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

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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?

I believe its a common trap to trick the test-taker to do a two-step solution (Well, I was tricked once before ). 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

I believe its a common trap to trick the test-taker to do a two-step solution (Well, I was tricked once before ). 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.

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.

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.

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.

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.

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?