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?

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: ${32 \choose 3 }-32\cdot28-32-16\cdot26=3616$

• @ mrprottolo : Have we assumed all objects are distinct or have we assumed a point of reference to imply a sense of top in which case all objects can be indistinguishable! I suppose without one of these assumptions, we cannot have ${32 \choose 3 }$ distinct cases. – KGhatak Jul 3 '17 at 15:54
• @KGhatak Yes I assumed we can fix a reference. We can also compute the possible configurations ignoring rotation simmetries. In this case I guess that the answer is just $3616/32=113$. – mrprottolo Jul 4 '17 at 9:42

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.