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What is the maximum number of arrangements in which N students can be 05 Feb 2015, 09:10
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What is the maximum number of arrangements in which N students can be seated in a row of N seats at a movie theater, if all students from the same college are to sit next to each other?

(1) All students come from three colleges, X, Y, and Z that sent 12, 10, and 9 students, respectively.
(2) N is a prime number between 30 and 40.


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What is the maximum number of arrangements in which N students can be Updated on: 09 Feb 2015, 05:04
Originally posted by gmat6nplus1 on 05 Feb 2015, 09:44.
Last edited by gmat6nplus1 on 09 Feb 2015, 05:04, edited 1 time in total.
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Bunuel
What is the maximum number of arrangements in which N students can be seated in a row of N seats at a movie theater, if all students from the same college are to sit next to each other?

(1) All students come from three colleges, X, Y, and Z that sent 12, 10, and 9 students, respectively.
(2) N is a prime number between 30 and 40.


Kudos for a correct solution.
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Statement 1: this statement is sufficient since we know the number of sits and the number of students in each college.

Statement 2: clearly insufficient.

Answer A.
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What is the maximum number of arrangements in which N students can be 09 Feb 2015, 01:03
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Bunuel
What is the maximum number of arrangements in which N students can be seated in a row of N seats at a movie theater, if all students from the same college are to sit next to each other?

(1) All students come from three colleges, X, Y, and Z that sent 12, 10, and 9 students, respectively.
(2) N is a prime number between 30 and 40.


Kudos for a correct solution.

Statement 1: assuming that the total number of chairs N is greater than the total number of students, this statement would be sufficient.

Statement 2: clearly insufficient.

Answer A.
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Why do we need to assume that? Doesn't the question say that the number of seats =number of students?
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What is the maximum number of arrangements in which N students can be 09 Feb 2015, 04:34
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Great. Even though I don't like combinations and permutations, in DS questions they are fine, as I know what I need to find the answer, I just don't always know how to find it.

In this sense, if this was a PS questions, and we were asked to find the number of possible seating arrangements would the answer b3 tht there are 325 possible arrangements?

We know that colleges X, Y, and Z sent 12, 10 and 9 students. We also know that the # of available seats are 31 and that every student from the same college should sit next to each other.

This would make the response 12*12 + 10*10 + 9*9 = 325?

Thank you.
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What is the maximum number of arrangements in which N students can be 09 Feb 2015, 05:35
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What is the maximum number of arrangements in which N students can be seated in a row of N seats at a movie theater, if all students from the same college are to sit next to each other?

(1) All students come from three colleges, X, Y, and Z that sent 12, 10, and 9 students, respectively.
(2) N is a prime number between 30 and 40.


Kudos for a correct solution.
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VERITAS PREP OFFICIAL SOLUTION:

Solution: A. While this problem is technically a combinatorics problem, it showcases one of the great things about Data Sufficiency - if you know that you will get an answer, you don't need to finish the math. Statement 1 tells you exactly what N is: it's 12 + 10 + 9 (which is 31). So the answer is 31!, which is a prohibitively large number but you know that it is one exact number, so statement 1 is sufficient.

Statement 2 is not sufficient, as there are two prime numbers between 30 and 40 (31 and 37).

Accordingly, the correct answer is A.
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What is the maximum number of arrangements in which N students can be 23 Mar 2015, 16:45
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Bunuel
Statement 1 tells you exactly what N is: it's 12 + 10 + 9 (which is 31). So the answer is 31!, which is a prohibitively large number but you know that it is one exact number, so statement 1 is sufficient.
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Dear Bunuel, I agree that Ans is A.

But aren't the total number of arrangements = (12!)(10!)(9!)(3!) ?

Wouldn't solution = (31!) if we did not have the restriction of seating all students from same college together?


Thank you.
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What is the maximum number of arrangements in which N students can be 24 Mar 2015, 04:03
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Bunuel
Statement 1 tells you exactly what N is: it's 12 + 10 + 9 (which is 31). So the answer is 31!, which is a prohibitively large number but you know that it is one exact number, so statement 1 is sufficient.

Dear Bunuel, I agree that Ans is A.

But aren't the total number of arrangements = (12!)(10!)(9!)(3!) ?

Wouldn't solution = (31!) if we did not have the restriction of seating all students from same college together?


Thank you.
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Yes, that's correct. Veritas Prep solution is not correct. I'll contact them on this question.
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What is the maximum number of arrangements in which N students can be 24 Mar 2015, 04:49
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What is the maximum number of arrangements in which N students can be seated in a row of N seats at a movie theater, if all students from the same college are to sit next to each other?

(1) All students come from three colleges, X, Y, and Z that sent 12, 10, and 9 students, respectively.
(2) N is a prime number between 30 and 40.


Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Solution: A. While this problem is technically a combinatorics problem, it showcases one of the great things about Data Sufficiency - if you know that you will get an answer, you don't need to finish the math. Statement 1 tells you exactly what N is: it's 12 + 10 + 9 (which is 31). So the answer is 31!, which is a prohibitively large number but you know that it is one exact number, so statement 1 is sufficient.

Statement 2 is not sufficient, as there are two prime numbers between 30 and 40 (31 and 37).

Accordingly, the correct answer is A.
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Sorry i have few doubts .
Firstly, i could not understand how 31! will be total number of arrangements with the given restriction 'if all students from the same college are to sit next to each other'
Secondly, a typical row in movie theater has N seats which can accommodate N Students , now it is not said in question that a typical row of a movie theater is designed such a way that i will accommodate all students of X,Y and Z college. N could be a large number 100.

Veritas Kindly suggest.
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What is the maximum number of arrangements in which N students can be 30 Mar 2015, 21:01
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Bunuel
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What is the maximum number of arrangements in which N students can be seated in a row of N seats at a movie theater, if all students from the same college are to sit next to each other?

(1) All students come from three colleges, X, Y, and Z that sent 12, 10, and 9 students, respectively.
(2) N is a prime number between 30 and 40.


Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Solution: A. While this problem is technically a combinatorics problem, it showcases one of the great things about Data Sufficiency - if you know that you will get an answer, you don't need to finish the math. Statement 1 tells you exactly what N is: it's 12 + 10 + 9 (which is 31). So the answer is 31!, which is a prohibitively large number but you know that it is one exact number, so statement 1 is sufficient.

Statement 2 is not sufficient, as there are two prime numbers between 30 and 40 (31 and 37).

Accordingly, the correct answer is A.

Sorry i have few doubts .
Firstly, i could not understand how 31! will be total number of arrangements with the given restriction 'if all students from the same college are to sit next to each other'
Secondly, a typical row in movie theater has N seats which can accommodate N Students , now it is not said in question that a typical row of a movie theater is designed such a way that i will accommodate all students of X,Y and Z college. N could be a large number 100.

Veritas Kindly suggest.
Show more

Yes, the actual number of arrangements will be (12!)*(10!)*9!*3! and not 31!.

The question says "N students can be seated in a row of N seats at a movie theater"...
Statement 1 gives us that the number of students is 31 so we know that N is 31. This means that you need to make 31 students sit in a row of 31 seats.
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