A group consisting of N couples are going to see a movie.

Question

A group consisting of N couples are going to see a movie. The seats in each row of the theater is greater than 2N. If the group decides to all sit in the same row, each couple is indifferent to empty seats next to them, and each couple insists on sitting together, how many seating arrangements are possible?

(1) N = 5

(2) The group will all sit next to one another, starting with the first seat in the row.

(1) N = 5

(2) The group will all sit next to one another, starting with the first seat in the row.

gmat1220 · Answer

First I thought s1 is sufficient but on taking a closer look the statement said greater than 2N seats. Hence C is sufficient. You have only three answers- a,c or worst case e. If this question turns out to be your 31-37 question take a moment to think why answer may not be E.

If the seats would have been Exactly 2N the answer is A. Please correct if the reasoning has flaw

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If the seats would have been Exactly 2N the answer is A. Please correct if the reasoning has flaw

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bhandariavi · Answer

Seats in each row greater than 2N brings possibility to increase no be seats in a row up to 100 to 1000 to millions. However N could be 2, 4 , 6.
There could be 1 or 50 or 100 or .. seats gap between each couple.Any number of arrangement is possible. 
Please correct me !

fluke · Answer

You are right; if the seats were exactly 2N, the arrangements would be = 5!*(2)^5 and "A" would have sufficed. St2 tells us that they are all seating together without any gaps starting with the FIRST seat, which now means that they are indeed occupying first 10 seats of the row and the rest of the seats become immaterial and makes "C" sufficient.

Ans: "C"

mission2009 · Answer

@bhandariavi: Can you tell me the source of this question???

subhashghosh · Answer

@fluke - "if the seats were exactly 2N, the arrangements would be = 5!*(2)^5 "

Is it something like - 5 couples in 5 pairs of seats = 5!

and then 2 persons in each of the 5 couples can be arranged among themselves as 2!

So total 5! * (2!)^5

= 5! * (2)^5

fluke · Answer

Precisely!!!

If you adhere all the couples and make them one unit each; there are 5 units;

5 units can be arranged in 5! ways

And within the unit; every couple can rearrange in 2! ways.

12bhang · Answer

Please help.

If the seats are more than 2N, there is no upper limit.

So, if we use the counting principle, depending on the value of 2N, the number of options available to the first couple could be 100, 1000 or whatever. SUbesquent couples would be limited by where the first couple chose to sit.

In this case, shouldn.t the answer be E?

Bunuel · Answer

Take a closer look at the second statement: The group will  all sit next to one another ,  starting with the first seat in the row .

mvictor · Answer

1 alone is insufficient..the 5 couples can be arranged in 5!*2!^5 ways (2! because the 2 from the couples can be arranged in 2! ways, and ^5 because 5 pairs). but we are told that there are >2N seats..so this makes us additional problems..what if there are 11 seats? then the total number of ways would be 11!/5!2!...so different numbers...

2 says that the couple will sit starting the first seat from the row..meaning that if there are 11 seats, and there are 5 couples, the couples will sit in the first 10...so the last one is irrelevant, and does not need to be taken into account. this alone is insufficient.

1+2 -> we know that there are 5 couples, and that they will take the first seats in the row..so no additional possibilities...total 5!*2!^2

manhasnoname · Answer

"Each couple is indifferent to empty seats next to them" --> doesn't this mean that we need not worry about empty spaces between the couples? In which case, (A) should be the answer.

hadimadi · Answer

Hi,

(1) From N=5 we know that there area more than 10 seats in that row. It could also be a billion or just 11. So we can't name a fixed number for the requested seating arrangement

(2) We don't know how big the group is

(1) and (2): We can completely ignore the part which says "they are indifferent between empty seats" due to statement (2). So the question just becomes how many ways can we put 5 couples next to each other. We can start with 5!, as that is the maximum amount of ways to put them next to each other as couples, but then each couple could switch positions. Since there are 5 couples, we get 5!*2^5

pudu · Answer

1. if n=5 then the total number of seats is more than 10. if it is 11 then 11P5 and if it is 15 then 15P5 etc. Hence insufficient.

2.no information about n

1 & 2 means the 5 couples are sitting in the first 10 seats. So 5!*2! is possible because one couple can be arranged in 2 ways. Hence C

bumpbot · Answer

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A group consisting of N couples are going to see a movie.

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