Any 3 points from 6 will make a triangle, since no 3 points are collinear, then:

6C3=20

Answer: C.

TIPS ON RELATED ISSUES:
In a plane if there are n points of which no three are collinear, then
1. The number of straight lines that can be formed by joining them is nC2.
2. The number of triangles that can be formed by joining them is nC3.
3. The number of polygons with k sides that can be formed by joining them is nCk.
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Choosing any 3 points out of the 6 available points will form a triangle. Hence this turns into a combination question.

Choose 3 points out of 6: 6c3=20

Bunuel - In this case I am a bit confused.
F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.

Regular pentagon is a pentagon where all sides are equal. In such pentagon center is not collinear to any two vertex, so ANY three points (from 5 vertices and center point) WILL form the triangle.

You wrote that ACF won't form the triangle because they don't lie on straight line - that's not true. EXACTLY because these three points DON'T lie on the straight line they WILL form triangle. (In your other example you stated that ABF will form the triangle, but these three points also aren't collinear).

The question basically asks how many triangles can be formed from the six points on a plane with no three points being collinear.

Hope it's clear.
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Generally in a plane if there are \(n\) points of which no three are collinear, then:
1. The number of triangles that can be formed by joining them is \(C^3_n\).

2. The number of quadrilaterals that can be formed by joining them is \(C^4_n\).

3. The number of polygons with \(k\) sides that can be formed by joining them is \(C^k_n\).

Since ABCDE is a regular pentagon then no three point out of 6 (5 vertices + center) will be collinear, so the number of triangles possible is \(C^3_6=20\).

Answer: C.

Similar questions to practice:
http://gmatclub.com/forum/m03-71107.html
http://gmatclub.com/forum/the-sides-bc- ... 09690.html
http://gmatclub.com/forum/if-4-points-a ... 32677.html

Hope it helps.
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some body please clear this discrepancy

from 6 if we choose 3 , how do we denote it , all books and materials show this as 6C3

but here parrot man shows this as 6c3 and moderator shows this as 3c6 , leading me to becoming confused !!

according to me , the larger one goes at the top and the smaller sub group goes below

so from 6 if we choose 3 then 6C3.

so please if someone could confirm why is 3C6 denoted here , and does 3C6 mean the same as 6C3 ? are they same ?

so from n if if have to choose , 3 or 4 or 5 etc shouldn't it be nc3 , nc4 , nc5 etc .... Please clear my confusion
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There are many forms of denoting the combinations.

nCk or C(n,k) or Ckn as written by bunuel, all are different standards.

But the meaning is same. Choosing 'k' objects from a group of 'n' objects.

You should explain the logic behind your approach.[/quote]


ABCDE, 5 vertices of pentagon, we can select any two vertices two form a line, 5C2 ways....

Now any of the 5C2 ways have only one way(As per question) to interact with a centre point F = 5C2 X 1 = 10 Ways.

How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

So, we are interested in the number of triangles that can be formed by joining ANY three out of these 6 points (A, B, C, D, E and F), not necessarily F and 2 other points. For example, triangle ABC is also a valid one.

Hope it's clear.
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Similar questions to practice:
http://gmatclub.com/forum/if-4-points-a ... 32677.html
http://gmatclub.com/forum/abcde-is-a-re ... 86284.html
http://gmatclub.com/forum/abcde-is-a-re ... 33328.html
http://gmatclub.com/forum/the-sides-bc- ... 09690.html
http://gmatclub.com/forum/gmat-diagnost ... 79373.html
http://gmatclub.com/forum/m03-71107.html
http://gmatclub.com/forum/how-many-tria ... 98236.html
http://gmatclub.com/forum/right-triangl ... 88958.html
http://gmatclub.com/forum/how-many-circ ... 28149.html

Hope it helps.
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This is basically a combinatorics problem.
If we make anagram grid for this, it will like below
A B C D E F
Y Y Y N N N
That means at any point of time only three points are making the triangle while other three are not.
hence total no of combinations possible = 6!/3!x3! =20

How do we know for sure that none of the sets with any three chosen points will be collinear?

It says different triangles