ABCD is a square. One point on each of AB and CD; and two distinct points on each of BC and DA are chosen. How many distinct triangles can be drawn using any three points as vertices out of these six points?

Correct Option (3)

The problem specifies that a total of six distinct points are chosen: one point on side AB, one point on side CD, two distinct points on side BC, and two distinct points on side DA. To form a triangle, three non-collinear points are required. In the absence of any stated collinearity constraints among the chosen points, it is assumed that any three distinct points selected from these six will form a triangle.

The number of ways to choose 3 points from a set of 6 distinct points is determined using the combination formula ${}^n{C}_r=\frac{n!}{r!(n-r)!}$, where $n$ represents the total number of items available and $r$ is the number of items to be chosen.

In this specific case, \(n = 6\) (total points) and \(r = 3\) (points required for a triangle).

$${}^6{C}_3=\frac{6!}{3!(6-3)!}=\frac{6!}{3!3!}$$

$$=\frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}=\frac{6\times 5\times 4}{3\times 2\times 1}=\frac{120}{6}=20$$

Therefore, 20 distinct triangles can be formed using any three of these six points as vertices.

Incorrect Options:

Options 1 (16), 2 (18), and 4 (24) are incorrect. The direct combinatorial calculation for selecting 3 points from 6 distinct points, under the standard assumption of no three points being collinear, yields 20 triangles. These alternative values do not align with the correct mathematical derivation for the given problem parameters.