The number of triangles whose vertices are th | vedclass

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(C) The total number of points $(x, y)$ such that $0 \leq x \leq 4$ and $0 \leq y \leq 4$ is $(4+1) 	imes (4+1) = 25$ points.To form a triangle,we ne

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$2160$

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(C) The total number of points $(x, y)$ such that $0 \leq x \leq 4$ and $0 \leq y \leq 4$ is $(4+1) 	imes (4+1) = 25$ points.To form a triangle,we need to select $3$ non-collinear points from these $25$ points.The total number of ways to select $3$ points from $25$ is $\binom{25}{3} = \frac{25 	imes 24 	imes 23}{3 	imes 2 	imes 1} = 2300$.Now,we must subtract the number of sets of $3$ collinear points.Points are collinear if they lie on the same row,column,or diagonal.$1$. Rows: There are $5$ rows,each with $5$ points. Number of ways to choose $3$ points from a row is $5 	imes \binom{5}{3} = 5 	imes 10 = 50$.$2$. Columns: There are $5$ columns,each with $5$ points. Number of ways to choose $3$ points from a column is $5 	imes \binom{5}{3} = 5 	imes 10 = 50$.$3$. Diagonals: - Main diagonals: $2$ diagonals of length $5$ $(\binom{5}{3} 	imes 2 = 20)$.- Other diagonals of length $4$: $4$ diagonals $(\binom{4}{3} 	imes 4 = 16)$.- Other diagonals of length $3$: $4$ diagonals $(\binom{3}{3} 	imes 4 = 4)$.Total collinear sets = $50 + 50 + 20 + 16 + 4 = 140$.Number of triangles = $2300 - 140 = 2160$.

The number of triangles whose vertices are the points $(x, y)$ in the $XY$-plane with integer coordinates satisfying $0 \leq x \leq 4$ and $0 \leq y \leq 4$ is:

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