Choosing $3$ objects from $32$ equidistant objects on a circle with restrictions given. Suppose $32$ objects are placed along a circle at equal distances. In how many ways can $3$ objects be chosen from among them so no two of the three chosen objects are adjacent nor diametrically opposite.
This is again a problem in a math contest in India and this is how I tried it:
Number of ways of choosing 3 objects from 32 objects$={32 \choose 3 }$
Number of ways to choose 3 points such that they are adjacent$=32$
Number of ways to choose $3$ points such that two of them are adjacent$=(32×2×30)$.......
for each point there are two ways to choose an adjacent point. For each choice there are $30$ options to choose the third point.
Number of ways to choose $3$ points such that two of them are diametrically opposite$=(32×30)$
Number of ways to choose $3$ points from $32$ equidistant points on a circle with the restrictions placed by the problem
$={32 \choose 3}-(32+(32×2×30)+(32×30)$
Am I correct in my approach?
 A: Your argument is not completely correct, as you counted more than once some combination. This a way to fix it:
Number of ways to choose $3$ adjacent points : $32$
Number of ways to choose $3$ points such that $\textbf{only}$ two of them are adjacent : $32\cdot28$. Because there are $32$ couples of adjacent points and the third point must not be adjacent the other two.
Number of ways to choose $3$ points such that two of them are diametrically opposite but $\textbf{not adjacent}$: $16\cdot26$. Because the are $16$ total diameters and you can choose the third point in $26$ ways, as it must not be near the other two.
Answer: ${32 \choose 3 }-32\cdot28-32-16\cdot26=3616$
A: Suppose  I   denote  the   objects  as   1-32.
Then suppose I select object 1 first. Then there would be 2 cases.
CASE 1 :-If I choose 3 or 31 (as I can't select 32 or 2 but I can choose those adjacent to 2 and 32) then there would be 25 possibilities for each .
Thus there being a total of 25×2 = 50 possibilities.
CASE 2 :-Then again leaving those numbers that are considered in first case  (3 and 31) I am left with 26 possibilities for 2nd object and for third I will have 23 chances.(excluding the adjacent and the opposite ones that are 3 in number) and excluding the very same digit I took in the  second place.
Hence in 2nd  case total possibilities are 26 × 23 = 598
So,Total in both cases = 598 + 50 = 648 
But I completely ignored those numbers adjacent to 1 and diametrically opposite to 1 that are 2,32,17 .
So again each of them will have **648 ** total possibilities thus making final answer:-
=> 648 × 4 = 2592.
Please rectify me if I'm wrong.
