A regular polygon of $10$ sides is constructed. The number of ways in which $3$ vertices can be selected so that no two vertices are consecutive is

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Solution

The required number of selections
$=$ The number of selections without restrictions -(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
$=^{10} C_{3}$
The number of selections of $3$ consecutive vertices
$= 10$
(By observation, $A_{1} A_{2} A_{3}, A_{2} A_{3} A_{4}, \dots A_{10} A_{1} A_{2})$

The number of selections when $2$ vertices are consecutive $= 10 \times^{6} C_{1}$
(After selecting two consecutive vertices in $10$ ways, the third can be selected from remaining $6$ vertices)

Therefore, the required number of selections
$=^{10} C_{3} - 10 - 10 \times^{6} C_{1} = \frac{10 \times 9 \times 8}{6} - 70 = 50$

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