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m06 Q 37

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Intern
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m06 Q 37 [#permalink] New post 31 Jul 2010, 15:12
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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)?
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Re: m06 Q 37 [#permalink] New post 15 Aug 2010, 05:54
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
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Re: m06 Q 37 [#permalink] New post 22 Aug 2010, 05:10
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...
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Re: m06 Q 37 [#permalink] New post 30 May 2011, 10:37
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
:oops:

help me
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Re: m06 Q 37 [#permalink] New post 30 May 2011, 10:51
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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
:oops:

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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Manager
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Affiliations: The Earth organization, India
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Re: m06 Q 37 [#permalink] New post 30 May 2011, 22:00
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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Re: m06 Q 37   [#permalink] 30 May 2011, 22:00
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42 Experts publish their posts in the topic M06 Q9 mbaobsessed 20 12 Nov 2008, 13:52
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