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.

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]

Sorry for replying late..
OA for this is C 1440

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

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 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?

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 )

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)

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