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PS - Movies (m05q27) [#permalink]
 21 Aug 2007, 21:25
Question Stats:
53% (01:22) correct
46% (00:58) wrong based on 2 sessions
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? (A) 6 (B) 20 (C) 24 (D) 60 (E) 120 Source: GMAT Club Tests - hardest GMAT questions Please Explain.
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Fistail 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?
1) 6
2) 20
3) 24
4) 60
5) 120
Please Explain. = 5!/3! =20 Could you be a bit more specific? I'm not sure how to get to the answer.
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leeye84 wrote: Fistail 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?
1) 6
2) 20
3) 24
4) 60
5) 120
Please Explain. = 5!/3! =20 Could you be a bit more specific? I'm not sure how to get to the answer. it is 5! but when there is a repetition you need to divide total by the repetitive number's factorial.
so it is 5!/3!.
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leeye84 wrote: Fistail 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?
1) 6
2) 20
3) 24
4) 60
5) 120
Please Explain. = 5!/3! =20 Could you be a bit more specific? I'm not sure how to get to the answer. Its clear form question that he has to watch 3 drama movie,1 Action movie and 1 Comedy in 5 movie times..
We juss have to find out the ways of arranging it.
When these type of questions come in which u have to arrange n things on n places where m are similar things.. Ways =n!/m!
So 5!/3! is the answer, 5 movies, 3 are same
B is the answer
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Re: PS - Movies (m05q27) [#permalink]
28 Jul 2010, 06:20
I think rewording this question is appropriate because it is part of the quantitative section not the verbal. The verb "intends" should be changed to "must"
Other than that I came up with the possible solutions thinking about it this way.
The are 5 movie slots available
The choices (in the bag) are Drama^1, Drama^2, Drama^3, Action, and Comedy.
For the first slot you have 5 choices
second slot there are 4 choices and so on.
So we know we have 5 x 4 x 3 x 2 x 1 (5!) ways to arrange the movies.
5! = 120
Then we have to find a way to account for duplication. As far as the movie watcher is concerned watchingDrama^1 is the same as Drama^2
I knew E was incorrect because of duplication and A seemed too few combinations considering there are 5 time slots. So figuring I was nearing the perpetual 2 minute time limit, I guessed at B. My thought was 120 / x = 20 x=6 sounds about right.
Others have suggested you can divide by the duplication (3!) to get a sure answer.
At this point I made the guess for B and I am crossing my fingers
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Re: PS - Movies (m05q27) [#permalink]
28 Jul 2010, 07:12
IMO B. She wants to watch drama 3 times.. so she has 5C3 slots for drama movies = 10 slots. The other 2 slots can be chosen in 2 * 1 ways = 2 ways.. Therefore total ways = 10 * 2 = 20...
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Re: PS - Movies (m05q27) [#permalink]
28 Jul 2010, 11:04
The answer is 20. Explanation: She has to watch 1 Action movie(a), 1 Comedy(c), and 1 Drama(d). However the drama movie needs to be watched thrice which is like watching the same movie 3 times. We can start working out like this... acddd adcdd addcd and so on.. The total number of movies to be watched is 5, of which 3 are repeated. Therefore the answer works out to 5!/3! Which is equal to 20 Posted from my mobile device 
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Re: PS - Movies (m05q27) [#permalink]
28 Jul 2010, 11:05
The answer is 20. Explanation: She is required to watch 1 Action movie(a), 1 Comedy movie(c), and 1 Drama movie(d). However the drama movie needs to be watched thrice which is like watching the same movie 3 times. We can start working out like this... acddd adcdd addcd and so on.. The total number of movies to be watched is 5, of which 3 are repeated. Therefore the answer works out to 5!/3! Which is equal to 20 Posted from my mobile device
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Re: PS - Movies (m05q27) [#permalink]
28 Jul 2010, 18:36
I quickly guessed at the question and instead of using the formula thought intuitively:
DDDCA - How many different ways can I arrange these 5 letters? 5! however the 3 D's are all the same so 3! needs to be deducted due to double counting. so the answer is 5!/3! or 20.
