GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Aug 2018, 23:12

### 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

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

# How many triangles can be formed using 8 points in a given p

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 20 Aug 2009
Posts: 103
How many triangles can be formed using 8 points in a given p  [#permalink]

### Show Tags

23 Aug 2009, 05:46
6
20
00:00

Difficulty:

25% (medium)

Question Stats:

64% (00:32) correct 36% (00:45) wrong based on 958 sessions

### HideShow timer Statistics

How many triangles can be formed using 8 points in a given plane?

(1) A triangle is formed by joining 3 distinct points in the plane
(2) Out of 8 given points, three are collinear
Manager
Joined: 27 May 2009
Posts: 239
Re: Permutations 1  [#permalink]

### Show Tags

23 Aug 2009, 06:08
5
2
IMO B. 8c3 - 1

A is the general stmt. we know that Triangle is fromed using only 3 points.
##### General Discussion
Manager
Joined: 14 Aug 2009
Posts: 121
Re: Permutations 1  [#permalink]

### Show Tags

23 Aug 2009, 18:19
1
The meaning of "8C3-1" is really brief, clear and precise. That is the beauty of math. I like it and you deserve a kudos for this.
_________________

Kudos me if my reply helps!

Senior Manager
Joined: 20 Mar 2008
Posts: 436
Re: Permutations 1  [#permalink]

### Show Tags

Updated on: 23 Aug 2009, 21:22
rohansherry wrote:
IMO B. 8c3 - 1

A is the general stmt. we know that Triangle is fromed using only 3 points.

Agree with B. But I think the combination formula might be wrong.

IMHO, only 3 points are collinear, all the other 5 points are not on the same line. Hence, each of the 3 points can be combined with 2 of the 5 points to create a triangle.

Therefore, the # of triangles = 3 x 5C2 = 3 x 10 = 30.

Originally posted by Jivana on 23 Aug 2009, 21:09.
Last edited by Jivana on 23 Aug 2009, 21:22, edited 1 time in total.
Manager
Joined: 14 Aug 2009
Posts: 121
Re: Permutations 1  [#permalink]

### Show Tags

23 Aug 2009, 21:20
Jivana wrote:
rohansherry wrote:
IMO B. 8c3 - 1

A is the general stmt. we know that Triangle is fromed using only 3 points.

Agree with B. But I think the combination formula might be wrong.

IMHO, only 3 points are collinear, all the other 5 points are not on the same line. Hence, each of the 3 points can be combined with 2 of the 5 points to create a triangle.

Therefore, the # of triangles = 3 x 5C3 = 3 x 10 = 30.

You forgot 2 of the 3 points (collinear) and 1 of the 5 points to form a triangle.
_________________

Kudos me if my reply helps!

Senior Manager
Joined: 20 Mar 2008
Posts: 436
Re: Permutations 1  [#permalink]

### Show Tags

23 Aug 2009, 21:26
Yep, I did indeed forget that.

# should be: 3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45.
Manager
Joined: 27 May 2009
Posts: 239
Re: Permutations 1  [#permalink]

### Show Tags

24 Aug 2009, 00:32
Jivana wrote:
Yep, I did indeed forget that.

# should be: 3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45.

I just feel you are missing here something.... Can yo explain how you comingup with this "3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45"
Director
Joined: 01 Apr 2008
Posts: 821
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Re: Permutations 1  [#permalink]

### Show Tags

24 Aug 2009, 07:48
rohansherry wrote:
Jivana wrote:
Yep, I did indeed forget that.

# should be: 3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45.

I just feel you are missing here something.... Can yo explain how you comingup with this "3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45"

Yup. It is not necessary to select ATLEAST one point from the three collinear points (as per the above equation)
SO we have to add 5C3 = 10 too.

So answer is 45+10 = 55 (Using Jivana's conventional method) and 8C3-1=56-1=55 (Using rohansherry's method faster method)
Intern
Joined: 30 Jun 2009
Posts: 48
Re: Permutations 1  [#permalink]

### Show Tags

25 Aug 2009, 06:35
HI Guys, My answer would also be B.
But, I am not clear in the possibilities to form triangles.
1. using the 3 non collinear : 5C3
2. using one collinear and 2 non collinear : 3C1x5C2

I am sure that I am missing something here.
Would you please be of any help to understand the 55 possibilities?

Rgds
Intern
Joined: 30 Jun 2009
Posts: 48
Re: Permutations 1  [#permalink]

### Show Tags

25 Aug 2009, 06:39
Guys,

I just noticed that I was missing the 2 of the 3 points (collinear) and 1 of the 5 points

Thx
Intern
Joined: 30 Jun 2009
Posts: 48
Re: Permutations 1  [#permalink]

### Show Tags

25 Aug 2009, 06:41
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

Rgds
Manager
Joined: 14 Aug 2009
Posts: 121
Re: Permutations 1  [#permalink]

### Show Tags

25 Aug 2009, 07:16
defoue wrote:
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

Rgds

the triangle by three points that are collinear.
_________________

Kudos me if my reply helps!

