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# m06 Q 37

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Intern
Joined: 13 Jan 2010
Posts: 23
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Kudos [?]: 6 [0], given: 10

m06 Q 37 [#permalink]

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31 Jul 2010, 16:12
There are 6 points on the plain. Any 3 points of these 6 don't lie on the same line. How many unique triangles can be drawn using these 6 points as vertices?

5
10
20
30
60

Why isn't the answer 60. We can choose 1 vertice and there are 5C2 = 10 ways to 2 choose the remain 2 vertices. 10*6 choices = 60 (10 choices each for each vertex)?
Intern
Joined: 17 Jul 2010
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Kudos [?]: 4 [0], given: 23

Re: m06 Q 37 [#permalink]

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15 Aug 2010, 06:54
suhasrao wrote:
There are 6 points on the plain. Any 3 points of these 6 don't lie on the same line. How many unique triangles can be drawn using these 6 points as vertices?

5
10
20
30
60

Why isn't the answer 60. We can choose 1 vertice and there are 5C2 = 10 ways to 2 choose the remain 2 vertices. 10*6 choices = 60 (10 choices each for each vertex)?

Trick here is to chose 3 points at a time out of 6 available, and order does not matter.

So number of triangles = 6C3 = 10
Manager
Joined: 27 May 2010
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Kudos [?]: 37 [0], given: 3

Re: m06 Q 37 [#permalink]

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22 Aug 2010, 06:10
This question tricked me too. Any how if they didn't mention "Any 3 points of these 6 don't lie on the same line" would the answer stay the same...
Manager
Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 193
WE 1: SAP consultant-IT 2 years
WE 2: Entrepreneur-family business 2 years
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Kudos [?]: 11 [0], given: 12

Re: m06 Q 37 [#permalink]

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30 May 2011, 11:37
why not pick 2 points out of 6 ?
=6C2

now since every 2 point will form a triangle with the 4 remaining points.

hence answer = 6C2*4 = 60

help me
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Verbal 34-Striving for 40

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
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Kudos [?]: 1440 [2] , given: 376

Re: m06 Q 37 [#permalink]

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30 May 2011, 11:51
2
KUDOS
bblast wrote:
why not pick 2 points out of 6 ?
=6C2

now since every 2 point will form a triangle with the 4 remaining points.

hence answer = 6C2*4 = 60

help me

Consider only 3 non-collinear points:

A B C
As per your formula:

Number of triangles = $$C^{3}_{2}*1=3$$. But, in reality it is 3/3 = 1.

Because your formula counts same triangle thrice. Thus, at the end you will have to divide by 3.

How so:

A B C are 3 points.
It will choose one line segment using two points:
AB and connect AB to C to form a triangle.
Then,
BC and connect BC to A to form a triangle.
AC and connect AC to B to form a triangle.
You see three counts for the same triangle ABC.

Divide your result by 3 and you will get the answer.

Or simply; select 3 points out of n points to know the number.

If there are n non-collinear points, where n>=3, we can make
$$C^{n}_{3}$$ distinct triangles.
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Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 193
WE 1: SAP consultant-IT 2 years
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Kudos [?]: 11 [0], given: 12

Re: m06 Q 37 [#permalink]

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30 May 2011, 23:00
kudos fluke for polishing this up,

actually my logic is applicable when there are two parallel lines with points and we have to draw triangles among them. Math 2 has this question I guess,

This was a relatively easier question which I messed up.
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Verbal 34-Striving for 40

Re: m06 Q 37   [#permalink] 30 May 2011, 23:00
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# m06 Q 37

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