AAPL wrote: ↑

Thu Jun 04, 2020 2:04 am

GMAT Paper Tests

There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points whereas CD contains 8 marked points. How many triangles can be formed by using these marked points as vertices?

A. (12!∗28)+(8!∗66)
B. 12!∗8!
C. 20C3
D. 864
E. 1024

OA D

Solution:

Since 3 non-collinear points form a triangle, a triangle can be form by either picking two points from AB and one point from CD OR two points from CD and one point from AB (notice that, for example, we can’t pick 3 points from AB to form a triangle since these 3 points would be linear).

Case 1: two points from AB and one point from CD

The number of ways this can be done is: 12C2 x 8C1 = (12 x 11)/2 x 8 = 66 x 8 = 528.

Case 2: two points from CD and one point from AB

The number of ways this can be done is: 8C2 x 12C1 = (8 x 7)/2 x 12 = 28 x 12 = 336.

Therefore, the total number ways a triangle can be formed is 528 + 336 = 864.

Answer: D