RMO question on combinatorics given in 2015

This is RMO question given in December 7, 2015. Suppose 28 objects are placed around a circle at equal distances. In how many ways I can choose 3 objects from among them so that no 2 of the 3 chosen are adjacent nor diametrically opposite? My try:- I can chose 3 out of 28 objects in $28C3$ ways. Among these choices all would be together in 28 cases; exactly 2 will be together in $28*24$ ways. Thus the answer would be $28C3-28-(28*24)$ but the answer given is $28C3-28-(28*24)-(14*22)$ where have I done mistake? Please clarify me.

The first object can be chosen in $28$ ways. Assume the object at site $0$ is chosen. It then blocks the four sites $27$, $0$, $1$, and $14$, and there are $24$ free sites for the second choice.
If the second object is chosen at the site $2$ (same for site $26$) this blocks three more sites, namely $2$, $3$, and $16$, leaving $21$ sites free for the third object. If the second object is chosen at the site $13$ (same for site $15$) this blocks two more sites, namely $12$ and $13$, leaving $22$ sites free for the third object.
If the second object is chosen at any of the $20$ free sites not mentioned so far this blocks four more sites, leaving $20$ sites free for the third object.
The total number $\tilde N$ of possible scenarios is therefore given by $$\tilde N=28(2\cdot 21+2\cdot 22+20\cdot 20)=13\,608\ .$$ Since in reality the order in which the three objects are chosen is irrelevant we have to divide by $3!$ in order to arrive at the end result $$N={\tilde N\over 6}=2268\ .$$ This value coincides with the "official" solution.