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 ?

(a) 16
(b) 18
(c) 20 ✓ UPSC's answer
(d) 24

Why the answer is (c)

• Total points chosen: 1 on AB, 1 on CD, 2 on BC, and 2 on DA, making a total of 6 distinct points.

• The total number of ways to choose any 3 points from 6 is given by the combination formula $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

• A triangle cannot be formed if the three chosen points are collinear (lie on the same straight line).

• We must check for collinear triples among the 6 points. The points are distributed on the four sides of the square: AB (1), BC (2), CD (1), DA (2).

• Since no side has 3 or more points, there are no three points lying on the same side of the square.

• Therefore, no three points are collinear, and all 20 combinations form valid triangles.

Why the other options are wrong

(a) 16: Option (a) is incorrect because it undercounts the total combinations, likely by incorrectly subtracting non-existent collinear cases or miscalculating $^6C_3$.
(b) 18: Option (b) is incorrect because it fails to account for all possible valid triangles, as all 20 combinations of 3 points from 6 non-collinear sets form triangles.
(d) 24: Option (d) is incorrect because it overcounts the possibilities, as the maximum number of triangles from 6 points is 20, not 24.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2023, held on 28 May 2023. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.