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

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18 Sep 2012, 01:24

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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?

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 \(C^2_{14}=91\).

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

Re: There are 15 points in a given plane, no three of which are [#permalink]

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18 Sep 2012, 04:27

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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. _________________

Re: There are 15 points in a given plane, no three of which are [#permalink]

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29 Oct 2014, 10:33

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Re: There are 15 points in a given plane, no three of which are [#permalink]

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10 Dec 2015, 08:49

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Re: There are 15 points in a given plane, no three of which are [#permalink]

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27 Aug 2017, 19:47

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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.

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

Answer: A
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