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Permutation & Comb.

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by priyankamishra11 » Tue Aug 19, 2008 6:56 am
Can any one solve this?

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?

360
720
1,440
5,040
10,080
Regards,
Priyanka
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by California4jx » Tue Aug 19, 2008 7:36 am
here is my attempt -

1 out 7 seats is reserved for one of the friend left with 6 seats
1 out of 6 friends wants to sit in the aisle left with 5 friends.

so the permutation will take place for 6 seats among 5 friends.

6x5x4x3x2 = 720

B is the answer.
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by California4jx » Tue Aug 19, 2008 7:43 am
just missed one portion of the problem

-[the rows are 7 seats across with aisles on either side]

which means there are total 14 seats and 2 seats on aisle -

5 friends can be seated on either side of the aisle with remaining 6 seats. As before, on one side, there are 720 possibilities for 5 of those friends. So, in total, there are 720 x 2 = 1440

C is the right answer. Well, let me know what the right answer, i might not have given a good explanation.[/quote]
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by priyankamishra11 » Sat Sep 06, 2008 8:54 am
Sorry for replying late..
OA for this is C 1440
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Priyanka
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by 4meonly » Sat Sep 06, 2008 10:28 am
California4jx wrote: -[the rows are 7 seats across with aisles on either side]
which means there are total 14 seats and 2 seats on aisle -
Can it be that there are 3 rows with 7 seats? 4 rows with 7 seats?
I am not English-native speaker but for me this question is not accurate
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by Stuart@KaplanGMAT » Sat Sep 06, 2008 11:03 am
California4jx wrote:here is my attempt -

1 out 7 seats is reserved for one of the friend left with 6 seats
1 out of 6 friends wants to sit in the aisle left with 5 friends.

so the permutation will take place for 6 seats among 5 friends.

6x5x4x3x2 = 720

B is the answer.
Close!

The only thing you forgot to take into account is that the "aisle" person can sit in either aisle seat, so you need to multiply by 2.

So the correct answer is 2*6! = 1440

(There's only one row - we can't interpret "rows" as "exactly 2 rows" - that's just describing the overall theatre. The friends find "an empty row" and it's at that point that we start to calcluate the different permutations. There is no "rows". There is no spoon! 8) )
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by Stockmoose16 » Wed Sep 10, 2008 10:13 am
Stuart Kovinsky wrote:
California4jx wrote:here is my attempt -

1 out 7 seats is reserved for one of the friend left with 6 seats
1 out of 6 friends wants to sit in the aisle left with 5 friends.

so the permutation will take place for 6 seats among 5 friends.

6x5x4x3x2 = 720

B is the answer.
Close!

The only thing you forgot to take into account is that the "aisle" person can sit in either aisle seat, so you need to multiply by 2.

So the correct answer is 2*6! = 1440

(There's only one row - we can't interpret "rows" as "exactly 2 rows" - that's just describing the overall theatre. The friends find "an empty row" and it's at that point that we start to calcluate the different permutations. There is no "rows". There is no spoon! 8) )
Stuart,

Why wouldn't it be:

2P1 (2 end seats, 1 person picking)

+

6P5
(6 seats left [after first guy picks], 5 people left to select from those 6 seats)

Answer: 6*2 = 30

What's wrong with this logic?
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by Stuart@KaplanGMAT » Wed Sep 10, 2008 10:49 am
Stockmoose16 wrote: Why wouldn't it be:

2P1 (2 end seats, 1 person picking)

+

6P5
(6 seats left [after first guy picks], 5 people left to select from those 6 seats)

Answer: 6*2 = 30

What's wrong with this logic?
Mostly becasue 6P5 isn't 6

(and as an aside, 6*2 doesn't equal 30 :o )

nPk = n!/(n-k)!

6P5 = 6!/(6-5)! = 6!/1! = 6*5*4*3*2 = 720

So, if you use 720 instead of 6 you get:

720*2 = 1440

which is just fine 8)
Last edited by Stuart@KaplanGMAT on Wed Sep 10, 2008 10:51 am, edited 1 time in total.
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by Stockmoose16 » Wed Sep 10, 2008 10:51 am
Stuart Kovinsky wrote:
Stockmoose16 wrote: Why wouldn't it be:

2P1 (2 end seats, 1 person picking)

+

6P5
(6 seats left [after first guy picks], 5 people left to select from those 6 seats)

Answer: 6*2 = 30

What's wrong with this logic?
Mostly becasue 6P5 isn't 6

nPk = n!/(n-k)!

6P5 = 6!/(6-5)! = 6!/1! = 6*5*4*3*2 = 720

So, if you use 720 instead of 6 you get:

720*2 = 1440

which is just fine 8)
Sometimes I make stupid errors :)
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