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

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

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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
[Reveal] Spoiler: OA

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

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New post 11 May 2016, 08:55
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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 [#permalink]

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New post 11 May 2016, 11:14
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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.)

Add up the two scenarios: \(60+145 = 205\)

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

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New post 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   [#permalink] 03 Jan 2018, 03:09
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