Suppose $28$ 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?
Problem
Source: CRMO 2015 region 1 p4
Tags: combinatorics, combinatorial geometry, circle
Pluto04
02.10.2019 14:39
aops29 wrote: Is this the answer?
$$(28\cdot 24 \cdot 20)/6 = 3360$$
No, it is wrong.
Pluto04
02.10.2019 14:47
We can choose 3 objects from 28 objects in $\binom{28}{3}$ All objects are adjacent in 28 cases. Two of the three objects are adjacent in (28×24) cases. Two will be dia!metrically opposite and the remaining object will be in a non adjacent position in 14×(28-6) cases. So, answer = $\binom{28}{3}$ - 28 - 28×24 - 14×22 = $\boxed{2268}$
MathGenius_
02.10.2019 15:54
I also got $2268$ and I used similar logic