It is currently 18 Nov 2017, 10:58

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m06 Q 37

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Intern
Joined: 13 Jan 2010
Posts: 23

Kudos [?]: 6 [0], given: 10

m06 Q 37 [#permalink]

### Show Tags

31 Jul 2010, 16:12
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)?

Kudos [?]: 6 [0], given: 10

Intern
Joined: 17 Jul 2010
Posts: 15

Kudos [?]: 4 [0], given: 23

Re: m06 Q 37 [#permalink]

### Show Tags

15 Aug 2010, 06:54
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)?

Trick here is to chose 3 points at a time out of 6 available, and order does not matter.

So number of triangles = 6C3 = 10

Kudos [?]: 4 [0], given: 23

Manager
Joined: 27 May 2010
Posts: 197

Kudos [?]: 71 [0], given: 3

Re: m06 Q 37 [#permalink]

### Show Tags

22 Aug 2010, 06:10
This question tricked me too. Any how if they didn't mention "Any 3 points of these 6 don't lie on the same line" would the answer stay the same...

Kudos [?]: 71 [0], given: 3

Manager
Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 190

Kudos [?]: 15 [0], given: 12

WE 1: SAP consultant-IT 2 years
WE 2: Entrepreneur-family business 2 years
Re: m06 Q 37 [#permalink]

### Show Tags

30 May 2011, 11:37
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
_________________

Cheers !!

Quant 47-Striving for 50
Verbal 34-Striving for 40

Kudos [?]: 15 [0], given: 12

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1965

Kudos [?]: 2089 [2], given: 376

Re: m06 Q 37 [#permalink]

### Show Tags

30 May 2011, 11:51
2
This post received
KUDOS
bblast wrote:
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

Consider only 3 non-collinear points:

A B C
As per your formula:

Number of triangles = $$C^{3}_{2}*1=3$$. But, in reality it is 3/3 = 1.

Because your formula counts same triangle thrice. Thus, at the end you will have to divide by 3.

How so:

A B C are 3 points.
It will choose one line segment using two points:
AB and connect AB to C to form a triangle.
Then,
BC and connect BC to A to form a triangle.
AC and connect AC to B to form a triangle.
You see three counts for the same triangle ABC.

Divide your result by 3 and you will get the answer.

Or simply; select 3 points out of n points to know the number.

If there are n non-collinear points, where n>=3, we can make
$$C^{n}_{3}$$ distinct triangles.
_________________

Kudos [?]: 2089 [2], given: 376

Manager
Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 190

Kudos [?]: 15 [0], given: 12

WE 1: SAP consultant-IT 2 years
WE 2: Entrepreneur-family business 2 years
Re: m06 Q 37 [#permalink]

### Show Tags

30 May 2011, 23:00
kudos fluke for polishing this up,

actually my logic is applicable when there are two parallel lines with points and we have to draw triangles among them. Math 2 has this question I guess,

This was a relatively easier question which I messed up.
_________________

Cheers !!

Quant 47-Striving for 50
Verbal 34-Striving for 40

Kudos [?]: 15 [0], given: 12

Re: m06 Q 37   [#permalink] 30 May 2011, 23:00
Display posts from previous: Sort by

# m06 Q 37

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.