In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _________.

This problem can be solved using the concept of derangements, which is a permutation of objects where no object appears in its original position. In this case, we have 5 students who should not sit in their allotted seats. <br/><br/>The formula for calculating the number of derangements (also known as subfactorials) is : <br/><br/>D(n) $$ =n !\left[\frac{1}{0!}-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots \ldots . .+(-1)^n \frac{1}{n !}\right] $$ <br/><br/>Where n is the number of students, in this case, 5. <br/><br/>Using the formula, let's calculate the derangements for 5 students : <br/><br/>D(5) = $5! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right)$ <br/><br/>D(5) = $120 \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$ <br/><br/>D(5) = $120 \left(0 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$ <br/><br/>D(5) = $120 \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$ <br/><br/>D(5) = $120 \left(\frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120}\right)$ <br/><br/>D(5) = $120 \left(\frac{44}{120}\right)$ <br/><br/>D(5) = 44 <br/><br/>So, there are 44 ways in which none of the students sits on the allotted seat.