Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 06 Jun 2012
Posts: 140

There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
18 Sep 2012, 00:24
1
This post received KUDOS
13
This post was BOOKMARKED
Question Stats:
50% (01:01) correct 50% (01:16) wrong based on 308 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Please give Kudos if you like the post
Last edited by Bunuel on 18 Sep 2012, 00:32, edited 1 time in total.
Renamed the topic and edited the question.



Math Expert
Joined: 02 Sep 2009
Posts: 43804

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
18 Sep 2012, 00:36
6
This post received KUDOS
Expert's post
4
This post was BOOKMARKED



Manager
Joined: 25 Jun 2012
Posts: 67
Location: India
WE: General Management (Energy and Utilities)

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
18 Sep 2012, 02:40
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



Director
Joined: 22 Mar 2011
Posts: 608
WE: Science (Education)

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
18 Sep 2012, 03:27
2
This post received KUDOS
bhavinshah5685 wrote: 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 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.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Senior Manager
Joined: 13 Aug 2012
Posts: 456
Concentration: Marketing, Finance
GPA: 3.23

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
05 Jan 2013, 00:43
summer101 wrote: 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 \(=\frac{14!}{2!12!} = 91\) 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.
_________________
Impossible is nothing to God.



NonHuman User
Joined: 09 Sep 2013
Posts: 13832

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
29 Oct 2014, 09:33
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



NonHuman User
Joined: 09 Sep 2013
Posts: 13832

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
10 Dec 2015, 07:49
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



NonHuman User
Joined: 09 Sep 2013
Posts: 13832

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
27 Aug 2017, 18:47
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Intern
Joined: 26 Mar 2017
Posts: 29

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
29 Aug 2017, 08:02
summer101 wrote: 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 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



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1975

Re: There are 15 points in a given plane, no three of which are [#permalink]
Show Tags
31 Aug 2017, 09:23
summer101 wrote: 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 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!/[2!(142)!] = 14!/(2!12!) = (14 x 13)/2! = 7 x 13 = 91 Answer: A
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: There are 15 points in a given plane, no three of which are
[#permalink]
31 Aug 2017, 09:23






