Leo2008 wrote:Two variants of a test paper are distributed among 12 students. How many ways are there of seating the students in two rows so that students sitting side by side do not have identical papers and those sitting in the same column have the same paper ?
A) 2*(6!)*(6!)
B) (12C6)*2*(6!)*(6!)
My suggestion to this question is to do it with 4 or 6 students and then use the same logic for 12 students.
The wording of this question is not particularly good. I've seen it posted here before, both times with only two answer choices (which is odd) so I'm wondering where it's from.
First, you might notice that answer choice B is exactly equal to 2*12!. I'm guessing that this is the intended answer, but that's because I'm guessing the question actually means to ask 'in how many ways could the students be seated *and* the tests distributed?'. The way the question is worded, the answer should really just be 12!; you can seat 12 people in 12! ways, and it makes no difference how you distribute the tests. If we also need to count how many ways to distribute the tests, then once you've seated the students (12! ways), we need to multiply by 2, because students in the first column could either get Test A or Test B. Not a well-worded question, however, and really not worth worrying about.
