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P&C Yet Again!!

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by parallel_chase » Wed Aug 20, 2008 5:30 am
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!)

Since there are only two options available, I am more interested in the method as compared to the answer.
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Re: P&C Yet Again!!

by sudhir3127 » Wed Aug 20, 2008 5:44 am
parallel_chase 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!)

Since there are only two options available, I am more interested in the method as compared to the answer.
I go with B

heres how i did it...

we have 2 rows.. thus we can arrange 6 students per row.

Arranging 6 students in first row is 12P6
arranging 6 students in the next row is 6!


We have 2 sets of question papers. Hence total no of arrangement become = 2 * 12P6 * 6! ( because we have 2 arrangements on which set is to be handed out to the first student )

thuus its 2*12P6*6!

which can be written as (12C6)*2*(6!)*(6!)

Hope it helps..
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by pepeprepa » Wed Aug 20, 2008 5:53 am
We disagree on this one... I am for A)

Here is my logic:
We have two possibilities of organizing the students.
XYXYXY
XYXYXY
or
YXYXYX
YXYXYX

for the X we have 6! possible combinations and for the Y we have the same so 6!*6!

Given we have 2 possibilities: 2*6!*6!
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by parallel_chase » Wed Aug 20, 2008 6:55 am
Thanks for your inputs.

The answer is B. I think the way Sudhir did is the right way but definitely not the fastest.

Pepeprepa I did exactly the way you did initially but A is incorrect.

I request Ian/Stuart kindly give some insights........
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by gabriel » Wed Aug 20, 2008 10:24 am
pepeprepa wrote:We disagree on this one... I am for A)

Here is my logic:
We have two possibilities of organizing the students.
XYXYXY
XYXYXY
or
YXYXYX
YXYXYX

for the X we have 6! possible combinations and for the Y we have the same so 6!*6!

Given we have 2 possibilities: 2*6!*6!
Well this just takes care of the sitting arrangement. You have to still distribute the papers and that is where you select 6 student out of 12 = 12c6 and the rest 6 gets the other set.
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by pepeprepa » Wed Aug 20, 2008 10:40 am
Thank you for this precision, I missed this point... indeed what bothered me is that I couldn't totally represent me the situation and I don't know why, I will do it again later.
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by nervesofsteel » Wed Aug 20, 2008 5:26 pm
first select 6 students from 12 who needs to be seated in first column..
it can be done in 12C6 ways.. Now arrange them in 6! ways..

Now the remaining 6 students can be seated in 6! ways in second column

But each column can have first set or second set
so there are 2 ways papers can be distributed.

so answer is 2*12C6* 6!*6!
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