Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

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

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

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

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!.

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?

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]
05 Aug 2012, 23:35