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# There are four points in line L and five points in line P,

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Director
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There are four points in line L and five points in line P, [#permalink]

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12 Jun 2007, 02:32
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There are four points in line L and five points in line P, where L is parallel to P. By linking the 9 points, how many triangles can be formed?

1. 76
2. 20
3. 70
4. 66
5. 80
Manager
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12 Jun 2007, 03:18
Option E, 80.
Each point on Line L can be combined in 20 ways with other 8 points to make different triangles. So 20*4=80. Lets say points on line L are 1,2,3,4 and points on line P are a,b,c,d,e. Now taking point 1 and any two points on line P, we can make 5C2 or 10 triangles. Similarly taking point 1, any other point on line L and one point on line P, we can still make 10 triangles. So, using point 1, we can make 20 different triangles. since there are 4 points, so 20*4=80.
VP
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12 Jun 2007, 03:27
vshaunak@gmail.com wrote:
There are four points in line L and five points in line P, where L is parallel to P. By linking the 9 points, how many triangles can be formed?

1. 76
2. 20
3. 70
4. 66
5. 80

3. 70

--x---x---x---x----x-- Line P

---x----x----x----x--- Line L

From Line P we have to link 2 points plus 1 of line L to form a tringle, so we have from Line P a Combination of 5,2 times 4 possibilities in line L, sorry if the notations are not that the one you guys know:

C(5,2)*4=( 5!/ ( (5-2)!*2!) ) *4 = ( 5*4 / 2 )*4 =40

Similarly from Line L

C(4,2)*5 = ( 4*3/2 )*5 = 30

Summing the two lines 30+40 = 70

Last edited by kwam on 12 Jun 2007, 04:29, edited 1 time in total.
CEO
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12 Jun 2007, 06:04
Hi Kwam I third 70 !! .. Exactly the way I did it.

What is the OA ?
Manager
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12 Jun 2007, 08:42
Line L: ---------x------x------x------x

Line P:---x-----x------x------x------x

We need to link those points. I chunked the problem into two separate types of triangles, those with the base on either line. If the base is on P, all triangles possible are 5C2*4C1; if on line L, 4C2*5C1.

Director
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12 Jun 2007, 12:18
Well done guys. OA is 70.
12 Jun 2007, 12:18
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