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In a certain movie theater, the rows are 7 seats across with aisles on

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Bunuel
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In a certain movie theater, the rows are 7 seats across with aisles on [#permalink] New post   17 Jul 2023, 01:55
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In a certain movie theater, the rows are 7 seats across with aisles on either side. A group of 6 friends goes to the movies and finds an empty row to sit in. One of them needs to sit in a seat along the aisle. Which of the following calculations represents the number of possible different seating arrangements for the 6 friends?

A. 360
B. 720
C. 1,440
D. 5,040
E. 10,080
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In a certain movie theater, the rows are 7 seats across with aisles on [#permalink] New post   19 Jul 2023, 13:41
Bunuel wrote:
In a certain movie theater, the rows are 7 seats across with aisles on either side. A group of 6 friends goes to the movies and finds an empty row to sit in. One of them needs to sit in a seat along the aisle. Which of the following calculations represents the number of possible different seating arrangements for the 6 friends?

A. 360
B. 720
C. 1,440
D. 5,040
E. 10,080


\(2_C_1\)*\(6_C_5 \)* \(5!\)\(= 2*6*120 =1440\)

\(2_C_1= \)Choosing one seat from the two aisle seats for that one person.
\(6_C_5=\) Choosing \(5\) seats from the remaining \(6\) seats
\(5! =\) # of ways to arrange those five seats.

Hope it helps.
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Re: In a certain movie theater, the rows are 7 seats across with aisles on [#permalink] New post   19 Jul 2023, 15:21
First, we must select which side the person sitting along the aisle is on. There are 2 sides, so there are 2 options.

Next, of the remaining 6 seats, we need to choose 5 for the seats that will be filled. 6C5 = 6!/5! = 6 total ways.

Within those 5 seats, the total number of ways to arrange the 5 friends = 5! = 5*4*3*2*1 = 120.

So, the total number of arrangements is 2*6*120 = 1,440. Answer C.
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Re: In a certain movie theater, the rows are 7 seats across with aisles on [#permalink] New post   19 Jul 2023, 19:20
ninjaintraining wrote:
First, we must select which side the person sitting along the aisle is on. There are 2 sides, so there are 2 options.

Next, of the remaining 6 seats, we need to choose 5 for the seats that will be filled. 6C5 = 6!/5! = 6 total ways.

Within those 5 seats, the total number of ways to arrange the 5 friends = 5! = 5*4*3*2*1 = 120.

So, the total number of arrangements is 2*6*120 = 1,440. Answer C.



Since it is a seating arrangement where order matters, shouldn't it be permutation?
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Re: In a certain movie theater, the rows are 7 seats across with aisles on [#permalink] New post   19 Jul 2023, 19:27
sapphirenm wrote:
ninjaintraining wrote:
First, we must select which side the person sitting along the aisle is on. There are 2 sides, so there are 2 options.

Next, of the remaining 6 seats, we need to choose 5 for the seats that will be filled. 6C5 = 6!/5! = 6 total ways.

Within those 5 seats, the total number of ways to arrange the 5 friends = 5! = 5*4*3*2*1 = 120.

So, the total number of arrangements is 2*6*120 = 1,440. Answer C.



Since it is a seating arrangement where order matters, shouldn't it be permutation?


Yes, you are right that order matters. Once we choose the 5 seats that the friends are sitting, then we do 5! to find all the possible permutations with the 5 friends sitting in those 5 seats. Hope this makes sense.
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Re: In a certain movie theater, the rows are 7 seats across with aisles on [#permalink] New post   19 Jul 2023, 19:35
ninjaintraining wrote:
sapphirenm wrote:
ninjaintraining wrote:
First, we must select which side the person sitting along the aisle is on. There are 2 sides, so there are 2 options.

Next, of the remaining 6 seats, we need to choose 5 for the seats that will be filled. 6C5 = 6!/5! = 6 total ways.

Within those 5 seats, the total number of ways to arrange the 5 friends = 5! = 5*4*3*2*1 = 120.

So, the total number of arrangements is 2*6*120 = 1,440. Answer C.



Since it is a seating arrangement where order matters, shouldn't it be permutation?


Yes, you are right that order matters. Once we choose the 5 seats that the friends are sitting, then we do 5! to find all the possible permutations with the 5 friends sitting in those 5 seats. Hope this makes sense.



That makes sense. I solved it in a different way.

1 person - one side Aisle
To seat 5 people in the remaining 6 seats - 6P5 = 720 ways

Total = 2x720 = 1440 ways
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Re: In a certain movie theater, the rows are 7 seats across with aisles on [#permalink]
19 Jul 2023, 19:35
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