Aug 18 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes. Aug 19 08:00 AM PDT  09:00 AM PDT Join a 4day FREE online boot camp to kick off your GMAT preparation and get you into your dream bschool in R1.**Limited for the first 99 registrants. Register today! Aug 20 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Aug 20 09:00 PM PDT  10:00 PM PDT Take 20% off the plan of your choice, now through midnight on Tuesday, 8/20 Aug 22 09:00 PM PDT  10:00 PM PDT What you'll gain: Strategies and techniques for approaching featured GMAT topics, and much more. Thursday, August 22nd at 9 PM EDT
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 19 Aug 2010
Posts: 61

The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
19 Feb 2011, 11:35
Question Stats:
69% (02:26) correct 31% (02:37) wrong based on 226 sessions
HideShow timer Statistics
The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points respectively on them. How many triangles can be formed using these points as vertices. A. 200 B. 205 C. 400 D. 410 E. 415
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 57022

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
19 Feb 2011, 11:57
medanova wrote: The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices. a.200 b.205 c.400 d.410 Total points: 3+4+5=12; Any 3 points out of 12 but those which are collinear will form a triangle, so \(C^3_{12}(C^3_{3}+C^3_{4}+C^3_{5})=220(1+4+10)=205\). Answer: B.
_________________




Retired Thread Master
Joined: 27 Jan 2010
Posts: 137
Concentration: Strategy, Other
WE: Business Development (Consulting)

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
25 Feb 2011, 10:55
medanova wrote: The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices. a.200 b.205 c.400 d.410 Using 1 point from each side of the triangle  3C1 * 4C1 * 5C1 = 3 * 4 * 5 = 60 Using 1 point on 1 side and 2 points on the other side  (3C1 * 4C2) + (3C1 * 5C2) + (4C1 * 3C2) + (4C1 * 5C2) + (5C1 * 4C2) + (5C1 * 3C2) = 145 Adding these we get = 60 + 145 = 205. I was a little bit confused at first  I was not sure if we could use 2 of the interior points and 1 of the vertices of the existing triangle ABC (as the 3rd vertice) to form a new triangle. I needed to look at the question this way  How many triangles can be formed using only these points as vertices.




Manager
Joined: 19 Aug 2010
Posts: 61

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
19 Feb 2011, 12:01
Thank you ,
I initially included also the points A B C in the calculation but now I read the question again and found my mistake.



Manager
Joined: 26 Sep 2010
Posts: 123
Nationality: Indian
Concentration: Entrepreneurship, General Management

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
25 Feb 2011, 23:20
Thanks for the question and great explanations.
_________________
You have to have a darkness...for the dawn to come.



Manager
Joined: 19 Aug 2010
Posts: 61

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
26 Feb 2011, 08:34
Thanks also from me.



Manager
Joined: 20 Jul 2012
Posts: 115
Location: India
WE: Information Technology (Computer Software)

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
10 Sep 2013, 10:57
medanova wrote: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points respectively on them. How many triangles can be formed using these points as vertices.
A. 200 B. 205 C. 400 D. 410 [1*3*4]*5=60 when all the three points are on different sides.. [1*3+1*6]*5 +[1*10+1*3]*4+[1*10+1*6]*3=45+52+48when two points on one side total=60+45+52+48=205



Intern
Joined: 17 Oct 2015
Posts: 17
Concentration: Technology, Leadership

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
09 Jan 2016, 06:35
Bunuel, is there a easier way to do the math, instead of multiplying 12 x 11 x 10 x 9 x 8 x 7.....? As far I understood your solution, I would need to do that at the C 3 12 combination. Regards and have a great year! Bunuel wrote: Bumping for review and further discussion.



Math Expert
Joined: 02 Aug 2009
Posts: 7756

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
09 Jan 2016, 06:51
mestrec wrote: Bunuel, is there a easier way to do the math, instead of multiplying 12 x 11 x 10 x 9 x 8 x 7.....? As far I understood your solution, I would need to do that at the C 3 12 combination. Regards and have a great year! Bunuel wrote: Bumping for review and further discussion. Hi, Iwillhelp you out on that .. the Q basically says there are three lines with 3,4,5 points respectively.. lets find ways triangle can be formed.. triangle requires three points,.. 1) 2 points from line containing 3 points and one from remaining 4+5 points=3C2*(4+5)=27.. 2) 2 points from line containing 4 points and one from remaining 3+5 points=4C2*(3+5)=48.. 3) 2 points from line containing 5 points and one from remaining 4+3 points=5C2*(4+3)=70.. 4) 1 point from each line containing 3,4,5 points =3*4*5=60.. total 27+48+70+60=205.. also you do not have to do the entire calculation in 12C3.. 12C3=12!/3!9!= 12*11*10*9!/3!9!=12*11*10/3*2... so you see you just have to multiply 12,11,and 10..
_________________



