OG 2016 Problem 73

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illuminati5288: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Sat Sep 24, 2016 1:32 pm

by illuminati5288 » Mon Oct 17, 2016 6:44 pm

Hi,

I am having a hard time understanding the problem and the solution.

S is set of points in a plane. How many distinct triangles can be drawn that have three of the points in S as vertices?

(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.

My solution:

(1) The distinct points in S could be in a straight line. So A and D are ruled out.
(2) Does not tell us how many points are there. I could have only one triangle or 10 from the 2nd statement. So B is ruled out

(1)+(2) Statement one says there are 5 distinct points. Does not say how many repeating points I have, Eg: S = {a,b,c,d,e,x,x,y,y...}. So C is ruled out and the answer is E.

The OG says the answer is C.

Could someone please explain.

Thank you.

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Source: Beat The GMAT — Data Sufficiency | Mon Oct 17, 2016 6:44 pm

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by [email protected] » Mon Oct 17, 2016 7:17 pm

Hi illuminati5288,

The word "distinct" in the prompt, and in Fact 1, means DIFFERENT.

When combining the two Facts, there are only 5 distinct points in the plane. Even if there were "duplicate points" (e.g.. on a co-ordinate plane, point A is at (1,1) and point B is also at (1,1)), triangle ACD and triangle BCD would be the SAME triangle, so you would not count it twice.

Since no group of three points are collinear, that means we can select ANY group of 3 (of the 5 points) to make a triangle. Thus, there are 5!/3!2! = 10 possible distinct triangles that can be formed.

If you're having trouble with the conceptual ideas behind this prompt, then you could try drawing 5 random points on a piece of paper. As long as you don't have 3 of them on the same line, then you'll end up with 10 distinct triangles.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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illuminati5288: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Sat Sep 24, 2016 1:32 pm

by illuminati5288 » Mon Oct 17, 2016 7:20 pm

Thanks Rich,

That makes sense now.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8149; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Oct 19, 2016 4:55 am

illuminati5288 wrote:
S is set of points in a plane. How many distinct triangles can be drawn that have three of the points in S as vertices?

(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.

We are given that S is a set of points in the plane and we must determine how many distinct triangles can be drawn with three of the points in S as vertices. So essentially, we must determine how many distinct triangles can be drawn with the points provided.

Statement One Alone:

The number of distinct points in S is 5.

Using the information in statement one, it may be tempting to conclude that the number of triangles that can be drawn is 5C3 = (5 x 4 x 3)/3! = 10 triangles. However, because we do not know the positioning of the points, we cannot actually say that 10 distinct triangles can be created. Let's say, for instance, that all the points were collinear, which means that they are all located on one line. If that were the case, we would not be able to create any triangles. Thus, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Note: We were able to determine that 10 triangles could be formed with 5 points if and only if no 3 points are collinear. Only then would the number of triangles be 5C3 = 10.

Statement Two Alone:

No three of the points in S are collinear.

Using the information in statement two, we cannot answer the question because we do not know how many points are in S. We can eliminate answer choice B.

Statements One and Two Together:

Using the information from statements one and two we know that we have 5 points in the plane and that no three points are collinear. Thus, we can determine that the number of triangles that can be created in the plane is

5C3 = 10.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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