A regular polygon of 10 sides is constructed. No. of ways 3 vertices be selected so that no two vertices are consecutive is

No of ways of selecting 3 from 10 vertices${=}^{10}{c}_3=120$

Two consecutive$=10{\times }^6{c}_1=60$

Three consecutive$=10$

$=120-60-10=50$

                  $or$

$=\frac{n\left(n-4\right)\left(n-5\right)}{6}=50$