In how many ways can 13 question papers be arranged so that the best and the worst are never come together?
A. $12!\, \times 11$
B. $\,11 \times 12!$
C. $12!{\,^2}$
D. None of these

Hint: First we will find the number of arrangements of 13 question papers and then we’ll find the number of arrangements so that the best and the worst come together. Now, we can say that the number of arranging 13 question papers so that the best and the worst are never come together will be equal to the difference in the above two arrangements, therefore we’ll get the required answer.

Complete step by step answer:

Given data: Number of question papers $ = \,13$
The number of arrangements of n elements is given by $n!$
We can say that the number of arranging 13 question papers so that the best and the worst are never come together will be equal to the difference in the total number of arrangements and the number of arrangements so that the best and the worst come together.
Total number of arrangements of 13 papers $ = 13!$
Now, suppose that the worst and the best paper come together and combined as one
Number of arrangements so that the best and the worst come together is $ = 12!\, \times 2$
It is multiplied by 2 as the best and the worst can interchange their position as well.
Therefore the required number of arrangement $ = 13!\, - \,(2)12!\,$
Using $n!\, = \,n(n - 1)!$
 $ = (13)12!\, - (2)12!\,$
Taking $12!$ common, we get,
$ = 12!(\,13 - 2)$
On simplification we get,
$ = 12!\, \times \,11$
Option(A) and (B) are correct.

Note: While writing the Number of arrangements so that the best and the worst come together most of the students misses the point that they can interchange their position 2 ways, so keep it in mind to get the correct answer.