A regular polygon has 20 sides. The number of | vedclass

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(B) The total number of ways to select $3$ vertices out of $20$ to form a triangle is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$.F

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$800$

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(B) The total number of ways to select $3$ vertices out of $20$ to form a triangle is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$.For $n=20$ and $r=3$,the total number of triangles is $^{20}C_3 = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 1140$.To find the number of triangles that do not use any sides of the polygon,we subtract the triangles that use $1$ side and $2$ sides from the total.Number of triangles with exactly $2$ sides (adjacent vertices) is equal to the number of vertices,which is $20$.Number of triangles with exactly $1$ side is calculated by choosing $1$ side out of $20$ ($20$ ways) and then choosing a third vertex that is not adjacent to the chosen side. There are $20 - 4 = 16$ such vertices.So,triangles with $1$ side = $20 	imes 16 = 320$.Total triangles not using any sides = $1140 - 320 - 20 = 800$.

$A$ regular polygon has $20$ sides. The number of triangles that can be drawn by using the vertices but not using the sides are

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