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31 Jul 2010, 15:58
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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)?
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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31 Jul 2010, 16:23
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suhasrao
\(6C_3 =\) 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.
That means your actual answer should 1/3 of your present answer. i.e. 1/3 * 60 = 20.
I advice you to drop your method and follow the one that I have stated previously.
Read this : math-combinatorics-87345.html
which is a part of gmat-math-book-87417.html
30 May 2011, 11:36
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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
=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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