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7 people (A, B, C, D, E, F and G) go to a movie and sit

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7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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7 people (A, B, C, D, E, F and G) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater.

How many different arrangements are possible? If A will not sit to the left of F and F will not sit to the left of E. How many different arrangements are possible.

A) 7!/2
B) 7!/3
C) 7!/4
D) 7!/5
E) 7!/6
[Reveal] Spoiler: OA

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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 05 Sep 2014, 13:28
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megatron13 wrote:
7 people (A, B, C, D, E, F and G) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater.

How many different arrangements are possible? If A will not sit to the left of F and F will not sit to the left of E. How many different arrangements are possible.

A) 7!/2
B) 7!/3
C) 7!/4
D) 7!/5
E) 7!/6


7 people can be arranged in a row in 7! ways.

Now, three people among those 7 can be arranged in 3! = 6 ways:
AFE
AEF
EAF
EFA
FAE
FEA

From the 6 arrangements above only EFA is possible (A is not to the left of F and F is not to the left of E), so out of total 7! ways only in 1/6th of the arrangements they are sitting as they want.

Answer: E (7!/6).
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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 05 Sep 2014, 13:30
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megatron13 wrote:
7 people (A, B, C, D, E, F and G) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater.

How many different arrangements are possible? If A will not sit to the left of F and F will not sit to the left of E. How many different arrangements are possible.

A) 7!/2
B) 7!/3
C) 7!/4
D) 7!/5
E) 7!/6


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goldenrod-and-no-hope-are-in-a-horse-race-with-6-contestants-82214.html
meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
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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

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Kudos [?]: 135204 [0], given: 12673

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Kudos [?]: 39 [0], given: 19

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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 06 Sep 2014, 02:35
Hi Bunuel,

I am having difficulty understanding this approach since there can be combinations in which other people might sit between A, F and E. For this question, I was trying to count all the invalid combinations and subtract from total combinations. But how to count invalid cases like below

A,_,_,F,_E,_ or _,F,_,E,_,_,A

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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 06 Sep 2014, 09:23
megatron13 wrote:
Hi Bunuel,

I am having difficulty understanding this approach since there can be combinations in which other people might sit between A, F and E. For this question, I was trying to count all the invalid combinations and subtract from total combinations. But how to count invalid cases like below

A,_,_,F,_E,_ or _,F,_,E,_,_,A


When I say AFE or AEF I don't mean that they are necessarily adjacent, there might be other people between them.
_________________

New to the Math Forum?
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Resources:
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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.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 07 Sep 2014, 08:20
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BOOKMARKED
Bunuel wrote:
megatron13 wrote:
7 people (A, B, C, D, E, F and G) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater.

How many different arrangements are possible? If A will not sit to the left of F and F will not sit to the left of E. How many different arrangements are possible.

A) 7!/2
B) 7!/3
C) 7!/4
D) 7!/5
E) 7!/6


7 people can be arranged in a row in 7! ways.

Now, three people among those 7 can be arranged in 3! = 6 ways:
AFE
AEF
EAF
EFA
FAE
FEA

From the 6 arrangements above only EFA is possible (A is not to the left of F and F is not to the left of E), so out of total 7! ways only in 1/6th of the arrangements they are sitting as they want.

Answer: E (7!/6).





Hi Bunuel.
Can you please explain how to solve such problems when there are higher nos. involved? To clarify, if there are constraints imposed on 5 out of 10 total people attending the event, then it is practically impossible to list 5! arrangements in 2 mins. What technique should be used in such situation? Any help on this matter will be appreciated.

Thanks.

Kudos [?]: 16 [0], given: 40

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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 07 Sep 2014, 08:33
p2bhokie wrote:
Bunuel wrote:
megatron13 wrote:
7 people (A, B, C, D, E, F and G) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater.

How many different arrangements are possible? If A will not sit to the left of F and F will not sit to the left of E. How many different arrangements are possible.

