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PS - Movies (m05q27)

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Re: PS - Movies (m05q27) [#permalink] New post 06 Aug 2012, 04:23
Bunuel wrote:
leeye84 wrote:
Carly has 3 movies that she can watch during the weekend: 1 Action movie, 1 Comedy, and 1 Drama. However, she needs to watch the Drama 3 times. Assuming Carly has time for 5 movies and intends to watch all of them, in how many ways can she do so?

Below is a revised version of this question:

Carly has three movies that she can watch during the weekend: an action movie, a comedy, or a drama. However, she wants to watch the same drama movie three times, an action movie once and a comedy movie also once. In how many different ways can she arrange these five screenings?

A. 6
B. 20
C. 24
D. 60
E. 120

The number of different ways Carly can watch Drama, Drama, Drama, Action, Comedy (DDDAC) is basically the number of arrangements of 5 letters DDDAC out of which 3 D's are identical, so it's \(\frac{5!}{3!}=20\).

Answer: B.

This is the way the question should be phrased to avoid ambiguity.

Otherwise the original question can also lead to the following solution:
(3D1C1A )5!/3!+ (3D2C+3D2A)2*5!/3!2! +(4D1A+4D1C)2*5!/4! +(5D)5!/5! = 46

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Re: PS - Movies (m05q27) [#permalink] New post 06 Aug 2012, 05:31
Easy one if one can remember the arrangement of letters DDDAC. Since all Ds are indistinguishable Ans is 5!/3! = 20

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Re: PS - Movies (m05q27) [#permalink] New post 23 Aug 2012, 23:22
That's so cool! I always had difficulty in solving this type of question. After going thru the book veritas prep combinatorics and probability I feel huuuuge progress and easily solve these questions!!!
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Re: PS - Movies (m05q27) [#permalink] New post 11 Jun 2013, 17:08
The duplication effect occurs here; all three viewings of the Drama movie are equal - not independent of each other. Moreover, it's not logical to think that the first viewing of drama can go after the third viewing of the drama, and the third viewing of the drama to come before the first.
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Re: PS - Movies (m05q27) [#permalink] New post 05 Aug 2013, 04:25
This is my plan: in total there are 5 slots for all movies
For the 2 movies that watched only one time:
the first movie: there are 5 slots
the 2nd movie: there are 4 slots (5 minus one picked for the first movie)
after selecting slots for 2 movies, there are to slot for the drama and any order are the same ---> 1 way

==> in total there are 5x4=20 ways
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PS - Movies (m05q27) [#permalink] New post 21 Jul 2014, 21:30

five things can be arranged in 5!ways. = 120
3 of them are similar , hence 5!/3! = 20 ways

P.S. - In case the three drama movies had been different , I mean, she had the choice to watch D1, D2 and D3 then the arrangements would have been 5!=120 only. But since the drama movie is the same D,D,D, HENCE 5!/3!
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Re: PS - Movies (m05q27) [#permalink] New post 24 Jul 2014, 13:01
_ _ _ _ _
5 movies

3 have to be drama, they can fit into our scheme in 5C3 ways.

remaining two movies, we have two slots left. therefore, 2C1.

the total number of ways to watch the movies : 5C3 * 2C1 = 20.
Re: PS - Movies (m05q27)   [#permalink] 24 Jul 2014, 13:01

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PS - Movies (m05q27)

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