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Bill is renting DVDs from the video store. He must choose three DVDs

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Bill is renting DVDs from the video store. He must choose three DVDs  [#permalink]

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New post 25 Jun 2018, 07:32
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Question Stats:

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Bill is renting DVDs from the video store. He must choose three DVDs from a list of eight new releases and then decide in what order to watch them. How many different schedules of DVDs can he create, if he cannot watch the same movie more than once?

(A) 24
(B) 56
(C) 72
(D) 336
(E) 512

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Re: Bill is renting DVDs from the video store. He must choose three DVDs  [#permalink]

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New post 25 Jun 2018, 07:39
CAMANISHPARMAR wrote:
Bill is renting DVDs from the video store. He must choose three DVDs from a list of eight new releases and then decide in what order to watch them. How
many different schedules of DVDs can he create, if he cannot watch the same movie more than once?
(A) 24
(B) 56
(C) 72
(D) 336
(E) 512


It is permutation question.. 8P3=8!/5!=8*7*6=56*6=336

Or you can work on combinations..

First find the possible combinations..
8C5=8!/(5!3!)=8*7*6/6=8*7=56
But these 3 can be arranged in 3! Ways
So total ways =56*3!=336

D
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Re: Bill is renting DVDs from the video store. He must choose three DVDs  [#permalink]

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New post 25 Jun 2018, 08:43
Total no of DVDs = 8
No of DVDs Bill has to select at a time = 3

He can selecting 3 from 8 in \(8c3\) ways.

Selected 3 DVDs can be seen in 3! ways.

So total no of ways = \(8c3\) * 3! = (\(\frac{8!}{3!*5!}\)) * 3! = 8*7*6 = 336
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Re: Bill is renting DVDs from the video store. He must choose three DVDs  [#permalink]

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New post 26 Jun 2018, 16:55
CAMANISHPARMAR wrote:
Bill is renting DVDs from the video store. He must choose three DVDs from a list of eight new releases and then decide in what order to watch them. How many different schedules of DVDs can he create, if he cannot watch the same movie more than once?

(A) 24
(B) 56
(C) 72
(D) 336
(E) 512


Because the order in which Bill watches the DVDs is important, we use permutations. The number of ways to select and watch 3 DVD’s is 8P3 = 8!/(8 - 3)! = 8!/5! = 8 x 7 x 6 = 336.

Answer: D
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Bill is renting DVDs from the video store. He must choose three DVDs  [#permalink]

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New post 28 Jun 2018, 13:33
The key word in this question is order. For this type of question, use Permutation.

A helpful hint I use when doing this type of problem is to think of the made up word "POM". I think of it as an acronym I made up to remind me to use Permutation when order matters:
POM = Permutation when Order Matters.

8P3 = 8!/(8-3)! = 8!/5! = 336

Answer D
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Bill is renting DVDs from the video store. He must choose three DVDs &nbs [#permalink] 28 Jun 2018, 13:33
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