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There are 2 parallel line segments AB and CD in a plane.

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TheOutlawTorn
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There are 2 parallel line segments AB and CD in a plane. [#permalink] New post   14 Sep 2016, 06:40
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There are 2 parallel line segments AB and CD in a plane. AB contains 12 marked points whereas CD contains 8 marked points.
How many triangles can be formed by using these marked points as vertices?

a. \((12!*28) + (8!*66)\)
b. \(12!*8!\)
c. \(^{20}C_3\)
d. \(864\)
e. \(1024\)
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Bunuel
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Re: There are 2 parallel line segments AB and CD in a plane. [#permalink] New post   14 Sep 2016, 07:20
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TheOutlawTorn wrote:
There are 2 parallel line segments AB and CD in a plane. AB contains 12 marked points whereas CD contains 8 marked points.
How many triangles can be formed by using these marked points as vertices?

a. \((12!*28) + (8!*66)\)
b. \(12!*8!\)
c. \(^{20}C_3\)
d. \(864\)
e. \(1024\)


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Hope it helps.
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Re: There are 2 parallel line segments AB and CD in a plane. [#permalink] New post   14 Sep 2016, 07:05
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TheOutlawTorn wrote:
There are 2 parallel line segments AB and CD in a plane. AB contains 12 marked points whereas CD contains 8 marked points.
How many triangles can be formed by using these marked points as vertices?

a. \((12!*28) + (8!*66)\)
b. \(12!*8!\)
c. \(^{20}C_3\)
d. \(864\)
e. \(1024\)



Combinations Problem :
Pick any 2 from AB and any one from CD, so \(12C_2 * 8\)
+
Pick any 2 from CD and any one from AB, so \(8C_2 * 12\)

= \(12C_2 * 8 + 8C_2 *12 = 864\)
Answer is D
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There are 2 parallel line segments AB and CD in a plane. [#permalink] New post   Updated on: 26 Jan 2024, 05:08
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We need 3 points in total to make a triangle.

You can select either 2 points from AB and 1 from CD or 2 points from CD and 1 from AB. Thus there are 2 possible scenarios
1. Choosing from AB first
12C2 AND 8C1 = 66*8
= 528

OR
2. Choosing from CD first
8C2 AND 12C1 = 336

Total up = 864

Originally posted by dcummins on 15 Sep 2019, 20:17.
Last edited by dcummins on 26 Jan 2024, 05:08, edited 1 time in total.
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Re: There are 2 parallel line segments AB and CD in a plane. [#permalink] New post   10 Jun 2020, 13:05
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TheOutlawTorn wrote:
There are 2 parallel line segments AB and CD in a plane. AB contains 12 marked points whereas CD contains 8 marked points.
How many triangles can be formed by using these marked points as vertices?

a. \((12!*28) + (8!*66)\)
b. \(12!*8!\)
c. \(^{20}C_3\)
d. \(864\)
e. \(1024\)


Solution:

Since 3 non-collinear points form a triangle, a triangle can be form by either picking two points from AB and one point from CD OR two points from CD and one point from AB (notice that, for example, we can’t pick 3 points from AB to form a triangle since these 3 points would be linear).

Case 1: two points from AB and one point from CD

The number of ways this can be done is: 12C2 x 8C1 = (12 x 11)/2 x 8 = 66 x 8 = 528.

Case 2: two points from CD and one point from AB

The number of ways this can be done is: 8C2 x 12C1 = (8 x 7)/2 x 12 = 28 x 12 = 336.

Therefore, the total number ways a triangle can be formed is 528 + 336 = 864.

Answer: D
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Re: There are 2 parallel line segments AB and CD in a plane. [#permalink] New post   06 Oct 2020, 07:53
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An easier way is to get all the triangles with the 20 points - 20C3
And then remove triangles formed from a single line - 12C3
Same goes for the other line - 8C3
Final answer - 20C3 - 12C3 - 8C3 = 1140 -220 -56 = 864.

Thanks.
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Pranayyerra
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Re: There are 2 parallel line segments AB and CD in a plane. [#permalink] New post   03 Jan 2022, 03:20
Choosing 3 points at once 20C3 out of all these we need to remove collinear points from both the lines

so 20C3 - 12C3 - 8C3

1140-220-56 = 864

Option D
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Re: There are 2 parallel line segments AB and CD in a plane. [#permalink] New post   10 Jan 2024, 15:04
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Re: There are 2 parallel line segments AB and CD in a plane. [#permalink]
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