On Friday, four people sat one behind the other in a movie theater.

Question

On Friday, four people sat one behind the other in a movie theater. How many ways are there to rearrange the order of the people for Saturday, so that no person is directly behind someone they were directly behind on Friday?

A. 8
B. 9
C. 10
D. 11
E. 12

A. 8
B. 9
C. 10
D. 11
E. 12

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KarishmaB · Accepted Answer

One can use Sets here. Friday's arrangement: A B C D We do not want AB, BC and CD in that order. No of ways of arranging 4 people = 4! Number of ways of arranging such that AB are together = 3! = 6 (same for BC and same for CD) There will be overlaps. When AB is together, BC is together too i.e. ABC is together in 2! cases = 2 cases (Two sets overlap) Same is true for BCD (2 sets overlap) and AB & CD (2 sets overlap) In 1 case (Friday's case) all AB, BC and CD are together (all three sets overlap in 1 case). Hence, this is how the overlap looks like: Screenshot 2024-05-03 at 11.14.55 AM.png Since AB and BC are together in 2 cases and in 1 case all 3 overlap, only AB and BC are together in 1 case. Similarly for others too. Hence, number of acceptable cases with none of these = 24 - 13 = 11 Answer (D) *Added some details for clarity.

8Harshitsharma · Answer

Since the answer choices are near the number 10 and no straightforward method comes to mind within 30 seconds, we can go with manually counting favorable cases.

Assume the order of people on Friday from 1st seat to the last seat was A B C D, where A,B,C,D are the people.

So to count total seatings on Saturday, let's start with arrangements starting from A

A C B D
A D C B

Total = 2
Now arrangements starting with B,

B A D C
B D A C
B D C A

Total = 3
Arrangements starting with C,

C A D B
C B A D
C B D A

Total = 3

Arrangments starting with D,

D A C B
D B A C
D C B A

Total = 3

So total count = 2+3+3+3 = 11.  Choice D

gmatophobia · Answer

This is a tricky one ...

Let's assume that A, B, C, and D have the following seating arrangement on Friday

A
B
C
D

B sit behind A, C sits behind B, and D sits behind C.

On Saturday
 
 B cannot sit behind A 
 C cannot sit behind B 
 D cannot sit behind C

All possible arrangements : The four people can sit in 4 ! ways, or 24 ways

Case 1 : Let's assume B sits behind A

A
B

A & B can be considered a single unit, and we can have 3!, or 6 ways in which the four people can sit.

Case 2 : Let's assume C sits behind B

B
C

B & C can be considered a single unit, and we can have 3!, or 6 ways in which the four people can sit.

Case 3 : Let's assume D sits behind C

C
D

D & C can be considered a single unit, and we can have 3!, or 6 ways in which the four people can sit.

Hence, of the 24 possible arrangements, 6 * 3 = 18 are invalid.

However, we have double counted few cases which we need to add back

Cases to add back -

A. When B sits behind A and C sits behind B

A
B
C

We have counted this situation in case 1 and case 2.

There are two ways the four friends can sit.

D
A
B
C

or

A
B
C
D

B. When C sits behind B and D sits behind C

B
C
D

We have counted this situation in case 2 and case 3.

There are two ways the four friends can sit.

A
B
C
D

or

B
C
D
A

C. When B sits behind A and C sits behind D

There are two ways the four friends can sit.

A
B
C
D

or

C
D
A
B

As the case, A B C D is common across all three cases, we need to retain one of them (else we would be ignoring that case altogether)

24 - 18 + 6 - 1 = 24 - 13 = 11

Option D

Nidzo · Answer

Let the four friends each be represented by A, B, C and D.
Let their Friday night seating arrangement be:
 A
B
C
D

From here we see that there are three pairs of people where one of the people has someone sitting directly behind them:  AB, BC and CD .

The approach will be to find the total arrangements possible and subtract from that the number of ways in which someone has the same person seated behind them.

Total arrangements possible:    4! = 24

AB:   As we are looking for the number of ways in which A is seated behind B, we can group them together as a single entity, with C and D left independent. As we are now dealing with essentially 3 slots which we are filling, the number of ways in which A will be behind B is:  3! = 6 .

BC:   Doing the same with BC as we did with AB, once again we will have  3! = 6  ways. However, we need to consider the fact that we have also included arrangements where A is before B which we counted for in the previous section. For A to come before B when we have grouped BC together, it can occur only when BC is in a slot which has a slot available above it. This can only occur when BC is in the middle of the three slots or in the very last slot. For this reason we subtract 2 from 6 to get  4  unique ways.

CD:   As previously, we group CD together and will find  3! = 6  ways of arranging it with an independent A and B.

We now have to subtract the arrangements which have already been counted in the two previous cases.

When CD are together in a slot, AB will only occur when CD is in the first slot or the last slot.
When CD are together in a slot, BC will only occur when B is ahead. We have already counted one case while considering when CD and AB occur. Another occurance is when B is first. 
This leaves us with  6 - 3 = 3  unique cases.

How many ways are there to rearrange the order of the people for Saturday, so that no person is directly behind someone they were directly behind on Friday?

24 - (6+4+3)  
 24 - 13 = 11

ANSWER D

sayan640 · Answer

KarishmaB  Any quick method , considering we need to get the answer in 2/3 mins ?

Posted from my mobile device

sayan640 · Answer

KarishmaB  , Thank you so much , maa'm for responding. No thanks is enough for the way you are helping us. 
However , I have got few doubts here(in the highlighted portions).
Don't you think the "Number of ways of arranging such that of AB (or BC or CD) together = 3! *2! because (AB) , C , D ; For each of these three units,  A and B can arrange themselves in 2! ways. Hence not just 3! ( as mentioned by you in the solution) , it should be 3!*2! , I think.  Please correct me if I am wrong.  Is it because the order does not matter in this case hence you are writing only 3! and not 3!*2! ?

Also , I dint understand completely  the way you arranged overlapped portion of 1 s ( highlighted in your solution ).
Can you help a bit more ?

sayan640 · Answer

KarishmaB  I think I got it now. Please correct me if I am wrong.
Number of ways of arranging such that  AB (or BC or CD)  are together = 3! only because (AB) , C , D and we are concerned about only this arrangement and not arrangements such as (BA) , C , D and hence dint multiply 3! with 2! for including the ways A and B can arrange between themselves. Hence only one order of A and B matters. And that's why 3! only.  Bunuel  , Am I correct in my understanding ? Can we have  links of questions of this type so that we can practise and master ?

Adarsh_24 · Answer

How many ways to rearrange-> that means from a particular source how many different combos?
Max rearrange is when all changes -> 4!=24

Let's see how many are valid.

It may be too complex to calculate which are possible for not directly behind.
Better to find 
total - directly behind

Let's say ABCD last Friday.
then directly behind similar to last Friday will be
AB
BC
CD

AB together
Combinations=3! (no need for AB & BA because only 1 way is directly ehind)
(AB)CD
C(AB)D
CD(AB)
(AB)DC
D(AB)C
DC(AB)

BC
(BC)AD
(BC)DA
A(BC)D
D(BC)A
AD(BC)
DA(BC)

CD
(CD)AB
(CD)BA
A(CD)B
B(CD)A
AB(CD)
BA(CD)

3!*3=18 (3 for AB, BC,CD)

So 24-18+2(ABCD)+3(CDAB,DABC,BCDA)
We have ABCD 3 times above. In reality we should have deleted only 1 since one item. So to compensate we add 2 back.
We have CDAB, DABC, BCDA each twice. So we deleted 6. But, since they are only 3. We add 3 back.
Ans 24-18+2+3=11

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