# 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.

You have not taken care of the condition that no two chosen objects (or sites) may be opposite to each other.

Assume for the moment that the order in which the three objects are chosen matters.

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.