Official Solution:

If one of two parallel lines has 7 points on it and the other has 8, how many unique triangles can be drawn using the points?

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


Approach #1:

There are two types of triangles possible:

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

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: \(196+168=364\).

Approach #2:

All different 3 points out of total \(8+7=15\) points will create a triangle EXCEPT those 3 points which are collinear.

\(C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364\) (where \(C^3_8\) and \(C^3_7\) are # of different 3 collinear points possible from the line with 8 points and the line with 7 points, respectively).


Answer: D
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D

C315−(C38+C37)=455−(56+35)=364
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C28∗C17=28∗7=196 What formula is the first C (which equals 28)?

C stands for combinations. Check for more here: math-combinatorics-87345.html
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I am very familiar with the combinatoric method. My approach to this problem was (7*8!/2! + 8*&7!/2!) but this was clearly wrong.

I have read the combinatorics section of the gmatclub site but still don't understand how to derive the proper combinatoric equation from the C with the two small numbers that is in the answer explanation above.

Daft, I know... Thanks for any help.

Yeah me too, I am completely confused with the explanation of this problem! Could anyone please, please help? Thank you!

Hi acarpio1 and Anonamy,

The Q talks of two parallel lines with 7points and 8 points on each...
Now we have to form triangles..

We cannot form a triangle from all three points on the same line..
so we have to take one from 1 line and 2 from second line = 7C1*8C2..
and in second case 2 from ONE and 1 from SECOND = 7C2*8C1...
so 7C1 *8C2 +8C1 * 7C2 = 7*7*8/2 + 8*7*6/2 = 49*4 + 8*21 = 364

This post is just to say that this question is categorized as 600 level in the Quizzs.
Thank you

Hard for a 600 Level

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)
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It has been marked as a 700 level question now. If that helps?
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I feel like this question is worded in a slightly confusing manner. I feel like indirect triangles can be made via lines that connect and overlap each other. In terms of lines that use the points as the vertices, I agree that the answer is clearly 364. Maybe I'm overlooking something or am overthinking the question.

Posted from my mobile device

Do we really need the word "unique" in the stem? Unique in terms of what?

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?

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Hope it helps.
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Two ways:

(1) 1 point one line with 7 points and 2 points from the other line with 8 points; \(C^7_1*C^8_2 = 196\)

(2) 1 point one line with 8 points and 1 point from the other line with 8 points; \(C^8_1*C^7_2 = 168\)

The total triangles will be \(196+168=364\)

The answer is \(D\)
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