Combi-triangle

On each side of the square is marked 10 different points outside the vertices of the square. How many triangles can be constructed from this set of points, where each vertex of the triangle lie on the other side of the square?

Result

n =  4000

Solution:

n=(410)(310)(210)/3!=4000n = (4\cdot 10)\cdot (3\cdot 10)\cdot (2\cdot 10)/3! = 4000



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