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# The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point

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Manager
Joined: 01 Jun 2015
Posts: 216
Location: India
GMAT 1: 620 Q48 V26
The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point  [#permalink]

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11 May 2016, 08:11
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00:00

Difficulty:

75% (hard)

Question Stats:

57% (02:31) correct 43% (02:23) wrong based on 76 sessions

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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
Math Expert
Joined: 02 Aug 2009
Posts: 7199
Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point  [#permalink]

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11 May 2016, 08:55
1
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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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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]

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11 May 2016, 11:14
2
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: $$60+145 = 205$$

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Dave de Koos

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Posts: 9423
Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point  [#permalink]

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03 Jan 2018, 03:09
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Re: The sides AB, BC & CA of a triangle ABC have 3, 4 and 5 interior point &nbs [#permalink] 03 Jan 2018, 03:09
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