Explain the following combination question in deep details Suppose $32$ objects are placed along a circle at equal distances. In how many ways can $3$ objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite $?$
 A: There are $32$ ways to place the first object. After that there are $28$ places for the second object. That is, there are $32\cdot28=896$ ways to place the first two objects.
If the first two objects are $2$ places apart, then there are $25$ places for the third; if the first two places are $15$ apart, there are $26$ places for the third; otherwise, there are $24$ places.

Of the $896$ ways to place the first two, $64$ have them being two apart and $64$ have them being $15$ apart. Thus, there are
$$
\overbrace{\ 64\cdot26\ }^{\text{first two $15$ apart}}+\overbrace{\ 64\cdot25\ }^{\text{first two $2$ apart}}+\overbrace{768\cdot24}^{\text{otherwise}}=21696
$$
ways to pick three objects where we care about their order. For each of the ways to arrange without caring about order, there are $3!=6$ ways to reorder the objects; therefore, if we don't care about the order, there are
$$
\frac16\cdot21696=3616
$$
ways to arrange three objects.
A: I am considering objects to be identical, and  arrangements to be identical under rotation. 
Number the objects $1-32$ clockwise, and let the first object chosen be $1$  We can't now choose $32, 2$ and $17$ (the two adjacent, and the one opposite).
Let us choose the second object in the right semi-circle from the $14$ positions available, and the third from the left semi-circle from which 
we can now only  choose from $(14-1)$ positions, except that number $16$ can be combined with $14$ from the other semicircle, since number $32$ was already barred when we started with $1$, thus yielding $ 14\cdot13 +1$ patterns.
But we could also choose the second and third object from the same semicircle, From $14$ such in one semicircle, there are $\binom{14}{2}$ combos of two from which $13$ adjacent ($3-4$ through $15-16$) need to be subtracted, yielding $78$ patterns for each semicircle.
Finally, we get $(14\cdot13) + 1 +  (2\cdot78) = 339$
NOTE
The formula this yields for $2n$ points, $n\ge4$ is:
$(n-2)(n-3) + 1 + 2\left[\binom{n-2}{2} - (n-3)\right]$
This yields $51$ patterns for $16$ points, which can easily be verified by enumeration
[You could start with $8$ points, which has only $3$ patterns, $1-3-6, 1-4-6$ and $1-4-7$ . Note that I am always fixing one point as $1$ ]
