No. of ways of selecting 3 people out of n people sitting in a circle such that no two are consecutive What I did was selecting three gaps out of n-3 objects sitting in a circle but it appears like I am doing something wrong 
Selecting 3 gaps(total n-3 gaps in a circular arrangement of n-3 people) should work as there are same number of ways to select people from the circular arrangement as there are to put them in the circle consisting of $n-3$ people such that no two are consecutive
 A: Another approach would be to subtract the number of arrangements where two people are consecutive:
There are $\displaystyle\binom{n}{3}$ ways to select 3 of the people, and
1) there are $n$ ways to select all 3 people consecutive and
2) there are $n$ ways to select 2 people who are consecutive and then $n-4$ ways to select the 3rd person 
$\;\;\;$so that this person is not next to either of the first 2.
This gives a total of $\displaystyle\binom{n}{3}-n-n(n-4)=\binom{n}{3}-n(n-3)=\color{blue}{\frac{n(n-4)(n-5)}{6}}\;$ possibilities.

$\textit{Alternate solution:}$
Assume first that the $n$ people are in a row.
Then there are $\dbinom{n-2}{3}$ ways to select 3 people so that no two are consecutive,
since we can line up the $n-3$ people not selected and then choose 3 of the $n-2$ gaps they create.
Now we must subtract the $n-4$ possibilities where the two people at the end were chosen, 
which gives an answer of $\displaystyle\binom{n-2}{3}-(n-4)=\color{blue}{\frac{n(n-4)(n-5)}{6}}\;$ possibilities.
A: Select a fixed point on the circle to start with, and read out the choices in clockwise order.
In order to find the number of outcomes where the first position is not chosen, you need to find some combination of $3$ times "no, then yes" and $n-6$ times "no" -- that is, $\binom{n-3}{3}$ different outcomes.
Now for the number of outcomes where the first position is chosen. We can make every such outcome by taking one the the ones from before where we started with "no, then yes", and then rotating the entire pattern one position counterclockwise. This gives us $\binom{n-4}{2}$ different outcomes.
A: If you have any uncertainty about a method, try it on some examples
with small numbers.
Consider the example of $6$ people sitting in a circle.
Numbering them $1,2,3,4,5,6$ starting at an arbitrary point,
there are two possible subsets of three people you can select such
that no two are consecutive:
$\{1,3,5\}$ or $\{2,4,6\}.$
So the answer for $n=6$ should be $2.$
Selecting three gaps out of $n-3=6-3=3$ objects, do you get the answer $2$?
If you want a more detailed answer about where your mistake was,
please edit the question to add all the missing details
such as why you thought selecting three gaps out of $n-3$ objects
in a circle would work,
what formulas you used to compute the result,
and what you know about what the correct answer should have been.
