There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points...

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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

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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

[swerve]

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

We need 3 points in total to make a triangle.

You can select either 2 points from AB or 2 points from CD. Thus there are 2 possible scenarios
1. Choosing from AB first
12C2 AND 8C1 = 66*8
= 528

OR
2. Choosing from CD first
8C2 AND 12C1 = 336

Total up = 864 \(\Longrightarrow\) D