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# There are 15 points in a given plane, no three of which are

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Manager
Joined: 06 Jun 2012
Posts: 122
There are 15 points in a given plane, no three of which are  [#permalink]

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Updated on: 18 Sep 2012, 01:32
2
11
00:00

Difficulty:

65% (hard)

Question Stats:

51% (01:22) correct 49% (01:40) wrong based on 327 sessions

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There are 15 points in a given plane, no three of which are on the same line. If one of the points is represented as 'A', then how many triangles can be determined with the 15 points that contain the point A?

A. 91
B. 105
C. 182
D. 210
E. 455

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Originally posted by summer101 on 18 Sep 2012, 01:24.
Last edited by Bunuel on 18 Sep 2012, 01:32, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 02 Sep 2009
Posts: 54434
Re: There are 15 points in a given plane, no three of which are  [#permalink]

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18 Sep 2012, 01:36
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3
summer101 wrote:
There are 15 points in a given plane, no three of which are on the same line. If one of the points is represented as 'A', then how many triangles can be determined with the 15 points that contain the point A?

A. 91
B. 105
C. 182
D. 210
E. 455

Any 2 points out 14 points will create triangle with third point A, so the answer is $$C^2_{14}=91$$.

Similar question to practice:
if-4-points-are-indicated-on-a-line-and-5-points-are-132677.html

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##### General Discussion
Manager
Joined: 25 Jun 2012
Posts: 62
Location: India
WE: General Management (Energy and Utilities)
Re: There are 15 points in a given plane, no three of which are  [#permalink]

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18 Sep 2012, 03:40
Fist point is A which is fixed so can be selected in 1 way
Second point can be selected in 14 ways
Third point can be selected in 13 ways
so total ways = 1x14x13 = 182
but answer is 91 which is 182/2
I m confused, where am I wrong
Director
Joined: 22 Mar 2011
Posts: 599
WE: Science (Education)
Re: There are 15 points in a given plane, no three of which are  [#permalink]

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18 Sep 2012, 04:27
2
bhavinshah5685 wrote:
Fist point is A which is fixed so can be selected in 1 way
Second point can be selected in 14 ways
Third point can be selected in 13 ways
so total ways = 1x14x13 = 182
but answer is 91 which is 182/2
I m confused, where am I wrong

Order of choosing the two other points doesn't matter: ABC and ACB is the same triangle.
Therefore, you should divide 182 by 2, because you counted each triangle twice.
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Re: There are 15 points in a given plane, no three of which are  [#permalink]

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05 Jan 2013, 01:43
summer101 wrote:
There are 15 points in a given plane, no three of which are on the same line. If one of the points is represented as 'A', then how many triangles can be determined with the 15 points that contain the point A?

A. 91
B. 105
C. 182
D. 210
E. 455

$$=\frac{14!}{2!12!} = 91$$
Note: The question specified that no three points lie on the same line. If that is true, the number of triangles would be less. But luckily, no such three points will form a line so rest assured, all will become triangles.
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Re: There are 15 points in a given plane, no three of which are  [#permalink]

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29 Aug 2017, 09:02
summer101 wrote:
There are 15 points in a given plane, no three of which are on the same line. If one of the points is represented as 'A', then how many triangles can be determined with the 15 points that contain the point A?

A. 91
B. 105
C. 182
D. 210
E. 455

Hi summer101,
Do you think you can change the wording "that contain the point A" to "that has A as one of its vertices" ? Otherwise the problem becomes a bit complex and confusing. The current wording says the formed triangle will contain the point A either on its perimeter or inside it.

Thanks
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Joined: 04 Mar 2011
Posts: 2825
Re: There are 15 points in a given plane, no three of which are  [#permalink]

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31 Aug 2017, 10:23
summer101 wrote:
There are 15 points in a given plane, no three of which are on the same line. If one of the points is represented as 'A', then how many triangles can be determined with the 15 points that contain the point A?

A. 91
B. 105
C. 182
D. 210
E. 455

A triangle has 3 vertices. However, since 1 of the 3 vertices has been predetermined to be A, we have to choose 2 points from the remaining 14 points as the remaining 2 vertices of the triangle. Thus, the number of triangles that can be created with A as 1 of the vertices is:

14C2 = 14!/[2!(14-2)!] = 14!/(2!12!) = (14 x 13)/2! = 7 x 13 = 91

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Re: There are 15 points in a given plane, no three of which are   [#permalink] 31 Aug 2017, 10:23
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