Bunuel wrote:
Bunuel wrote:
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.