A regular polygon of 10 sides is constructed. In how many way can 3 vertices be selected so that no two vertices are consecutive?

Explanation

The required number of selections is given as

The number of selections without restriction (the number of selections when 3 vertices are consecutive)-(the number of selections when 2 vertices are consecutive)
Now, the number of selections of 3 vertices without restriction is `.^(10)C_(3)`.
The number of selections of 3 consecutive vertices is 10 (by observation: `A_(1)A_(2)A_(3),A_(2)A_(3)A_(4),..,A_(10)A_(1)A_(2))`.
The number of selections when two vertices are consecutive is `10xx .^(6)C_(1)`.
(After selecting two consecutive vertices in 10 ways, the third can be selected from 6 vertices.)
Therefore, the required number of selections is
`.^(10)C_(3)-10-10xx .^(6)C_(1)=(10xx9xx8)/(6)-10-60`
=120-70
=50