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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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megatron13
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.
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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
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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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megatron13
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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Zizad
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:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/10 ... s-part-ii/
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Understood for the range of ways for AFE, only 1 out of 6 options work.

How does this translate to the answer though? I saw that you said later in the posts that you have also included e.g. A _ _ F _ E. How does that work?
Bunuel
megatron13
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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unicornilove
Understood for the range of ways for AFE, only 1 out of 6 options work.

How does this translate to the answer though? I saw that you said later in the posts that you have also included e.g. A _ _ F _ E. How does that work?
Bunuel
megatron13
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).
­
­
I'll try once more. Three people, A, F, and E, can be arranged in 3! = 6 ways. Out of these 6 arrangements, only one, EFA, satisfies the restriction, making it 1/6. Thus, out of the total 7! arrangements, only one-sixth of them satisfies the restriction.
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Hi Bunuel, thanks for the explanation. I understand that. However, aren't you missing the cases where they fit the requirement but not directly next to each other?
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Hi Bunuel, thanks for the explanation. I understand that. However, aren't you missing the cases where they fit the requirement but not directly next to each other?

­No, we haven't overlooked any cases there. Each of the six arrangements of A, F, and E is relative to each other. However, within each arrangement, there might be additional people positioned among them.
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7 seats.

For any 3 seats E, F, A can sit in only 1 way.

Out of 7 seats, choose 3 seats and arrange E, F, A.
Then arrange B,C,D,G on other 4 seats.

(7C3 * 1) * (4!)
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