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S is a set of points in the plane. How many distinct triangles can be
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Author:  GK002 [ 06 Dec 2021, 18:26 ]
Post subject:  Re: S is a set of points in the plane. How many distinct triangles can be
im baffled as to how this is a low difficulty problem. It's classified as Hard in the OG.
Author:  Bunuel [ 07 Dec 2021, 01:23 ]
Post subject:  Re: S is a set of points in the plane. How many distinct triangles can be
GK002 wrote:
im baffled as to how this is a low difficulty problem. It's classified as Hard in the OG.


You can check difficulty level of a question along with the stats on it in the first post. For this question Difficulty = 700 Level. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.
Author:  Schachfreizeit [ 10 Dec 2022, 11:32 ]
Post subject:  Re: S is a set of points in the plane. How many distinct triangles can be
GGUY wrote:
S is a set of points in the 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.



I don't whether it is bc I am not a native speaker, but I am having huge problems with understanding the question.
"How many distinct triangles can be drawn that have three of the points in S as vertices?"
Can someone explain to me what we are looking for here?
Author:  Schachfreizeit [ 10 Dec 2022, 11:34 ]
Post subject:  Re: S is a set of points in the plane. How many distinct triangles can be
walker wrote:
C

1. we cannot create triangles for 5 points of a line but can do that for points that are not collinear. insuff.
2. we don't know the number of points insuff.

1&2 \(C^5_3=\frac{5*4*3*2}{3*2*2}=10\)



can you explain why you calculated 5C3?
Author:  egmat [ 15 Jan 2023, 09:02 ]
Post subject:  Re: S is a set of points in the plane. How many distinct triangles can be
Schachfreizeit wrote:
walker wrote:
C

1. we cannot create triangles for 5 points of a line but can do that for points that are not collinear. insuff.
2. we don't know the number of points insuff.

1&2 \(C^5_3=\frac{5*4*3*2}{3*2*2}=10\)



can you explain why you calculated 5C3?


Hello Schachfreizeit,



I will respond to both your queries here so please bear with me as the response may be long. 😊

Let me begin by telling you that just like you, I am also not a native speaker. I can understand your pain and will try to help you understand the question and the solution in as simple terms as possible.



UNDERSTANDING THE QUESTION (Not solving yet)

  • Question Stem: We will take small chunks of the stem and understand it as we go.
    • “S is a set of points in the plane.”
      • This sentence tells us that there are some points in a plane (a plane is any 2D surface, say a wall). All these points form a group represented as S.

    • “How many distinct triangles can be drawn that have three of the points in S as vertices?”
      • This sentence asks us how many triangles we can draw using the points from set S only.
      • Now, to draw a triangle, we always need 3 points. These 3 points act as the vertices of our triangle.
      • So, the question basically asks, “in how many ways we can select 3 points from group S”. Because each new combination of 3 points will form a unique triangle. However, there is an *EXCEPTION to this that we will cover later.

  • Statement 1: “The number of distinct points in S is 5.”
    • This gives us the total number of points in set S.
  • Statement 2: “No three of the points in S are collinear.”
    • To understand this statement, you need to understand the term “collinear”. Collinear means “lying on the same line”. So, this statement says that no three points in set S lie on a straight line.
    • Now, why is this information relevant? Let’s see. Suppose we select 3 points and these 3 points are collinear. Then, when you join these 3 points, you would form a line instead of a triangle as needed. So, such a collection of 3 points will not be valid for us. *This is the exception that we had referred to earlier. That is, in general, each combination of 3 points will not give us a triangle. But each combination of 3 non-collinear points will definitely give us a triangle!
    • Now that we understand the term ‘collinear’, let’s come back to Statement 2. This statement confirms that every collection of 3 points of set S will give us a triangle. Hence, we must consider every possible combination.

I hope you now understand the question clearly.


SOLVING THE QUESTION AND COMING TO 5C3

We need to find number of all possible combinations of 3 points. Each such combination would form a triangle.

To find number of ways in which we can select 3 points from set S, we must know:

  1. The total number of points in set S.
  2. And whether any 3 points are collinear. (Because if they are, they would not form a triangle and hence that combination of 3 points must be rejected)

From Statement 1: “The number of distinct points in S is 5”
This gave us (i) as we needed. But we still do not know about collinearity of these points – that is, we have no idea about (ii). So, statement 1 alone is insufficient.

From Statement 2: “No three of the points in S are collinear”
This tells us all about (ii) but nothing about (i). (Make sure you never drag anything from statement 1 into 2) So, statement 2 alone is insufficient.

Since, individually each statement is insufficient, we must combine the statements. Now, Statement 1 and 2 together give us both - (i) and (ii) - that we needed. Thus, the answer is C.

We will still do the Math and show you the final calculation.
  • The total number of triangles that can be formed = Number of ways in which we can select 3 points from a total of 5 points.
  • From our conceptual understanding, we know that we can select ‘r’ things from a total of ‘n’ things in nCr ways.
  • Using the same, we can select 3 points from a total of 5 points in 5C3 ways. This is precisely how you get 5C3 as the number of distinct triangles that can be drawn using 3 points in the set as vertices.


Hope this helps!


Best Regards,

Ashish

Quant Expert, e-GMAT
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