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Tough GMAT Probability Question ?

Expert replies
by Leo2008 » Tue Mar 25, 2008 1:57 pm
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.
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Source: — Problem Solving |

by ikant » Tue Mar 25, 2008 9:02 pm
The answer is B.

The logic is that we have 2 rows and hence we need to accomodate 6 students per row.

So permutaions of 6 students for first row = 12P6= 12! / 6!

Now the permutaions for back row = 6!.

Now we have two sets of paers :: A and B. Hence we will have two distinct arrangements depending on which paper is handed out to the first student. So the total no of arrangement become = 2 * 12P6 * 6!

If we rearrange the answer => 2 * 12! / 6! * 6!.
=> 2 * 12C6 * 6! * 6!
"To do is to be"
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by Stockmoose16 » Tue Oct 07, 2008 10:43 pm
ikant wrote:The answer is B.

The logic is that we have 2 rows and hence we need to accomodate 6 students per row.

So permutaions of 6 students for first row = 12P6= 12! / 6!

Now the permutaions for back row = 6!.

Now we have two sets of paers :: A and B. Hence we will have two distinct arrangements depending on which paper is handed out to the first student. So the total no of arrangement become = 2 * 12P6 * 6!

If we rearrange the answer => 2 * 12! / 6! * 6!.
=> 2 * 12C6 * 6! * 6!
I disagree with your answer, I think the correct answer is A, not B. Here's a diagram to explain my logic:
#1 #2 #1 #2 #1 #2
12 * 6 * 4 * 4 * 2 *2

5 * 5 *3 * 3 * 1 *1

Starting at the top left, any of the 12 students can sit in that seat. The person directly below him must have the same test, which means there are 5 students who could fill that spot. On to the second column, there are 6 students left who can receive the second test, then right below that choice, 5 (because the person in the same column, must get the same test). And so on...


if you split 12 into 6 *2, the answer comes out 6! * 6! *2


Can anyone confirm my logic to be correct?
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Re: Tough GMAT Probability Question ?

by Ian Stewart » Wed Oct 08, 2008 11:56 am
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.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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