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Three points are marked on a line, and five points are

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Manager
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Three points are marked on a line, and five points are [#permalink]  30 May 2007, 04:54
Three points are marked on a line, and five points are marked on another, parallel to the first. The number of triangles that can be drawn using any 3 of these 8 points is:

(A) 26
(B) 90
(C) 25
(D) 45
(E) 42
Manager
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[#permalink]  30 May 2007, 07:40
care to share how you came up with that answer?

some readers might not know how to set up a problem like this
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[#permalink]  30 May 2007, 08:08
I got D.

My way of solving the problem however is not very efficient. There's gotta be a faster way.

Basically we have this:

Line 1) --A-----B-----C--

Line 2) --D-----E-----F----G----H--

First, if A connects with any 2 other points from Line 2 there are exactly 10 possible triangles. Same for B and C. So far, 3x10 = 30 triangles

Then, if we "flip" the lines and have point D connect with any 2 other points from Line 1 exactly 3 triangles can be formed. Same for E, F, G and H. So, 5x3 = 15

Therefore, 15 + 30 = 45 triangles can be formed.
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[#permalink]  30 May 2007, 08:42
joroivanov wrote:
I got D.
Basically we have this:

Line 1) --A-----B-----C--

Line 2) --D-----E-----F----G----H--

First, if A connects with any 2 other points from Line 2 there are exactly 10 possible triangles. Same for B and C. So far, 3x10 = 30 triangles

Then, if we "flip" the lines and have point D connect with any 2 other points from Line 1 exactly 3 triangles can be formed. Same for E, F, G and H. So, 5x3 = 15

Therefore, 15 + 30 = 45 triangles can be formed.

Thanks again joroivanov. I myself couldn't solve this question qithin 2 mins... Your explanation is very clear to me.

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[#permalink]  30 May 2007, 10:32
no. of triangles = 8C3 - 5C3 -3C3 = (8*7*6/3*2*1 - 5*4/2*1 - 1)
=270/6
=45

WHERE 8C3 = TRIANGLES MADE BY TAKING 3 PTS OUT OF TOTAL 8 pts.

5C3 = No triangles for 5 collinear points.

3C3 = same.
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[#permalink]  30 May 2007, 11:06
joroivanov wrote:
I got D.

My way of solving the problem however is not very efficient. There's gotta be a faster way.

Basically we have this:

Line 1) --A-----B-----C--

Line 2) --D-----E-----F----G----H--

First, if A connects with any 2 other points from Line 2 there are exactly 10 possible triangles. Same for B and C. So far, 3x10 = 30 triangles

Then, if we "flip" the lines and have point D connect with any 2 other points from Line 1 exactly 3 triangles can be formed. Same for E, F, G and H. So, 5x3 = 15

Therefore, 15 + 30 = 45 triangles can be formed.

Nice approach! Kind of a thinking-on-your-feet approach but very effective.

Here's the approach I was taught:
We need to select 3 points to form a triangle.
We can't get all 3 from the 5 marked in a line because that won't form a triangle. Same for the 3 marked in a line.

So we can select 2 from the 5 points and the third from the 3 points or vice versa. In quant that translates to
(5C2)(3C1) + (3C2)(5C1) = 45
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[#permalink]  30 May 2007, 11:15
vidyasagar wrote:
no. of triangles = 8C3 - 5C3 -3C3

I think this is a great approach too but looks like this approach needs one to understand clearly what one is getting into.

Here's my attmept to put this to theory
Triangles = Total Triangle's - Triangle's in a straight line (or really 3 points in a line)
[#permalink] 30 May 2007, 11:15
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