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Three variants of exam paper are to be given to $12$ students. In how any ways can the students be placed in $2$ rows of $6$ each so that there should be no identical variants side by side and that the students sitting one behind the other should have the same variant. Find the number of ways it can be done.

Ths students can be placed in two rows as

$\dbinom{12}{6}\cdot 6!^{2}$

but I am not able to apply the $3$ varients.

I look for a short and simple way.

I have studied maths up to $12$th grade.

EDIT - I have uploaded the pic, I was asking que 21.

enter image description here

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    $\begingroup$ When you say "the student sitting behind should have the same variant" do you mean "the student sitting behind should not have the same variant"? $\endgroup$ – Henry Sep 22 '15 at 9:29
  • $\begingroup$ Question $20$ is very badly phrased. It appears from Question $21$ that the intended meaning is that each variant is given to the same number of students, but this is not stated in Question $20$. You also did not state it in your original question. $\endgroup$ – joriki Sep 22 '15 at 10:58
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This is a completely new answer, replacing my original answer to the misleadingly phrased original question.

The question (as now provided in a raster image) is still very badly phrased, but can be interpreted to mean that an assignment of variants to students is given and the possible assignments of students to seats under the given constraints are to be counted.

First we need to choose two seat pairs (consisting of one seat in the front row and the one behind it in the back row) for each variant. Number the seat pairs from $1$ to $6$. We have $3$ choices for the variant assigned to seat pair $1$. Then if that variant is also assigned to seat pair $3$ or $5$, the remaining seat pairs can be assigned the remaining variants in $2$ different ways; if it is also assigned to seat pair $4$, then that leaves $4$ different ways, and if it is also assigned to seat pair $6$, then that leaves $2$ different ways, for a total of $3(2\cdot2+4+2)=30$ different assignments. Then we also have $4!$ ways to assign the $4$ students who receive a variant to the $4$ seats assigned to that variant, for a total of $30\cdot4!^3=6!\cdot4!^2$ possible assignments, which corresponds to none of the options provided.

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