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M05-27

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M05-27  [#permalink]

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New post 15 Sep 2014, 23:26
4
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

66% (00:40) correct 34% (00:59) wrong based on 247 sessions

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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

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Re M05-27  [#permalink]

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New post 15 Sep 2014, 23:26
1
5
Official Solution:

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: M05-27  [#permalink]

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New post 16 Sep 2014, 11:19
Bunuel wrote:
Official Solution:

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


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! - ?
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Re: M05-27  [#permalink]

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New post 16 Sep 2014, 13:39
2
2
earnit wrote:
Bunuel wrote:
Official Solution:

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


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! - ?


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.
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Re: M05-27  [#permalink]

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New post 21 Sep 2015, 07:58
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?
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Re: M05-27  [#permalink]

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New post 25 Sep 2015, 12:35
The answer is D
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M05-27  [#permalink]

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New post 28 Sep 2015, 10:38
1
patrickmhoy wrote:
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
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Re: M05-27  [#permalink]

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New post 11 Nov 2015, 08:08
I find this question vague because nowhere in the question does it state that she wants to watch the three drama movies after each other, and since we are taught not to assume anything in the GMAT handling of questions shouldnt we assume that she can watch DACDD , DDACD, ADCDD etc...

doesnt seem logical to treat {DDD} as a 'glued' set since it doesnt state she wants to watch them three times IN SEQUENCE...

Anyone agree with my reasoning?

thanks
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Re: M05-27  [#permalink]

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New post 20 Jul 2016, 06:27
The answer is for 3 Drama screenings and not 3 consecutive Drama screening.

If question had mentioned 3 consecutive drama screening then answer would have been 6. (consider 3 Ds as a single entity and A and C so total 3 entities which can be arranged in 3! ways).

Hope its clear !!!
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Re: M05-27  [#permalink]

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New post 11 Apr 2017, 07:59
Hey

Not sure i understand why we divide by 3!, what are we getting rid of? What is the general idea behind "repetitions allowed"? Can someone explain this problem according to filling spaces method?
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Re: M05-27  [#permalink]

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New post 15 Jan 2018, 06:26
As we have 5 different movies (DDDAC), I applied the formula (n-1)!, which results in (5-1)!=4!=1*2*3*4=24.

Why is this wrong? Where am I going in the wrong direction?
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Re: M05-27  [#permalink]

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New post 15 Jan 2018, 06:35
Gregsterh wrote:
As we have 5 different movies (DDDAC), I applied the formula (n-1)!, which results in (5-1)!=4!=1*2*3*4=24.

Why is this wrong? Where am I going in the wrong direction?


(n-1)! is used in a circular combination .... say n person sitting around a circular table. (n-1)!

but here there are 5 items out of which three are of on ekind..
formula is 5!/3!.. div by 3! is because these three can be arranged among themselves in 3! ways and so to avoid repetitions, we div by 3!
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Re M05-27  [#permalink]

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New post 15 Aug 2018, 11:43
I think this is a poor-quality question and I don't agree with the explanation. The question states Carly wants to watch the same drama movie (3) times over the course of the weekend.

Carly can watch the SAME drama movie (3) times over the weekend without watching three times in a row: (DADCD)

Where in the question is it clearly stated she watches the drama consecutively because all the D's in DADCD are the SAME.

Thanks
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Re: M05-27  [#permalink]

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New post 15 Aug 2018, 12:32
NickBrady wrote:
I think this is a poor-quality question and I don't agree with the explanation. The question states Carly wants to watch the same drama movie (3) times over the course of the weekend.

Carly can watch the SAME drama movie (3) times over the weekend without watching three times in a row: (DADCD)

Where in the question is it clearly stated she watches the drama consecutively because all the D's in DADCD are the SAME.

Thanks


You did not understand the solution.

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

5!/3! gives ALL premutations possible, not necessarily those in which three D's are in a row.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M05-27 &nbs [#permalink] 15 Aug 2018, 12:32
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