Counting non-degenerate triangles in a square lattice
A 4 × 4 square is composed of 16 unit squares. The points formed from the unit squares are used to create non-degenerate dank triangles. Call a triangle dank if none of its vertices lie on the same row or column of points. How many triangles are dank?
My thoughts are first there will be $25*16*9$ cases in total, if we dont count when the triangle is degenerated. It is the permutation so we divide by $3!$ to get 600. A straight line will make the triangle degenerate, so we need to minus those cases. There are 5 points on one diagonal of the square, 4 on its left and right side, and 3 on its left side. So the total case of degenerate triangles are $(\binom53+(\binom43+\binom33)\cdot2)\cdot2=40$. However the answer is not 560? Can someone help me with this?
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$\begingroup$There are indeed 600 dank triangles; the problem is in counting the degenerate cases. There are 52 of them and not 40:
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. . * . . 2 (orientation) * 9 (position) = 18 cases
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. . . . . 4 * 4 = 16
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. . . . . 4 * 1 = 4
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. . * . . 2 * 1 = 2
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. . . . * 4 * 3 = 12
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. . . . .This leaves the number of non-degenerate dank triangles as $600-52=548$.
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