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