Let ABC be a triangle. Consider four points p | vedclass

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(C) To form a pentagon using the given points, we must select $5$ points such that no three points are collinear. Since the points are on the sides of

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$660$

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(C) To form a pentagon using the given points, we must select $5$ points such that no three points are collinear. Since the points are on the sides of a triangle, we can select points from the sides $AB$, $BC$, and $AC$. Case $1$: $2$ points from $AB$, $2$ points from $BC$, and $1$ point from $AC$. Number of ways = $\binom{4}{2} 	imes \binom{5}{2} 	imes \binom{4}{1} = 6 	imes 10 	imes 4 = 240$. Case $2$: $2$ points from $AB$, $1$ point from $BC$, and $2$ points from $AC$. Number of ways = $\binom{4}{2} 	imes \binom{5}{1} 	imes \binom{4}{2} = 6 	imes 5 	imes 6 = 180$. Case $3$: $1$ point from $AB$, $2$ points from $BC$, and $2$ points from $AC$. Number of ways = $\binom{4}{1} 	imes \binom{5}{2} 	imes \binom{4}{2} = 4 	imes 10 	imes 6 = 240$. Total number of pentagons = $240 + 180 + 240 = 660$.

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