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.

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.htmlwhich is a part of

gmat-math-book-87417.html
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