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Five points lie on a straight line

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Junior | Next Rank: 30 Posts
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by imago » Tue Sep 18, 2007 1:19 pm
Five points lie on a straight line and another 4 points lie on another straight line that is parallel to the first one, how many different triangles can be made by linking these 9 points?

a.7
b.20
c.70
d.140
e.400

OA: C
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by kajcha » Tue Sep 18, 2007 1:30 pm
It will be 70.. but I am thinking how to type my scratch work here :?
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by kajcha » Tue Sep 18, 2007 2:24 pm
To explain it

L1 has 5 points (a,b,c,d,e)
L2 has 4 points (w,x,y,z)

Consider point a on L1 and w on L2 - there are 4 triangles awb, awc, awd, awe. Similarly there will be 4 triangles for each point on L2

Now, think of this, there would be 6 more traingles for point a awx, awy, awz, axy, axz, ayz

In a nutshell No of traingles from each point on L1

= [(no of points on L2)*(no of remaining points on L1)] + 6

So no of triangles
from a = 22
from b = 18
from c = 14
from d = 10
from e = 6

Total = 70
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by givemeanid » Tue Sep 18, 2007 5:18 pm
(4 * 5C2) + (5 * 4C2) = 4*10 + 5*6 = 70
So It Goes
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by kajcha » Tue Sep 18, 2007 5:23 pm
givemeanid wrote:(4 * 5C2) + (5 * 4C2) = 4*10 + 5*6 = 70
Thanks givemeanid
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by imago » Wed Sep 19, 2007 8:08 am
Thanks Givemeanid & Kajcha!!!!
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