There are 15 points in a given plane, no three of which are

Question

There are 15 points in a given plane, no three of which are on the same line. If one of the points is represented as 'A', then how many triangles can be determined with the 15 points that contain the point A?

A. 91
B. 105
C. 182
D. 210
E. 455

A. 91
B. 105
C. 182
D. 210
E. 455

Bunuel · Accepted Answer

Any 2 points out 14 points will create triangle with third point A, so the answer is C^2_{14}=91 . Answer: A. Similar question to practice: if-4-points-are-indicated-on-a-line-and-5-points-are-132677.html Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html

bhavinshah5685 · Answer

Fist point is A which is fixed so can be selected in 1 way
Second point can be selected in 14 ways
Third point can be selected in 13 ways
so total ways = 1x14x13 = 182
but answer is 91 which is 182/2
I m confused, where am I wrong :oops:

EvaJager · Answer

Order of choosing the two other points doesn't matter: ABC and ACB is the same triangle.
Therefore, you should divide 182 by 2, because you counted each triangle twice.

mbaiseasy · Answer

=\frac{14!}{2!12!} = 91 
 Note: The question specified that no three points lie on the same line. If that is true, the number of triangles would be less. But luckily, no such three points will form a line so rest assured, all will become triangles.

shakun · Answer

Hi  summer101 ,
Do you think you can change the wording "that contain the point A" to "that has A as one of its vertices" ? Otherwise the problem becomes a bit complex and confusing. The current wording says the formed triangle will contain the point A either on its perimeter or inside it.

Thanks

JeffTargetTestPrep · Answer

A triangle has 3 vertices. However, since 1 of the 3 vertices has been predetermined to be A, we have to choose 2 points from the remaining 14 points as the remaining 2 vertices of the triangle. Thus, the number of triangles that can be created with A as 1 of the vertices is:

14C2 = 14!/  = 14!/(2!12!) = (14 x 13)/2! = 7 x 13 = 91
 
Answer: A

Poki1729 · Answer

Can it be 15C3-14C3=91
 I.e total possibility-excluded A possibility

Posted from my mobile device

Gaurav07 · Answer

91 two-point lines can be created from 14 points in 14c2 ways. All those lines will be a triangle each when connected with 'A'. That's why I chose 91. Is that correct approach?

There are 15 points in a given plane, no three of which are

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

There are 15 points in a given plane, no three of which are

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards