Question

$8$ different things are arranged around a circle. In how many ways can $3$ objects be selected when no two of the selected objects are consecutive.

Answer 1

Considering n different objects are to be arranged in a circle. The no of ways of selecting 3 such that no 2 of them are consecutive is
$=\frac{n}{3}{}^{n-4}{C}_2$
$=\frac{n}{3}\left(\frac{(n-4)(n-5)}{2}\right)$
Substituting $n=8$ gives us
$\frac{8}{3}\left(\frac{(8-4)(8-5)}{2}\right)$
$=\frac{8}{6}(4\times 3)$
$=8\times 2$
$=16$ ways.