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In a triangle ABC, there are 5, 6 and 4 points, different from vertice

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In a triangle ABC, there are 5, 6 and 4 points, different from vertice [#permalink]

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New post 21 Jan 2018, 10:23
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In a triangle ABC, there are 5, 6 and 4 points, different from the vertices A, B and C of the triangle, on the three sides AB, BC and CA respectively. How many distinct triangles can be drawn using the 18 points (including the vertices A, B and C)?
A) 111
B) 705
C) 816
D) 910
E) 955
[Reveal] Spoiler: OA

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Re: In a triangle ABC, there are 5, 6 and 4 points, different from vertice [#permalink]

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souvonik2k wrote:
In a triangle ABC, there are 5, 6 and 4 points, different from the vertices A, B and C of the triangle, on the three sides AB, BC and CA respectively. How many distinct triangles can be drawn using the 18 points (including the vertices A, B and C)?
A) 111
B) 705
C) 816
D) 910
E) 955


As direct calculation involves many possible cases, we'll solve this by calculating "1 - the probability".
This is a Logical approach.

There are 18C3 ways to choose 3 vertices from 18 points.
This is 18!/(15!*3!) = (18*17*16)/(3*2*1) = 3*17*16 = 51*16 = 510+306=816
(C),(D),(E) are eliminated as they are too large (we know we're going to have less than 816).
And since (A) looks way too small, we should feel comfortable guessing (B).

Let's calculate: of all the ways to choose vertices, the only 'illegal' choice is when all 3 points are on the same side.
All 3 on AB (including points A,B): 7C3 = (7*6*5)/(3*2*1) = 35
All 3 on BC (including points B,C): 8C3 = (8*7*6)/(3*2*1) = 56
All 3 on AB (including points A,B): 6C3 = (6*5*4)/(3*2*1) = 20

35+56+20 =111
816 - 111 = 705
(B) is our answer.
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Re: In a triangle ABC, there are 5, 6 and 4 points, different from vertice   [#permalink] 21 Jan 2018, 10:50
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