Question

In an examination hall there are four rows of chairs. Each row has $6$ chairs one behind the other. There are two classes sitting for the examination with $12$ students in each class. It is desired that in each row, all students belong to the same class and that no two adjacent rows are allotted to the same class. In how many ways can these 24 students be seated?

Answer 1

Row-1 Row-2 Row-3 Row-4
Case1 Class 1 Class 2 Class 1 Class 2
Case2 Class 2 Class 1 Class 2 Class 1
Required seating arrangement can be done in two ways case 1 and case 2.
Total number of arrangements $=Number$ of arrangements in case 1 $+Number$ of arrangements in case 2
Now, 12 students of class1 can be seated in 12 chairs in ${}^{12}{P}_{12}$ ways and 12 students of a class 2 can be seated
in 12 chairs in ${}^{12}{P}_{12}$ ways.
Hence, the total number of arrangements
$=(12!\times 12!)+(12!\times 12!)$
(since ${}^{12}{P}_{12}=12!$)
$=2(12!{)}^2$