Board of Directors
Joined: 17 Jul 2014
Posts: 2531
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
06 Mar 2016, 19:56
very nice question...tried to analyze it first..then did not how to solve it..went with chetan2u 's method.. we can have 1 point on from 3, 1 point from 4, 1 point from 5: 3C1*4C1*5C1 = 60 we can have 1 point from 3, and 2 points from 5 = 3C1*5C2 = 30 we can have 1 point from 3, and 2 points from 4 = 3C1*4C2 = 18 we can have 1 point from 4, and 2 points from 3 = 4C1*3C2 = 12 we can have 1 point from 4, and 2 points from 5 = 4C1*5C2 = 40 we can have 1 point from 5, and 2 points from 3 = 5C1*3C2 = 15 we can have 1 point from 5, and 2 points from 4 = 5C1*4C2 = 30 now total: 60+30+18+12+40+15+30 = 205
_________________



Math Expert
Joined: 02 Aug 2009
Posts: 7756

Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point
[#permalink]
Show Tags
11 May 2016, 09:55
techiesam wrote: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices.
A. 220. B. 205. C.250 D.105. E.225 hi, different ways.. 1) EASIEST approach..lets see how many total can be made if none of the points were colliner that is NO three were in one line = 12C3 = 12!/9!3! = 220... Due to collinear, extra triangles considered above = 3C3 + 5C3 +4C3 = 1+10+4 = 15.. total = 22015 = 205 2) SECOND method a) ONE point from each. b) two points from one side and third from any of two other sides
_________________



Manager
Joined: 09 Jul 2013
Posts: 109

Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point
[#permalink]
Show Tags
11 May 2016, 12:14
To expand on Chetan4u's second method, each triangle can be formed with one point from each of the 3 lines AB (3 points), BC (4 points) and CA (5 points), or with two points from one line, and the third point from either of the other two lines. The calculations look like this: Scenario 1) One point from each line = \(3*4*5 = 60\) Scenario 2) Two points from one line and one from the other two lines = \(3c2*9 + 4c2*8 + 5c2*7 = 27+48+70=145\) (Quick explanation here in case I just lost some of you  Taking 2 points from the line AB, which has 3 points on it can be done in 3c2 ways  choosing two points from the 3 available on AB. Then we can choose the third point from any of the remaining points on the other two lines BC or CA, which have a total of 9 points among them. So the number of triangles that can be formed with two points on line AB is \(3c2*9 = 27\). Similarly for the other two lines.) Add up the two scenarios: \(60+145 = 205\) Answer: B
_________________
Dave de Koos GMAT aficionado



Manager
Joined: 01 Jan 2018
Posts: 78

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
04 Aug 2018, 21:42
Hi, Can someone please clear my doubt If we are joining three points ..one point from each side of original triangle then 4 triangles formed. Why we dont consider those triangles as well? Sent from my Redmi Note 3 using GMAT Club Forum mobile app
_________________
+1 Kudos if you find this post helpful



Manager
Joined: 19 Nov 2017
Posts: 190
Location: India
GPA: 4

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
04 Aug 2018, 22:41
dvishal387 wrote: Hi, Can someone please clear my doubt If we are joining three points ..one point from each side of original triangle then 4 triangles formed. Why we dont consider those triangles as well? Sent from my Redmi Note 3 using GMAT Club Forum mobile appThe triangles need to be formed from the interior points i.e. treating the interior points as vertices. This is mentioned in the question, just go back and have a look.
_________________
Regards,Vaibhav Sky is the limit. 800 is the limit.
~GMAC



Manager
Joined: 01 Jan 2018
Posts: 78

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
04 Aug 2018, 23:45
vaibhav1221 wrote: dvishal387 wrote: Hi, Can someone please clear my doubt If we are joining three points ..one point from each side of original triangle then 4 triangles formed. Why we dont consider those triangles as well? Sent from my Redmi Note 3 using GMAT Club Forum mobile appThe triangles need to be formed from the interior points i.e. treating the interior points as vertices. This is mentioned in the question, just go back and have a look. Okay..thanks Sent from my Redmi Note 3 using GMAT Club Forum mobile app
_________________
+1 Kudos if you find this post helpful



Manager
Joined: 20 Apr 2019
Posts: 81

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
10 Aug 2019, 02:15
Bunuel wrote: medanova wrote: The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices. a.200 b.205 c.400 d.410 Total points: 3+4+5=12; Any 3 points out of 12 but those which are collinear will form a triangle, so \(C^3_{12}(C^3_{3}+C^3_{4}+C^3_{5})=220(1+4+10)=205\). Answer: B. We substract (C^3_{3}+C^3_{4}+C^3_{5}) from C^3_{12} to exclude the triangles that would be possible by using the points from the same sides, correct?



Math Expert
Joined: 02 Sep 2009
Posts: 57022

Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
Show Tags
10 Aug 2019, 02:31
Luca1111111111111 wrote: Bunuel wrote: medanova wrote: The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices. a.200 b.205 c.400 d.410 Total points: 3+4+5=12; Any 3 points out of 12 but those which are collinear will form a triangle, so \(C^3_{12}(C^3_{3}+C^3_{4}+C^3_{5})=220(1+4+10)=205\). Answer: B. We substract (C^3_{3}+C^3_{4}+C^3_{5}) from C^3_{12} to exclude the triangles that would be possible by using the points from the same sides, correct? We subtract those combinations of points which are on a straight line (collinear) and thus do not form a triangle.
_________________




Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
[#permalink]
10 Aug 2019, 02:31






