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# M05-27

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Math Expert
Joined: 02 Sep 2009
Posts: 56304

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16 Sep 2014, 00:26
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35% (medium)

Question Stats:

66% (00:40) correct 34% (01:01) wrong based on 258 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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Math Expert
Joined: 02 Sep 2009
Posts: 56304

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16 Sep 2014, 00:26
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 $$\frac{5!}{3!}=20$$.

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Joined: 06 Mar 2014
Posts: 228
Location: India
GMAT Date: 04-30-2015

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16 Sep 2014, 12:19
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 $$\frac{5!}{3!}=20$$.

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! - ?
Math Expert
Joined: 02 Sep 2009
Posts: 56304

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16 Sep 2014, 14:39
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 $$\frac{5!}{3!}=20$$.

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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Joined: 24 Aug 2015
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21 Sep 2015, 08:58
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?
Intern
Joined: 01 Aug 2015
Posts: 1

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25 Sep 2015, 13:35
Intern
Joined: 14 Mar 2015
Posts: 33
Schools: ISB '18

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28 Sep 2015, 11:38
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
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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It ain’t about how hard you hit. It’s about how hard you can get hit and keep moving forward;
how much you can take and keep moving forward.

That’s how winning is done!
Intern
Joined: 07 Oct 2015
Posts: 1

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11 Nov 2015, 09:08
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
Intern
Joined: 25 Nov 2014
Posts: 3
Schools: Alberta'18

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20 Jul 2016, 07:27
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 !!!
Intern
Joined: 26 Sep 2016
Posts: 22

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11 Apr 2017, 08:59
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?
Intern
Joined: 01 Nov 2017
Posts: 11

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15 Jan 2018, 07:26
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?
Math Expert
Joined: 02 Aug 2009
Posts: 7764

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15 Jan 2018, 07:35
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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15 Aug 2018, 12:43
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
Math Expert
Joined: 02 Sep 2009
Posts: 56304

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15 Aug 2018, 13:32
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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Re: M05-27   [#permalink] 15 Aug 2018, 13:32
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# M05-27

Moderators: chetan2u, Bunuel