M03-23

Official Solution:

If one of two parallel lines has 7 points on it and the other has 8 points, how many distinct triangles can be formed using these points as vertices?

A. 168
B. 196
C. 316
D. 364
E. 455


Note that this is not a Geometry question. While it uses basic knowledge of lines and figures, it is actually a probability and combinatorics question. There are 8 questions within GMAT Prep Focus Edition that use similar principles. Here is one example.

Approach #1:

There are two types of triangles that can be formed:

1. Triangles with two vertices on the line with 8 points and the third vertex on the line with 7 points: \(C^2_8*C^1_7=28*7=196\);

2. Triangles with two vertices on the line with 7 points and the third vertex on the line with 8 points: \(C^2_7*C^1_8=21*8=168\);

Total number of triangles: \(196+168=364\).

Approach #2:

Any three distinct points chosen from the total of \(8+7=15\) points will form a triangle, except for the cases where the three points are collinear.

Therefore, the total number of triangles can be calculated as \(C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364\). Here, \(C^3_8\) and \(C^3_7\) represent the number of different sets of 3 collinear points possible from the line with 8 points and the line with 7 points, respectively.


Answer: D
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  • Questions
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For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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I would say visualize what is happening.

Imagine one line, which has 7 points and there is only 1 point on the opposite side. How many unique triangles we can make?

7C2 * 1 = 21

Now there are actually 8 points on the other side and not just 1 point.

So we get 21*8=168

Let's go on the other side of the table. Now we have 8 points out of which we have to select 2 and there are 7 points on the opposite side.

We get 8C2*7=28*7=196

Total Unique Triangles = 168+196=364 (Tip: Just add 200 to 168 and then subtract 4)

Bunuel I understand combinatorics but find it difficult to conceptually think of how to set up the question in the way you did. Do you have other similar combinatorics questions that require this kind of conceptual thinking and ways of getting around these?

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Huge thank you for this list! I'm curious, Is this list sorted in order of level? Bottom ones being tougher than the ones at the top of the list?

No, they are not in any order. By the way, you can check the difficulty level of a question in the tags just above the first post as well as in the timer.
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Okay, got it! The tags are by default hidden, and I had to click on the "Show tags" button. This is new to me, and I will use this feature. WHy not have the tags show up by default along with "Hide tags" button?

The default setting hides tags to ensure users aren't inadvertently influenced by them. In contexts like Sentence Correction, a tag could inadvertently hint at the error type, simplifying the challenge.
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