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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?

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!.

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

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 watching\(Drama^1\) 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

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

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

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.

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?

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.

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) _________________

KUDOS me if you feel my contribution has helped you.

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

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?

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 \(\frac{5!}{3!}=20\).

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?

I started with the combinations but got caught up in whether the sequence mattered or not. So then switched to listing out the movies as DDDAC. Worked on this approach for a while and was not getting anywhere. In the end I chose A which was wrong.

The above explanations have surely helped me understand where I was going wrong. The anagram explanation was particularly helpful and I will surely apply it the next time I face such a problem. _________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

I started with the combinations but got caught up in whether the sequence mattered or not. So then switched to listing out the movies as DDDAC. Worked on this approach for a while and was not getting anywhere. In the end I chose A which was wrong.

The above explanations have surely helped me understand where I was going wrong. The anagram explanation was particularly helpful and I will surely apply it the next time I face such a problem.

Try to use simple logic: Carly has to watch 5 movies. She wants to see the drama 3 times, and the other two movies, once each. She can watch them in different orders, and here definitely order matters.

To watch the action movie, she can chose from 5 possibilities - either first, second, third, fourth or fifth movie. Then, for the comedy, she can chose from 4 possibilities - any place left in the sequence, after the action movie was "placed". All the remaining 3 slots, will be given to the drama movie.

Therefore, a total of 5*4=20 possibilities for the orders in which she can watch the movies.

Answer B. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

gmatclubot

Re: PS - Movies (m05q27)
[#permalink]
06 Aug 2012, 00:35