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

Question

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

Bunuel · Accepted Answer

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 
 A E F 
E AF 
 EFA 
 F A E 
 FE A

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).

chetan2u · Answer

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....

Bunuel · Answer

Similar questions to practice: mother-mary-comes-to-me-86407.html (or: mary-and-joe-126407.html); six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html in-how-many-different-ways-can-the-letters-a-a-b-91460.html 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

megatron13 · Answer

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

Bunuel · Answer

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

Zizad · Answer

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

AryamaDuttaSaikia · Answer

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.

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

Hope this helps.

KarishmaB · Answer

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/

einstein801 · Answer

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 · Answer

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.

einstein801 · Answer

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?

Bunuel · Answer

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.

Dooperman · Answer

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!)

Akshat_verma_25 · Answer

Let me shade some light on this extreme beautiful question.

When a condition is only about  who sits to the left or right of whom , we don’t need to place people in exact seats.
 
 Pick the “marked” set (the people in the condition). 
 Among them, there are k! possible relative orders. 
 Each is equally likely. 
 Favorable fraction = valid orders/k!. 
 Multiply by n! total arrangements.  
This is called the  relative order method  (or symmetry in linear arrangements). I am not sure what it is called but I read it in my Engineering course.

Apply to this example
 
 Total = 7!. 
 Marked set = {A,F,E}. 
 Relative orders = 3!=6. 
 Condition: E<F<A. Only 1 of the 6 works. 
 Answer = 1/6×7!=7!/6.

Here are some tiny example around which you spindle your head. 
 
  8 people, A must sit to the left of B. 
  
  9 people, A must sit to the left of both B and C. 
  
  7 people, A between B and C. 
 
  6 people, A left of B and C left of D (independent).
   
Answers
1. Answer = 8!/2
2. (A is first among the three). Valid orders = 2/6.
Answer = 9!/3.
3. (A in middle → BAC or CAB). Valid orders = 2/6.
Answer = 7!/37!/37!/3.
4. Each pair has 1/2 chance, so total fraction = 1/4.
Answer = 6!/46!/46!/4.

bb   Bunuel   KarishmaB   chetan2u  Please check my solution — if I’ve messed up, feel free to roast me gently. And I’m desperate for kudos, so don’t be stingy... they’re calorie-free anyway!”

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards