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S is a set of points in the plane. How many distinct triangles can be

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S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 15 Mar 2008, 00:41
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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.
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 15 Mar 2008, 00:49
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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 \(C^5_3=\frac{5*4*3*2}{3*2*2}=10\)
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Re: S is a set of points in the plane. How many distinct triangles  [#permalink]

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New post Updated on: 23 Sep 2015, 01:02
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My observation about such kind of questions - you read it, and have no clue/idea how to solve it - read the options, and in the second option there is a hint about whether the points are collinear. As for me, it helped me to solve this question.

Official Explanation (I've used the same logic, but it's just well written here :)
1) the number of triangles can be 0 (if the points are collinear) and the number of triangles can be greater than 0 (if the points are not all collinear); NOT sufficient.

2) Given that no three points of S are collinear, the number of triangles can
be 1 (if S consists of 3 points) and the number of triangles can be 4 (if S
consists of 4 points); NOT sufficient.

Taking (1) and (2) together, the number of distinct triangles must be
5C3 = 10, which is the number of combinations of 5 points taken 3 at a time

Answer (C)
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Originally posted by BrainLab on 23 Sep 2015, 00:57.
Last edited by BrainLab on 23 Sep 2015, 01:02, edited 1 time in total.
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S is a set of points in the plane. How many distinct triangles  [#permalink]

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New post 23 Sep 2015, 00:51
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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.


Source: OG 2016

It's a an interesting question that I've rated it as 700 level, please correct the rating if appropriate.
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 26 May 2016, 00:10
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.


l got A. What does collinear mean? :oops:
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 26 May 2016, 00:54
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ranaazad wrote:
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.


l got A. What does collinear mean? :oops:


Collinear points are those that lie on the same straight line. BTW the correct answer is C, not A. Check the solutions above and ask if anything remains unclear.
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 27 May 2016, 00:11
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 \(C^5_3=\frac{5*4*3*2}{3*2*2}=10\)



Does distinct pint not mean that points are differently located.?
I marked the A assume the above.
please make it clear.
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 27 May 2016, 07:02
robu 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 \(C^5_3=\frac{5*4*3*2}{3*2*2}=10\)



Does distinct pint not mean that points are differently located.?
I marked the A assume the above.
please make it clear.


From (1) the points are distinct but 3 or more from them can be on the same line (collinear), thus they won't form a triangle.
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 28 May 2016, 13:02
Bunuel wrote:
robu 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 \(C^5_3=\frac{5*4*3*2}{3*2*2}=10\)



Does distinct pint not mean that points are differently located.?
I marked the A assume the above.
please make it clear.


From (1) the points are distinct but 3 or more from them can be on the same line (collinear), thus they won't form a triangle.


I marked E... considering how can we be sure of the distances between the points would make valid Triangles?? What am I missing??
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 29 May 2016, 02:26
rachitshah wrote:
Bunuel wrote:
robu wrote:


Does distinct pint not mean that points are differently located.?
I marked the A assume the above.
please make it clear.


From (1) the points are distinct but 3 or more from them can be on the same line (collinear), thus they won't form a triangle.


I marked E... considering how can we be sure of the distances between the points would make valid Triangles?? What am I missing??


ANY 3 points on a plane that are not collinear form a triangle.
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S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 29 May 2016, 05:43
Example: if the distances between the 3 points are 5,6 & 13? Here they won't form a triangle right? Since 5+6 < 13?
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 19 Oct 2016, 05:57
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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.


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
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 22 Jun 2017, 18:08
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Bunuel wrote:
rachitshah wrote:
Example: if the distances between the 3 points are 5,6 & 13? Here they won't form a triangle right? Since 5+6 < 13?


You cannot have this case. If the distance from A to B is 5 and the distance from B to C is 6, then the distance C to A cannot be more than 11.


I know that this is over a year late but I just want to clarify that if one were to connect 3 points (ABC) with the distances of 5(AB), 6(BC), and 13(AC), the only possible way is to make all 3 points collinear. Which would thus not form a triangle.
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 22 Jun 2017, 21:33
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.


St1: the position of the distinct points are not discussed
If all the five points lie in a straight line, then no triangle can be formed

NS
St 2 : No of points are not discussed NS

1+2
Collinear ( Not straight line)and 5 points
sufficient C
total triangles can be obtained by using combination
5C3=10
#5C3= selecting 3 points from total 5 points
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 09 Jul 2017, 00:03
Could you please explain how did you derive the PNc form. I am bit rusty in PNc so I am still not being able to understand how to arrive at this solution. Even-though I
don't need to solve in a DS.
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 \(C^5_3=\frac{5*4*3*2}{3*2*2}=10\)
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 14 Nov 2017, 21:53
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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.


Responding to a pm:

S is a set of points. To make a triangle, we need 3 distinct points such that the 3 do not lie in a straight line (i.e. are not collinear). When you join 3 points which are in a straight line, you get a line, not a triangle.

(1) The number of distinct points in S is 5.

We know that we have 5 points. but what if all 5 are in a straight line? We won't be able to make any triangles.
If they are not in a straight line, we will be able to make triangles. Hence, this statement alone is not sufficient.

(2) No three of the points in S are collinear.

We know that the points are not collinear but how many points do we have? The more the number of points, the more the number of triangles.

Using both statements, we know that we have 5 points, no 3 of which are collinear. So we can select any 3 points out of 5 and make a triangle out of them. No of triangles we can make = 5C3 = 10 triangles.
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 14 Jul 2018, 14:14
KarishmaB wrote:
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.


Responding to a pm:

S is a set of points. To make a triangle, we need 3 distinct points such that the 3 do not lie in a straight line (i.e. are not collinear). When you join 3 points which are in a straight line, you get a line, not a triangle.

(1) The number of distinct points in S is 5.

We know that we have 5 points. but what if all 5 are in a straight line? We won't be able to make any triangles.
If they are not in a straight line, we will be able to make triangles. Hence, this statement alone is not sufficient.

(2) No three of the points in S are collinear.

We know that the points are not collinear but how many points do we have? The more the number of points, the more the number of triangles.

Using both statements, we know that we have 5 points, no 3 of which are collinear. So we can select any 3 points out of 5 and make a triangle out of them. No of triangles we can make = 5C3 = 10 triangles.


What does the C in 5C3 = 10 triangles stand for?
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 16 Jul 2018, 05:33
cheyconnors wrote:
KarishmaB wrote:
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.


Responding to a pm:

S is a set of points. To make a triangle, we need 3 distinct points such that the 3 do not lie in a straight line (i.e. are not collinear). When you join 3 points which are in a straight line, you get a line, not a triangle.

(1) The number of distinct points in S is 5.

We know that we have 5 points. but what if all 5 are in a straight line? We won't be able to make any triangles.
If they are not in a straight line, we will be able to make triangles. Hence, this statement alone is not sufficient.

(2) No three of the points in S are collinear.

We know that the points are not collinear but how many points do we have? The more the number of points, the more the number of triangles.

Using both statements, we know that we have 5 points, no 3 of which are collinear. So we can select any 3 points out of 5 and make a triangle out of them. No of triangles we can make = 5C3 = 10 triangles.


What does the C in 5C3 = 10 triangles stand for?


This is the combinations formula.

\(nCr = \frac{n!}{r! * (n-r)!}\)

You use this formula when you need to choose r items from n distinct items without repetition.

Learn more about this formula in this post I wrote for Veritas Prep: https://www.veritasprep.com/blog/2011/1 ... binations/
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Re: S is a set of points in the plane. How many distinct triangles can be  [#permalink]

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New post 23 Jul 2018, 21:54
what does 'no three of the points in S are collinear' mean?
Also is this really a sub 600 level question? Shouldn't it be at least 600-700 level?
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Re: S is a set of points in the plane. How many distinct triangles can be   [#permalink] 23 Jul 2018, 21:54

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