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Re: PS - Movies (m05q27) [#permalink]
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 _________________

Re: PS - Movies (m05q27) [#permalink]
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!!!

Re: PS - Movies (m05q27) [#permalink]
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. _________________

+1 Kudos if my comment was helpful. Thanks!

Failure forges confidence, and confidence cultivates success. Proving the answer choices wrong is almost better than calculating what is right.

Re: PS - Movies (m05q27) [#permalink]
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

PS - Movies (m05q27) [#permalink]
21 Jul 2014, 21:30

D,D,D,A,C

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!