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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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05 Sep 2014, 13:02

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Difficulty:

95% (hard)

Question Stats:

30% (01:33) correct
70% (01:06) wrong based on 227 sessions

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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.

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.

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.

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

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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

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.
_________________

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

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07 Sep 2014, 08:20

1

This post was 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.

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.
_________________

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

Show Tags

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.

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)

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.

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....
_________________

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

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15 Oct 2017, 06:36

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