17) Given 9 points arranged in a square as shown in the figure next to it, how many non-degenerate triangles can be formed with vertices at three of the given points?
(A) 27
(B) 56
(C) 60
(D) 76
(E) 84.
17) The answer is (D). Each triangle is uniquely identified by its three vertices: so we must count the different ways to choose three non-aligned vertices among the nine points present. The ways to choose any three points among the nine present are equal to the combinations of three elements out of nine, that is,

$$
\left(\begin{array}{l}
9 \\
3
\end{array}\right)=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}=84
$$

However, there are 8 choices that correspond to aligned points (and thus degenerate triangles) in the diagram: the points aligned horizontally, vertically, or along one of the major diagonals. The actually constructible triangles are therefore $84-8=76$.