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The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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19 Feb 2011, 11:35
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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.
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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19 Feb 2011, 11:57
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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 \(C^3_{12}-(C^3_{3}+C^3_{4}+C^3_{5})=220-(1+4+10)=205\).
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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25 Feb 2011, 10:55
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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.
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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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+48-when two points on one side total=60+45+52+48=205
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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09 Jan 2016, 06:51
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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..
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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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
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Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point
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11 May 2016, 09:55
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2
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 = 220-15 = 205
2) SECOND method a) ONE point from each. b) two points from one side and third from any of two other sides
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Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point
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11 May 2016, 12:14
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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.)
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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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?
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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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?
The 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.
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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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?
The 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.
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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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?
Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points resp
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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.
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