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

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$^{n} C_{3} + n - n (n - 4)$

$^{n} C_{3} - n . n . (^{(n - 4)} C_{2})$

$^{n} C_{3} - n . (n - 4)$

$^{n} C_{3} - n - n (n - 4)$

answer is B.

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Detailed Solution

From n vertices

Total no of ways of selecting 3 vertices ${nC}_{3}$

No of ways in which all 3 of them are consecutive = n

No of ways in which only 2 are consecutive = (n-4)n

Required selection $=^{n} C_{3} - n - n (n - 4)$

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