objects to choose $2$ objects from. Again, I can select the second object in $n-3$ ways. The neighbours of this object cannot be selected. However from here onwards, I am unable to extend this argument as the selection of the third object is dependent on the position of the first and the second object. Is there any simpler method to prove the result?

REPLY [10 votes]: References: The number of ways to choose $d$ non-consecutive 
positions on a circle of size $n$ is 
$${n-d+1 \choose d} - {n-d-1 \choose d-2}.$$
Aryabhata explains the derivation in the answer here 
Consecutive birthdays probability.
The first term is the number of ways to choose $d$ non-consecutive 
positions on a line of size $n$, and the second term subtracts off
 those arrangements where positions $1$ and $n$ were both chosen. 
This expression can also be rewritten as 
$${n\over n-d}{n-d\choose d}.$$ This form, called $d_k$ in the following link, 
 is used in the solution of
 the menage problem, (with $2n$ instead of $n$, and $k$ instead of $d$).