Triangles from a set of points

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Triangles from a set of points

by cypherskull » Fri May 04, 2012 2:40 pm
7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?

A. 91

B. 105

C. 196

D. 21

E. 14

The ans is E.. Not sure how to approach.
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by Mike@Magoosh » Fri May 04, 2012 3:20 pm
Hi, there! I'm happy to help! :)

With all due respect, are you sure you have the answer choices copied correctly? I did this problem, and got an answer considerably bigger than any of the answers you list here. In particular, what you give as the OA, 14, is simply absurd. There is no way that is the answer. What is the source of this question?

Here's the solution.

Let's call the line with 7 points Line A, and the line with 8 points line B.

There are two scenarios for possible triangles:
Scenario #1 -- two vertices on Line A and one vertex on Line B
Scenario #2 -- one vertex on Line A and two vertices on Line B
I'll treat those two scenarios separately.

Scenario #1 -- two vertices on Line A and one vertex on Line B
On line A, there are 7 ways to pick the first point, and then six ways to pick the second point on that line --- 7*6 = 42. This counts every pair twice, once for each order (i.e. GH and HG), so we need to divide by 2 --- 42/2 = 21. That's the number of pairs we can pick from Line A.
For each one of those pairs, we could construct a triangle by connecting it to any of the 8 points on Line B. 21*8 = 168 possible triangles in scenario #1

Scenario #2 -- one vertex on Line A and two vertices on Line B
On line B, there are 8 ways to pick the first point, and then seven ways to pick the second point on that line --- 8*7 = 56. This counts every pair twice, once for each order (i.e. PQ and QP), so we need to divide by 2 --- 56/2 = 28. That's the number of pairs we can pick from Line B.
For each one of those pairs, we could construct a triangle by connecting it to any of the 7 points on Line A. 28*7 = 196 possible triangles in scenario #2

Total number of possible triangles = 168 + 196 = 364

So, as you see, that's considerably larger than the answer choices you gave --- I wonder if the last two answer choices were supposed to be bigger than (C), rather than smaller, and if choice (E) was supposed to be 364. Almost always, on the real GMAT, positive integer answers are listed from smallest to biggest.

Here's a free video you may find helpful.
https://gmat.magoosh.com/lessons/335-fun ... -principle

Let me know if you have any questions.

Mike :)
Magoosh GMAT Instructor
https://gmat.magoosh.com/

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by cypherskull » Fri May 04, 2012 11:50 pm
Hey Mike...Thanks for your help with this. I got the same answer as u did but since 364 is way greater than any of the answer choices here, I thought my approach to the problem was seriously flawed somewhere.

Well, these were the options listed. The source is "Free GMAT Quiz" on 4gmat.com.