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# Two tracks are parallel. The first track has 6 checkpoints and the sec

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Manager
Joined: 14 Dec 2015
Posts: 50
Concentration: Entrepreneurship, General Management
WE: Information Technology (Computer Software)
Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink]

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02 Jun 2016, 22:43
3
8
00:00

Difficulty:

45% (medium)

Question Stats:

62% (01:28) correct 38% (01:17) wrong based on 145 sessions

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Two tracks are parallel. The first track has 6 checkpoints and the second one has 10 checkpoints. In how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle?

(A)120
(B)150
(C)200
(D)270
(E)420

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India
Re: Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink]

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02 Jun 2016, 22:47
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5
snorkeler wrote:
Two tracks are parallel. The first track has 6 checkpoints and the second one has 10 checkpoints. In how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle?

(A)120
(B)150
(C)200
(D)270
(E)420

To make a triangle, you need 2 checkpoints from one track and 1 from the other. You cannot have all 3 from the same track since then the points will be in a line (assuming straight line of track)

You select 2 checkpoints from the first track and one from the second or two from the second track and one from the first.

6C2 * 10C1 + 10C2 * 6C1 = 150 + 270 = 420

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Senior Manager
Joined: 18 Jan 2010
Posts: 254
Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink]

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02 Jun 2016, 23:48
1
snorkeler wrote:
Two tracks are parallel. The first track has 6 checkpoints and the second one has 10 checkpoints. In how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle?

(A)120
(B)150
(C)200
(D)270
(E)420

Two ways this can be done:

a) select any two points on the first track (out of available 6 points) to form the base, and choose the vertex from track B

b) select any two points on the second track (out of available 10 points) to form the base, and choose the vertex from track A

(6C2 * 10C1) + (6C1*10C2)

[$$\frac{6!}{4!2!}$$ * 10]+ [$$\frac{6!}{5!1!}$$ * $$\frac{10!}{8!2!}$$]

[$$\frac{6 *5}{2}$$ * 10]+ 6 * [$$\frac{10 * 9}{2}$$]
= 150+ 270
=420

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Joined: 09 Sep 2013
Posts: 7003
Re: Two tracks are parallel. The first track has 6 checkpoints and the sec [#permalink]

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16 Sep 2017, 00:10
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Re: Two tracks are parallel. The first track has 6 checkpoints and the sec   [#permalink] 16 Sep 2017, 00:10
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