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# If ten students appear in an examination and 4 of them are

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Director
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If ten students appear in an examination and 4 of them are [#permalink]  06 Jan 2004, 15:14
If ten students appear in an examination and 4 of them are appearing for mathematics and rest for 6 different subjects, in how many ways can they be seated in a row so that no copying is possible?
Director
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Re: Exam copying [#permalink]  06 Jan 2004, 15:30
If ten students appear in an examination and 4 of them are appearing for mathematics and rest for 6 different subjects, in how many ways can they be seated in a row so that no copying is possible?

If it were a computer adaptive test then it would be 10!

If it were paper and pencil it would be 6!x4!
Senior Manager
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[#permalink]  06 Jan 2004, 15:33
I feel the answer should be 6! * 4! * 4
Director
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Re: Exam copying [#permalink]  06 Jan 2004, 15:57
If ten students appear in an examination and 4 of them are appearing for mathematics and rest for 6 different subjects, in how many ways can they be seated in a row so that no copying is possible?

I'm a little confused now, there are only 5 spaces between 6 students. I'm not in my right mind today - stressful day in the markets for me.
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Re: Exam copying [#permalink]  06 Jan 2004, 16:05
Titleist wrote:
If ten students appear in an examination and 4 of them are appearing for mathematics and rest for 6 different subjects, in how many ways can they be seated in a row so that no copying is possible?

I'm a little confused now, there are only 5 spaces between 6 students. I'm not in my right mind today - stressful day in the markets for me.

oops - misread the question
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[#permalink]  06 Jan 2004, 16:29
I agree with Geethu - I get 6!*4!*4
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[#permalink]  06 Jan 2004, 21:36
Explanation Titleist & Geethu for ur answers

Vivek
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"

Senior Manager
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[#permalink]  07 Jan 2004, 08:04
Lets assume S as six different subject students and M as Math students. There are four different ways to arrange them.

M S M S M S M S S S

S M S M S M S M S S

S S M S M S M S M S

S S S M S M S M S M

S can be arranged in 6! ways and
M can be arranged in 4! ways

So, the answer should be 6! X 4! X 4
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[#permalink]  07 Jan 2004, 08:52
Geethu wrote:
Lets assume S as six different subject students and M as Math students. There are four different ways to arrange them.

M S M S M S M S S S

S M S M S M S M S S

S S M S M S M S M S

S S S M S M S M S M

S can be arranged in 6! ways and
M can be arranged in 4! ways

So, the answer should be 6! X 4! X 4

After thinkin about this question more last night, I'm not quite sure if that really is the answer.

You can also have more staggered patterns which would mean that the multiplier can be more than 4:

M S S M S S M S M S

S M S M S M S M S S x 2

S S M S M S M S M S x 2

S S S M S M S M S M x 2

M S M S M S M S S S

M S S M S M S M S S

M S S S M S M S M S

M S S S S M S M S M

M S S S M S S M S M

As you can see there are quite a few patterns here.

You can arrange them as long as the math students are not sitting next to each other. There was no limit on the question.

hmmm...
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[#permalink]  07 Jan 2004, 09:25
Yes. You are right Titleist. So, what is the maximum number of possible combinations.
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[#permalink]  09 Jan 2004, 20:35
subtract number of ways in which atleast two are together from the total number of ways.

if 2 are together then we have 4C2 and treating this as one group we have 8 people left so number of ways = 4C2 * (8+1)!
smilarly 3 people together = 4C3 * (7+1)!
and 4 people together = 4C4 * (6+1)!

total ways in which copying takes place =
4C2 * 9! + 4C3 * 8! + 4C4 * 7! = 7! * 465
subtract this from 10!
so we get
7! ( 10*9*8 - 465 ) = 255 * 7!

what is the official answer ?
[#permalink] 09 Jan 2004, 20:35
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# If ten students appear in an examination and 4 of them are

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