A) 7!/2
B) 7!/3
C) 7!/4
D) 7!/5
E) 7!/6


7 people can be arranged in a row in 7! ways.

Now, three people among those 7 can be arranged in 3! = 6 ways:
AFE
AEF
EAF
EFA
FAE
FEA

From the 6 arrangements above only EFA is possible (A is not to the left of F and F is not to the left of E), so out of total 7! ways only in 1/6th of the arrangements they are sitting as they want.

Answer: E (7!/6).





Hi Bunuel.
Can you please explain how to solve such problems when there are higher nos. involved? To clarify, if there are constraints imposed on 5 out of 10 total people attending the event, then it is practically impossible to list 5! arrangements in 2 mins. What technique should be used in such situation? Any help on this matter will be appreciated.

Thanks.


The difficulty of this problem is as hard as it gets on the GMAT. So, I wouldn't worry about that.
_________________

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.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135204 [0], given: 12673

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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 07 Sep 2014, 08:43
Bunuel wrote:
megatron13 wrote:
Hi Bunuel,

I am having difficulty understanding this approach since there can be combinations in which other people might sit between A, F and E. For this question, I was trying to count all the invalid combinations and subtract from total combinations. But how to count invalid cases like below

A,_,_,F,_E,_ or _,F,_,E,_,_,A


When I say AFE or AEF I don't mean that they are necessarily adjacent, there might be other people between them.




Thanks for the response Bunuel.

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7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 25 Sep 2015, 14:37
Hi Bunuel,


When I apply symmetry here the answer becomes (1/2) * (1/2) * (7!) = 7!/4. What am I doing incorrectly here? Could you please help me out. Thank you.

Regards,
Zizad

Last edited by Zizad on 01 May 2016, 01:06, edited 1 time in total.

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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 29 Sep 2015, 23:17
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In this case, A not left of F (means A right of F) and F not left of E (means F right of E). There is a common element F and both the conditions need to be satisfied. (A right of F and F right of E). So, the concept of symmetry overlooks this common bit and thus different answer.

Out of the three elements(A,F,E). Only one of the arrangement sufficies this aspect so 7!*(1/6)

Hope this helps.
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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 17 May 2016, 20:57
megatron13 wrote:
7 people (A, B, C, D, E, F and G) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater.

How many different arrangements are possible? If A will not sit to the left of F and F will not sit to the left of E. How many different arrangements are possible.

A) 7!/2
B) 7!/3
C) 7!/4
D) 7!/5
E) 7!/6


Hi debbiem,

Refer your PM..

For doing any of these Qs, the MEANING is most important..
what does ---- A will not sit to the left of F and F will not sit to the left of E -- mean.
Keep in mind- it does not talk of immediate left..


so If I have a combination like
D,A,C,F,B,E,G.....
we are concerned about the position of A, F and E.. they have to be in ...E..F..A..
so If we keep the position of all other B,C,D, and G constant these 3- E,F and A can be arranged in 3! ways and ONLY one way ...E..F..A.. is correct

so ONLY 1 out of 3! or 6...OR 1/6 are VALID
Same will be the case for all other arrangements of B,C,D, and G ..
so our answer will be TOTAL *VALID = 7!*1/6 = 7!/6....
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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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New post 18 May 2016, 01:50
Zizad wrote:
Hi Bunuel,


When I apply symmetry here the answer becomes (1/2) * (1/2) * (7!) = 7!/4. What am I doing incorrectly here? Could you please help me out. Thank you.

Regards,
Zizad


Another way to explain why you cannot use symmetry:

A will be to the right of F and F will be to the right of E. So it looks like

E ---- F ---- A

Here, A also has a relation with E - it cannot be to the left of E. Symmetry works when dealing with independent cases.

Check this post for the symmetry discussion:
http://www.veritasprep.com/blog/2011/10 ... s-part-ii/
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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit [#permalink]

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Re: 7 people (A, B, C, D, E, F and G) go to a movie and sit   [#permalink] 15 Oct 2017, 06:36
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