Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 563 reviewsTarget Test Prep

PS Question # 1 -How many different triangles can be formed?

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Sat Aug 21, 2010 8:18 am
Location: India
Thanked: 5 times
by sumit.sinha » Sun Dec 19, 2010 9:43 am
Image

Answer Choices:
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Source - Princeton CAT5

OA - C
Last edited by sumit.sinha on Sun Dec 19, 2010 10:39 am, edited 2 times in total.
Cheers,
Sumit
Top
Join the discussion
Source: — Problem Solving |
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Sun Dec 19, 2010 9:50 am
sumit.sinha wrote:Image

Answer Choices:
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Source - Princeton CAT5

OA - Soon
Total number of available points = 6
Total number of points needed to form a triangle = 3

Thus, total number of triangles can be formed = C(6,3) = 6!/(3!*3!) = 4*5 = 20

As there in no 3 points co-linear we dont have to subtract anything.

IMO C
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Sat Aug 21, 2010 8:18 am
Location: India
Thanked: 5 times

by sumit.sinha » Sun Dec 19, 2010 9:55 am
shovan85 wrote:
sumit.sinha wrote:Image

Answer Choices:
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Source - Princeton CAT5

OA - Soon
Total number of available points = 6
Total number of points needed to form a triangle = 3

Thus, total number of triangles can be formed = C(6,3) = 6!/(3!*3!) = 4*5 = 20

As there in no 3 points co-linear we dont have to subtract anything.

IMO C
Have you made sure that you are not counting the same triangle twice?
Cheers,
Sumit
Top
Join the discussion
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Sun Dec 19, 2010 10:12 am
I don't know. According to formula it worked.

By considering AB as base AFB the red triangle, ADB the green one, ABC the yellow one, and ACF the blue-yellow one, I got 4 triangles.

Please help me see if I am recounting any triangle more than once.
You do not have the required permissions to view the files attached to this post.
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Sat Aug 21, 2010 8:18 am
Location: India
Thanked: 5 times

by sumit.sinha » Sun Dec 19, 2010 10:32 am
shovan85 wrote:I don't know. According to formula it worked.

By considering AB as base AFB the red triangle, ADB the green one, ABC the yellow one, and ACF the blue-yellow one, I got 4 triangles.

Please help me see if I am recounting any triangle more than once.
With base AB: you missed the triangle ABE and triangle ACF does not have base AB.

But more importantly my doubt has cleared, I was not counting the triangles like ACF.
With base as AB - 4 different triangles
With base as BC - 3 different triangles
With base as CD - 3 different triangles
With base as DE - 3 different triangles
With base as AE - 2 different triangles
the five i was missing were:
ACF, ADF,BEF,BDF and CEF

Ok. i guess i am going to use the formula next time around. Much faster. Thanks.
Cheers,
Sumit
Top
Join the discussion
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members
Thanked: 147 times

by anshumishra » Sun Dec 19, 2010 10:38 am
shovan85 wrote:I don't know. According to formula it worked.

By considering AB as base AFB the red triangle, ADB the green one, ABC the yellow one, and ACF the blue-yellow one, I got 4 triangles.

Please help me see if I am recounting any triangle more than once.
I don't think there should be any repetitions.
sumit probably is thinking whether there are multiple ways in which 3 points are counted twice (for e.g, [A,B,C] and [B,A,C]).
However, the combination formula takes care of that, so when we select r items out of n, all the nCr ways are unique sets.

If it would have been nPr, then the counting will make difference as the arrangements [A,B,C] and [B,A,C] would have been considered differently.

In our case [A,B,C] and [B,A,C] are same, they make same triangle. So, Shovan's approach looks good to me (unless I am missing something here).

Thanks
Top
Join the discussion
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Sun Dec 19, 2010 10:40 am
sumit.sinha wrote:
With base AB: you missed the triangle ABE and triangle ACF does not have base AB.
No man I did not miss ACF triangle. I have highlighted it in blue in the picture.

And the reason I did not consider ABE as I can see it is similar to ABC and plus it is using two sides of the polygon. Thus is bound to get repetitive. :)

Hope OA is 20 :)
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Sat Aug 21, 2010 8:18 am
Location: India
Thanked: 5 times

by sumit.sinha » Sun Dec 19, 2010 10:41 am
shovan85 wrote:
sumit.sinha wrote:
With base AB: you missed the triangle ABE and triangle ACF does not have base AB.
No man I did not miss ACF triangle. I have highlighted it in blue in the picture.

And the reason I did not consider ABE as I can see it is similar to ABC and plus it is using two sides of the polygon. Thus is bound to get repetitive. :)

Hope OA is 20 :)
yes, it is. :)
Cheers,
Sumit
Top
Join the discussion
Legendary Member
Posts: 1119
Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members
Thanked: 29 times

by diebeatsthegmat » Sun Dec 19, 2010 5:29 pm
shovan85 wrote:
sumit.sinha wrote:Image

Answer Choices:
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Source - Princeton CAT5

OA - Soon
Total number of available points = 6
Total number of points needed to form a triangle = 3

Thus, total number of triangles can be formed = C(6,3) = 6!/(3!*3!) = 4*5 = 20

As there in no 3 points co-linear we dont have to subtract anything.

IMO C[/quote
very nice, i love this solution. thanks! :)
Top
Join the discussion