Manager
Joined: 27 May 2009
Posts: 239
Re: Permutations 1  [#permalink]

### Show Tags

25 Aug 2009, 07:19
3
2
defoue wrote:
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

Rgds

hey see,

if have to make all triangles from n points...then it would be nc3.....because triangle has 3 sides.... so out of n any 3 sides...

now 3 point are colinear so we cant make one triangle...so we se subtract that from it..

hope its clear....now
Intern
Joined: 30 Jun 2009
Posts: 48
Re: Permutations 1  [#permalink]

### Show Tags

26 Aug 2009, 04:59
2
1
rohansherry wrote:
defoue wrote:
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

Rgds

hey see,

if have to make all triangles from n points...then it would be nc3.....because triangle has 3 sides.... so out of n any 3 sides...

now 3 point are colinear so we cant make one triangle...so we se subtract that from it..

hope its clear....now

Cristal clear
Thx very much
Manager
Joined: 20 Aug 2009
Posts: 103
Re: Permutations 1  [#permalink]

### Show Tags

26 Aug 2009, 05:27
Guys, I am sorry that it took me so long to post the OA. But here it is:

1. is insufficient because it just states a well known fact
2. is sufficient because in such case we can calculate the number of triangles that can be formed: 5C3+8*3C2+3C1*5C2

__________________________________
Please kudos me if you find my post useful
Intern
Joined: 19 Mar 2013
Posts: 22
Re: Problem: triangle in a plane Level: Medium How many  [#permalink]

### Show Tags

10 Dec 2013, 06:09
Why can't we just pick all possible combinations of 3 points out of 8 to answer the first question?
8C3 looks sufficient to me
Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re: Problem: triangle in a plane Level: Medium How many  [#permalink]

### Show Tags

10 Dec 2013, 06:20
Christy111 wrote:
Why can't we just pick all possible combinations of 3 points out of 8 to answer the first question?
8C3 looks sufficient to me

Ask yourself: how many triangles can be formed out of 8 collinear points? Is it 8C3 or 0?
_________________
Manager
Joined: 08 Nov 2014
Posts: 90
Location: India
GPA: 3
WE: Engineering (Manufacturing)
Re: How many triangles can be formed using 8 points in a given p  [#permalink]

### Show Tags

23 Dec 2014, 09:57
Question doesnt mention the orientation of 8 points. What if all points are collinear.
So, statement 1) is ruled out
Statement 2) Three points are collinear, they cannot form a triangle. By combination if all points are non-collinear 8C3 no.s of distinct triangle can be formed.
Hence, answer will be 8C3-1
_________________

"Arise, Awake and Stop not till the goal is reached"

Current Student
Joined: 18 Sep 2015
Posts: 93
GMAT 1: 610 Q43 V31
GMAT 2: 610 Q47 V27
GMAT 3: 650 Q48 V31
GMAT 4: 700 Q49 V35
WE: Project Management (Health Care)
Re: How many triangles can be formed using 8 points in a given p  [#permalink]

### Show Tags

29 May 2016, 06:15
CasperMonday wrote:
How many triangles can be formed using 8 points in a given plane?

(1) A triangle is formed by joining 3 distinct points in the plane
(2) Out of 8 given points, three are collinear

The answer here is not b but E.

Comment: point (x,y) and points (a,b) are identical if [x=a] and [y=b].

1. A is a general stetment, and in addition, we canno't form a trianle in which 2 or more points are identicle.
-> since we do not know how many points are identical:
-> not suff.
2. Exacly 3 points are co linear, this say nothing about how many points are identical.
2 points can be coliner but identical.
-> not suff

Hence, Choice E.
Intern
Joined: 08 Dec 2016
Posts: 40
Re: How many triangles can be formed using 8 points in a given p  [#permalink]

### Show Tags

19 Jun 2017, 14:32
1
We have 8 points, and 3 are collinear:
- Select 2 points among 5 non-collinear points (5C2) and 1 among the 3 collinear points ( x3 ): 3x5C2
- Select 2 points among 3 collinear points (3C2) and 1 among 5 non-collinear ( x5 ) : 5x3C2
- Select 3 points among the 5 non-collinear points: 5C3
Add all of them = 55 --> equivalent to 8C3 - 1
Re: How many triangles can be formed using 8 points in a given p &nbs [#permalink] 19 Jun 2017, 14:32

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by

# How many triangles can be formed using 8 points in a given p

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

# Events & Promotions

 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®.