(b) 806
Number of triangles that can be drawn with all 18 points in the plane = 18C3
But 5 points are collinear
∴ Required number of triangles = 18C3 – 5C3
= \(\frac{18!}{15!3!}-\frac{5!}{2!3!}=\frac{18\times17\times16}{3\times2}-\frac{5\times4}{2}\)
= 816 – 10 = 806.