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.