Combination of elements in a ring and selecting non adjacents Ok, suppose we have a clock, with the usual design of numbers ordered from 1 to 12 (so 1 and 12 are adjacents). The question is what is the number of possible combinations of four non adjacent numbers.
The order of the numbers chosen doesn't matter, that is, if I choose 1, 3, 6, 9, that would be the same as 6, 1, 3, 9, and so on... (anyway, it would be a difference of a factor 4! to multiply or divide by, if I am right).
So, what would be the answer to the question, and what is a method of reasoning for this kind of problems?
Thank you in advance.
 A: Let us first solve another problem. How many ways are there to choose $4$ non-adjacents, and paint one of the chosen numbers red?
The red number can be chosen in $12$ ways. For any such choice, the two neighbouring numbers are not allowed to be chosen. So we have $9$ numbers left, and we can think of them as forming a line.
We will choose $3$ non-adjacents from these. Represent the numbers not chosen abstractly by $6$ $\ast$, like this
$$\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast$$
These determine $7$ "gaps" ($5$ between consecutive $\ast$, and the $2$ endgaps).
We must choose $3$ of these to slip a $\square$ into, to represent the chosen numbers. This can be done in $\binom{7}{3}$ ways.  
So there are $(12)\binom{7}{3}$ ways to choose $4$ non-consecutives and paint one of them red. It follows that there are $\frac{(12)\binom{7}{3}}{4}$ ways to choose $4$ non-consecutives. 
A: We follow the suggestion of JMoravitz.
First, we solve the problem for a line.  
Place eight blue balls in a row, leaving spaces between them and at the ends of the row.  There are nine such spaces, seven between successive blue balls and two at the ends of the row.  Place a green ball in four of these nine spaces.  Now number the balls from left to right.  The numbers on the green balls form the desired four-element subset of $\{1, 2, 3, \ldots, 12\}$ in which no two of the numbers in the subset are consecutive.  Since we place the green balls by choosing four of the nine spaces, there are 
$$\binom{9}{4}$$
four-element subsets of $\{1, 2, 3, \ldots, 12\}$ in which no two of the numbers in the subset are consecutive.  
Alas, we have counted subsets that include both $1$ and $12$.  We must exclude these.  If $1$ is included, then $2$ cannot be.  Similarly, if $12$ is included, then $11$ cannot be.  That leaves the eight numbers in the subset $\{3, 4, 5, \ldots, 10\}$, of which we must choose two numbers that are not consecutive.  
Line up six blue balls again, leaving spaces between successive balls and at the ends of the row.  We choose two of these seven spaces in which to insert a green ball, then number these eight balls from $3$ to $10$ as we proceed from left to right.  The numbers on the two green balls are the other numbers besides $1$ and $12$ in the subset consisting of four numbers of which no two are consecutive.  Hence, there are $\binom{7}{2}$ four-element subsets of $\{1, 2, 3, \ldots, 12\}$ in which no two elements are consecutive that contain both $1$ and $12$.
Hence, there are 
$$\binom{9}{4} - \binom{7}{2}$$
four-element subsets of $\{1, 2, 3, \ldots, 12\}$ in which no two elements are consecutive that do not contain both $1$ and $12$.
