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Greg is renting movies from the video store. He must choose

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Greg is renting movies from the video store. He must choose [#permalink]

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New post 27 Nov 2005, 20:38
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Greg is renting movies from the video store. He must choose 3 videos from a list of 10 videos and decide in which order to watch them. How many schedules of videos can he create?

A. 30
B. 70
C. 700
D. 720
E. 1000

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 [#permalink]

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New post 27 Nov 2005, 23:26
I believe its a common trap to trick the test-taker to do a two-step solution (Well, I was tricked once before :oops: ). Assuming the videos are numbered 1,2,3,4,5,6,7,8,9,10. It doesn't matter if we pick 1,2,3 and then order them to get 1,2,3; 3,2,1; 2,3,1; 3,1,2; 2,1,3;1,3,2. It's the same as ordering 1,2,3 from the 10 videos.

So, to solve, just permute 3 videos from the selection of 10.
10P3 = 10!/3!7! = 720

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But my understanding was.... [#permalink]

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New post 28 Nov 2005, 04:29
ywilfred wrote:
I believe its a common trap to trick the test-taker to do a two-step solution (Well, I was tricked once before :oops: ). Assuming the videos are numbered 1,2,3,4,5,6,7,8,9,10. It doesn't matter if we pick 1,2,3 and then order them to get 1,2,3; 3,2,1; 2,3,1; 3,1,2; 2,1,3;1,3,2. It's the same as ordering 1,2,3 from the 10 videos.

So, to solve, just permute 3 videos from the selection of 10.
10P3 = 10!/3!7! = 720


My understanding was if the order doesnt matter than you are supposed to use combination. So according to your example, we should have used combination rather than permutation? I think according to the question, it does matter how you watch the first three and thats why you used permutation.

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Can you explain why you used combination? [#permalink]

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New post 28 Nov 2005, 04:31
gamjatang wrote:
10C3 = 120

3! = 6

120 * 6 = 720


Can you explain how you got the answer?

Thanks

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Re: Can you explain why you used combination? [#permalink]

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New post 28 Nov 2005, 04:39
faisalt wrote:
gamjatang wrote:
10C3 = 120

3! = 6

120 * 6 = 720


Can you explain how you got the answer?

Thanks


Cases of picking 3 movies out of 10 movies
= 10C3

Cases of arranging 3 movies
= 3!

Cases of picking 3 movies and arranging 3 movies
= 10C3 * 3!
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GMAT 1: 740 Q48 V42
 [#permalink]

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New post 28 Nov 2005, 06:30
I agree with Gamjatang.
Wilfred, Can you elaborate more on your explanation? Thanks

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I kinda disagree with Gamjatang... [#permalink]

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New post 28 Nov 2005, 15:30
anandsebastin wrote:
I agree with Gamjatang.
Wilfred, Can you elaborate more on your explanation? Thanks


Greg is renting movies from the video store. He must choose 3 videos from a list of 10 videos and decide in which order to watch them. How many schedules of videos can he create?

I think we are supposed to use permutation, as the order in which movies are being watched matters. It is not same as watching {1, 2, 3} as {3,2,1}. I agree with the Wilfred but his explanation is kinda conflicting.

Thanks,

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 [#permalink]

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New post 28 Nov 2005, 20:42
since order matters so its permutations ..
10P3 = 720

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Re: PS-Probability [#permalink]

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New post 07 May 2009, 11:49
The trick relies on the assumption that order does NOT matter. I mean, we think we have to choose any 3 movies out of 10 and after that arrange them in schedules.
From the question "How many different schedules of videos can he create?" we should notice that order DOES matter so we shortcut by picking the movies in the different orders (Permutation) we can schedule them since the beginning.

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Re: PS-Probability [#permalink]

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New post 13 May 2009, 22:22
I am a bit confused here....

see we are saying that out of 10 we need to select 3.

So we can have 720 sets (or ways) containing 3 movies each {S1},{S2},{S3}............{S720}.
Now when he play set S1 he have 6 choices as there are 3 movies in this set.

So the answer should be 720*6 = 4320

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Re: PS-Probability [#permalink]

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New post 13 May 2009, 23:23
mdfrahim wrote:
I am a bit confused here....

see we are saying that out of 10 we need to select 3.

So we can have 720 sets (or ways) containing 3 movies each {S1},{S2},{S3}............{S720}.
Now when he play set S1 he have 6 choices as there are 3 movies in this set.

So the answer should be 720*6 = 4320


You can only get 720 sets if you consider the order in each set. If you ignore the order, then you only get 120 sets.

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Re: PS-Probability [#permalink]

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New post 15 May 2009, 13:35
faisalt wrote:
Greg is renting movies from the video store. He must choose 3 videos from a list of 10 videos and decide in which order to watch them. How many schedules of videos can he create?

A. 30
B. 70
C. 700
D. 720
E. 1000


1. 10x9x8 = 720
2. 10c3 x 3p1 = (3!) (10x9x8x7!)/[(10-3)!3!] = 720
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Re: PS-Probability   [#permalink] 15 May 2009, 13:35
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Greg is renting movies from the video store. He must choose

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