Permutations

In an examination hall there are four rows of chairs. Each row has 8 chairs one behind the other.There are two classes sitting for the examination with 16 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 alloted to the same class.In how many ways these 32 students can be seated?
Plz explain in detail.

5 Answers

1
nkhlshd ·

for a particular selection for rows , students in each row can be arranged among themselves in 8! ways.
we can divide the 16 students in each class into two rows in (16 c 8) ways as there are 8 chairs in 1 row.
Once such a division is made for a row, note that set of students to be seated in the other row pertaining to this class is fixed.
so the number of ways 16 students of a class can sit is (16 c 8)(8!)(8!). (call this number p)

as each way of arrangement of students of one class can be combined with each way of arrangement of students of the other class, we get the total number of arrangements =
no of ways we can arrange two nonadjacent rows for each class * p^2
= (8)* p^2

1
rocky89 ·

Ans is 2 *(16! * 16!)

will this (8)* p^2 be equal to it?

7
Sigma ·

16! for any 1 class and 16! for the rest...x 2 because these 2 classes can exchange their rows between themselves. Am i right?


1
rocky89 ·

Are u saying 16! * 16! OR 16!+ 16! ??????????????????

7
Sigma ·

the process is multiplicative so its (16!)2 x 2..

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