M06-37

Official Solution:


If, out of the six points on the xy-plane, no three of them lie on the same straight line, how many distinct triangles can be formed by selecting three of the six points as vertices?


A. 5
B. 10
C. 20
D. 30
E. 60


Since no three of the six given points are collinear, any combination of three points will form a triangle. The number of different combinations of three points out of the six is given by \(C_6^3 = \frac{6!}{3! * 3!} = \frac{6 * 5 * 4}{3 * 2} = 20\)

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.


Answer: C
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I agree that using the combinations formula is the most efficient way to solve the problem. Like at least one other person who posted, though, I thought there might be some trick that I was missing, so I checked my work (since I was practicing, not taking a quiz). I did not draw anything, and I did not exactly count triangles. But I did organize information based on 6 unique points: call them A, B, C, D, E, and F. Just maintain alphabetical order, and the process is pretty quick.

AB | ABC, ABD, ABE, ABF (4)
AC | ACD, ACE, ACF (3)
AD | ADE, ADF (2)
AE | AEF (1)

That gets us to 10 valid triangles. Notice the 4-3-2-1 pattern. This sort of countdown will continue.

BC | BCD, BCE, BCF (3)
BD | BDE, BDF (2)
BE | BEF (1)

We can add another 6 valid triangles and move on to C.

CD | CDE, CDF (2)
CE | CEF (1)

Another 3 valid triangles to add to the tally. All that is left is the final valid combination.

DE | DEF (1)

10 + 6 + 3 + 1 = 20, so there must be 20 valid triangles within the given conditions. Once you pick up on the pattern with A, the rest goes quickly, and the question can still be solved in under 2 minutes. If you cannot remember which formula to use on test day, why not try logic?

Good luck with your studies.

- Andrew
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Hi,

I used the following approach -

Ways to Select any two points = 6C2 = 15
Each of these two points will need one more point to make a triangle, and we are left with 4 points, so total number of triangles = 15*4 = 60.

Please correct where my reasoning is flawed.

This way you get duplication. Say you get points A and B with 6C2 and the third point is C, so you'd get triangle ABC. But 6C2 gives also points A and C and in this case with point B, you'd get the same triangle ABC.
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Oh I guess thay any 3 points are not on the same line, means that there are no 3 points that lie on the same line. The way I read this sentence was that out of the 6 points, 3 of them are not on the same line. So, 2 or 3 of the rest of the points could be on the same line...

Anyway, selecting 3 out of 6 what does to the points?
Because for me 6!/3!3! would mean that out of 6 points we are choosing 3 and leaving out 3...

Hi Iw asws answering the question in the GMAT test and misunderstood the wording.

I was trying to understanding any 3 point in the line and 3 points are not not on the line.

Hi Bunuel,
As soon as i saw this question, i knew i had to use combinatorics. And i directly wrote \(6c3 = 20\) as the answer.
But then the statement 'Any 3 points of these 6 don't lie on the same line' took me. I tried to understand it by drawing different possibilities on the paper but could not decipher it.
Finally, i took it as a 'restriction' on the total number of triangles that are possible (20) and divided the answer by 2 and marked 10

If the question was worded like - 'There are 6 non-collinear points on the plain. How many unique triangles can be drawn using these 6 points as vertices?' Wouldnt the answer be 20?
Then what is the significance of the statement 'Any 3 points of these 6 don't lie on the same line' in the original question?

Please explain.
-Thanks

Say if one set of 3 points were collinear then those 3 points will not give a triangle and in this case the answer would be 20 - 1 = 19.

Hope it's clear.
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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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