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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

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  • $\begingroup$ I'm sorry to say this, but your edit explains nothing. The first sentence of the question set up two circles, one of $n$ people/objects and one of $n-3$ people/objects, so when you write "the circular arrangement" or "the circle" we don't know which circle you mean. If you mean the same circle both times the statement is obviously wrong: there are always more ways to select $3$ items from a circle of $k$ than to place $3$ items in non-adjacent places in a circle of $k$. You also don't define what you mean by "gap," and you have shown no formulas or calculations at all. $\endgroup$ – David K Dec 21 '16 at 15:56
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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.

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  • $\begingroup$ @bulbasaur This is a correct answer to the problem posed in your question's title, so perhaps if you compare it to the work you did you may find your error. Perhaps your error was you set up some set of arrangements analogous to the way this answer counts the first $\binom{n-3}{3}$ arrangements, but you overlooked some constraint in that scheme (like the constraint here that the first position is not chosen) which caused you not to count the other $\binom{n-4}{2}$ arrangements. $\endgroup$ – David K Dec 21 '16 at 16:04
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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.

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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.

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  • $\begingroup$ Selecting 3 gaps 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 such that no two are consecutive $\endgroup$ – bulbasaur Dec 21 '16 at 15:47

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