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Re: PS - Movies (m05q27) [#permalink]
02 Aug 2010, 08:00
One more way of solving:
(5C3x2C1)=20
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Re: PS - Movies (m05q27) [#permalink]
01 Aug 2011, 06:04
bddurgap wrote: One more way of solving:
(5C3x2C1)=20 Hey, can you talk about this way a bit more? I'm missing something here. 5C3 is choosing 3 spots out of 5 for the drama movies. Im missing the explanation for 2C1. thanks.
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Re: PS - Movies (m05q27) [#permalink]
01 Aug 2011, 06:39
Guys, I have done MGMAT Word Translation twice but I am still having problems with probability and combinatorics. Could you suggest anything to improve these two areas?
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Re: PS - Movies (m05q27) [#permalink]
01 Aug 2011, 06:41
For problems like these, I tend to go straight to the anagram method
ACDDD
5 total letters = 5!
Divide that by the product of the factorials of all the different letters = 1! (A) x 1! (C) x 3! (there are 3 D's)
5! / 3! = 20
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Re: PS - Movies (m05q27) [#permalink]
01 Aug 2011, 07:54
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? (A) 6 (B) 20 (C) 24 (D) 60 (E) 120 Source: GMAT Club Tests - hardest GMAT questions Please Explain. There are multiple ways to solve this problem: Q: There are 5 movie slots to fill, such that, D movie watched 3 time, and A and C movies once each. Sol 1: also known as Anagram M1 M2 M3 M4 M5 Total ways = 5! , if all different movies But D repeats 3 times and others once, so Total conditional ways = 5!/(3!*1!*1!) = 20 Sol 2: Total conditional ways = (number of ways to select movie slots where D will be played)*(number of ways to select movie slots where A will be played from the remaining places)*(number of ways to select movie slots where C will be played from the remaining places) = 5C3 * 2C1 * 1C1 = 10 * 2 * 1 = 20 There are few other ways but these are something easy to comprehend. Hope these helps!
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Re: PS - Movies (m05q27) [#permalink]
01 Aug 2011, 10:56
took some time in drawing a correlation between this question and the total number of ways of arrangements of items: ABDDD ABDDD: 5!/3! = 20: total 5 items but 3 identical. A and B represent different movies, and D represents drama. In satisfying the condition (ABDDD) for 5 different time slots we therefore require: 5! / 3!(repetitions)
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Re: PS - Movies (m05q27) [#permalink]
02 Aug 2011, 10:35
This problem can be thought of as a permutation problem.
Since we have 5 slots to fill with movies A,B,D,D,D, we can calculate number of ways A and B can be put into 5 positions.
Rest of the positions will be filled by D hence it does not matter how we distribute remaining 3 Ds in 3 positions.
Which is 5P2 = 5!/(5-2)! = 5!/3! = 20
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Re: PS - Movies (m05q27) [#permalink]
10 Aug 2011, 12:07
5C3 X 2C1 = 5!/(3!X2!) *2! = 5!/3! = 20
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Re: PS - Movies (m05q27) [#permalink]
03 Aug 2012, 05:10
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? (A) 6 (B) 20 (C) 24 (D) 60 (E) 120 Source: GMAT Club Tests - hardest GMAT questions Please Explain. 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.
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Re: PS - Movies (m05q27) [#permalink]
03 Aug 2012, 05:29
ACDDD 5!/3! = 20 B what percentile question is it? Did not require more than 1 min
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Re: PS - Movies (m05q27) [#permalink]
03 Aug 2012, 22:31
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? (A) 6 (B) 20 (C) 24 (D) 60 (E) 120 Source: GMAT Club Tests - hardest GMAT questions Please Explain. Carly has options to watch 5 movies, but she as 1 Action, 1 Comedy and 1 Drama Movie... for 5 options she has 3 Drama movies, she will watch the same...so DDD and Action A and Comedy C so 5 movies she has is DDDAC hence these can be arranged in 5! ways now in the above arrangement there can be arrangements of just DDD in 3! ..so the new arrangement for her requirement is 5!/3! = 5*4 = 20 which is B
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Re: PS - Movies (m05q27)
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03 Aug 2012, 22:31
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