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On Friday, four people sat one behind the other in a movie theater.
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02 May 2024, 01:38
02 May 2024, 01:38
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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
Quant Chat Moderator
Joined: 22 Dec 2016
Posts: 3134
Given Kudos: 1856
Location: India
Concentration: Strategy, Leadership
Re: On Friday, four people sat one behind the other in a movie theater.
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02 May 2024, 03:46
02 May 2024, 03:46
1
Kudos
Bunuel wrote:
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
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
Re: On Friday, four people sat one behind the other in a movie theater.
[#permalink]
02 May 2024, 05:33
02 May 2024, 05:33
3
Kudos
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
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
