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# m06 Q 37

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

m06 Q 37 [#permalink]  31 Jul 2010, 14:58
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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Kudos [?]: 1248 [0], given: 235

Re: m06 Q 37 [#permalink]  31 Jul 2010, 15:23
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)?

$$6C_3 =$$ Number of triangles that can be formed using 6 points = Number of ways in which you can select 3 points out of 6.
Since it is clearly stated that " Any 3 points of these 6 don't lie on the same line" , the above statement holds true.

thus answer is $$6C_3 = 20$$

Your logic is wrong as you are counting the items repeatedly.

Suppose a b c d e f are 6 points
you can select 1 out of 6 in 6 ways, and remaining 2 in 10 ways.
Let the first selection is a , and the later two are b and c.
Now when you will select b as first selection and remaining 2 in 10 ways, a and c are counted again.
The selection becomes b , a , c.
But the triangle formed by a b c should be counted once but you have counted it 3 times.

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Kudos [?]: 11 [0], given: 12

Re: m06 Q 37 [#permalink]  30 May 2011, 10:36
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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Re: m06 Q 37   [#permalink] 30 May 2011, 10:36
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# m06 Q 37

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