M05-27

Question

Carly has three different movies to watch during the weekend: an action movie, a comedy movie, and a drama movie. She plans to watch the drama movie three times, the action movie once, and the comedy movie once, for a total of five screenings. How many different ways can she arrange the order of these five movie screenings?

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

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

Bunuel · Accepted Answer

Official Solution:

Carly has three different movies to watch during the weekend: an action movie, a comedy movie, and a drama movie. She plans to watch the drama movie three times, the action movie once, and the comedy movie once, for a total of five screenings. How many different ways can she arrange the order of these five movie screenings?

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

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

The number of different ways Carly can arrange the order of the five movie screenings, which include Drama, Drama, Drama, Action, and Comedy (DDDAC), can be determined by calculating the number of permutations of 5 elements (DDDAC) where 3 D's are identical. This can be computed as  \frac{5!}{3!} = 20 .

Answer: B

Bunuel · Answer

In this case we should subtract from total ways, which is 20, the number of ways when 3 D's are together.

Consider 3 D's as one unit {DDD}. We'll have 3 units: {DDD}{A}{C}. The number of ways to arrange them is 3! = 6.

So, the answer in this case would be 20 - 6 = 14.

Hope it's clear.

earnit · Answer

Hi Bunuel,

The above explanation is perfect. Thank you.
However, if the question asks with respect to DDDAC, how many ways that 3Ds are NOT Together then how exactly should one approach?

I think: Total ways possible - when the 3Ds are together.
ie. 5!/3! - ?

patrickmhoy · Answer

What is the rule for this type of combination problem? It doesn't seem to follow the fundamental counting rule (n!)/ .... or does it?

vivekgautam1 · Answer

The first thing in Combinatorics is to figure out whether the problem is of Permutation or Combination.

This problem asks for the  "arrangement" , so this is clearly a Permutation problem. But the author also says that one type of movie (Drama) has to be watched thrice.

So possible arrangements are -

DDDAC
ADDDC
ACDDD
DDACD
...
..
and so on.

Now you should see here that the 3 Ds are same, so their arrangement, when they are placed together, doesn't really matter.

So, like Bunuel has mentioned in the official solution, the total number of arrangements for n objects with repetition, in which object-1 repeats n 1  times, object-2 repeats n 2  times, object-3 repeats n 3  times and so on, can be given from the formula -
n!/(n 1 ! *n 2 ! * n 3 ! .....)
= 5!/(3! * 1! * 1*)
= 5!/3!
= 20

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

MT1302 · Answer

Hey  Bunuel ,

Wanted to know that because it is not mentioned that Drama movies are distinct i.e. why we are doing   \frac{ 5!}{3! }    ?

If they had mentioned Distinct then solution would be 5! = 120 . Correct ?

Just wanted to know in general that if Gmat  does not mentions  in these type of Questions  Distinct  then, we would have to  assume  objects to be  Identical  . Like in this question . Correct ?

Bunuel · Answer

The question says that there are THREE DIFFERENT movies: an action movie, a comedy movie, and a drama movie. Carly plans to watch THE drama movie three times, THE action movie once, and THE comedy movie once, for a total of five screenings.

I think it's clear that Carly wants to watch the three movies mentioned in the setup of the question. So, THE drama movie three times, THE action movie once, and THE comedy movie once, for a total of five screenings.

GMAT questions are always precise, so by reading them correctly, you'll know whether the objects mentioned are distinct or not.

P.S. If the question simply were in how many ways one can watch 5 different movies, then the answer would be 5!

pierren · Answer

Hello, where can i find exercices on permutation combination please ?

Bunuel · Answer

21. Combinatorics/Counting Methods 
     Theory   
 Math: Combinatorics 
 Combinatorics Made Easy! 
 1 hour Combinatorics Video Lesson 
 Combinations while dealing with similar and dissimilar things 
 When Permutations & Combinations and Data Sufficiency Come Together 
     Questions   
 Advanced Topic: Probability and Combinations Questions With Solutions 
 Letter Arrangements in Words 
 Seating Arrangements in a Row and around a Table 
 Constructing Numbers, Codes and Passwords 
 DS Questions 
 PS Questions

For even more:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

MyNameisFritz · Answer

I made a mistake by multiplying (5 x 4 x 3) and dividing by 3!

Why do we divide 5! by 3! if Carly will watch 3 movies, not 5 movies?

Bunuel · Answer

Carly will have 5 screenings: 3 drama movies, 1 action movie, and 1 comedy movie, represented as DDDAC. The number of ways to arrange DDDAC is calculated as 5!/3!.

Hope it's clear.

RITESH1984 · Answer

This is what I did C(5, 3)+C(5, 1)+C(5, 1), I got the right answer but I am not comfortable with these questions. I am below average on combination, permutation and probability and I have just been avoiding these concepts I guess now it's time to invest in these topics.

Bunuel · Answer

Your approach C(5, 3) + C(5, 1) + C(5, 1) is incorrect because you're treating this as a sum of independent selections, which doesn’t reflect the structure of the problem.

If you were trying to solve it by selecting positions for each movie type, you'd need to use multiplication, not addition, because the cases are dependent. Specifically, you would do:

C(5, 3) * C(2, 1) * C(1, 1)

Where:

• C(5, 3): choose 3 slots for the drama movies
• C(2, 1): from the remaining 2 slots, choose 1 for the action movie
• C(1, 1): the last remaining slot goes to the comedy movie

Combinatorics/Counting Methods

Theory 
 
  Math: Combinatorics  
  Combinatorics Made Easy!  
  1 hour Combinatorics Video Lesson  
  Combinations while dealing with similar and dissimilar things  
  When Permutations & Combinations and Data Sufficiency Come Together

Questions 
 
  Advanced Topic: Probability and Combinations Questions With Solutions  
  Letter Arrangements in Words  
  Seating Arrangements in a Row and around a Table  
  Constructing Numbers, Codes and Passwords  
  DS Questions  
  PS Questions

For even more:
 
  ALL YOU NEED FOR QUANT ! ! !  
  Ultimate GMAT Quantitative Megathread

Hope it helps.

aynavya · Answer

I like the solution - it’s helpful.

agrasan · Answer

I like the solution - it’s helpful.

rakshitapoda · Answer

Why are we assuming that the Drama movie needs to be watched 3 times back to back? Since it is a permutation problem, changing the order will create a new order in which the movies can be watched. For example, DDDAC is different from DDACD, which is different from DADCD.

Bunuel · Answer

Nobody is assuming the 3 dramas are back to back. They are simply identical copies of the same movie. All different orders like DDDAC, DDACD, DADCD, etc., are already included in 5!/3!.

M05-27

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