M05-27

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! - ?

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

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 n1 times, object-2 repeats n2 times, object-3 repeats n3 times and so on, can be given from the formula -
n!/(n1! *n2! * n3! .....)
= 5!/(3! * 1! * 1*)
= 5!/3!
= 20

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

 
­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 ?­

 
­

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!­

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

­

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.­

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?

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.

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.

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

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.­