Counting - Permutations & Combinations Two series of a question booklet for an aptitude test are to be given to twelve students. In how many ways can the students be placed in two rows of six each so that there should be no identical series side by side and that the students sitting one behind the other should have the same series?
 A: Now you have to divide the 12 students in two series each of them have 6 student .
So No. of ways that the students be placed  is $12\cdot11\cdot 10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$  
Since  there should be no identical series side by side and that the students sitting one behind the other should have the same series , each series must have the same booklet. $\implies$ No. of ways to place the booklet equals $2$
$X:1,2,1,2,1,2$ & $Y:1,2,1,2,1,2$  OR $X:2,1,2,1,2,1$ &  $Y:2,1,2,1,2,1$
Therefore the total ways to place the student with the  conditions mentioned in the question is  $2\cdot12\cdot11\cdot 10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$ 
I'm NOT sure that my solution is true !
A: Notice that placing a test booklet in the first chair completely determines the correct test booklet for every other chair. And so, since there 2 test booklets, there are 2 ways of placing the test booklets. Thus the total number of arrangements is twice the number of ways to seat the students.
Now we need to know the number of ways to seat the students. For the first chair we have 12 students to choose from. For the second chair we have 11 students to choose from. Etc. This implies there are $12\cdot11\cdot10\cdot\cdots\cdot1 = 12!$ ways to seat the students.
And so the total number of arrangements is $2\cdot12!=958\,003\,200$.
A: If row A and row B has 6 seats each then the following sequence of two series will work.


*

*$A:1,2,1,2,1,2$

*$B:2,1,2,1,2,1$ and the reversed order for each row.

