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S is a set of points in the plane. How many distinct triangles can be https://gmatclub.com/forum/sisasetofpointsintheplanehowmanydistincttrianglescanbe6133720.html 
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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)
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:
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.
Hope this helps! Best Regards, Ashish Quant Expert, eGMAT 
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