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.

The formula for calculating the number of derangements (also known as subfactorials) is :

D(n) $$ =n !\left[\frac{1}{0!}-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots \ldots . .+(-1)^n \frac{1}{n !}\right] $$

Where n is the number of students, in this case, 5.

Using the formula, let's calculate the derangements for 5 students :

D(5) = $5! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right)$

D(5) = $120 \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$

D(5) = $120 \left(0 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$

D(5) = $120 \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$

D(5) = $120 \left(\frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120}\right)$

D(5) = $120 \left(\frac{44}{120}\right)$

D(5) = 44

So, there are 44 ways in which none of the students sits on the allotted seat.