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 _______

We need to find the number of derangements of 5 elements -- the number of permutations where no student sits in their allotted seat.

Define the problem formally.

A derangement of $$n$$ elements is a permutation $$\sigma$$ such that $$\sigma(i) \neq i$$ for all $$i$$. In this problem, $$n = 5$$ students, and we need no student in their correct seat.

State the derangement formula.

The number of derangements of $$n$$ elements is given by:

$$ D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} $$

This formula comes from the inclusion-exclusion principle. If $$A_i$$ is the set of permutations where element $$i$$ is fixed, then:

$$ D_n = |S| - |A_1 \cup A_2 \cup \cdots \cup A_n| $$

By inclusion-exclusion, $$|A_{i_1} \cap \cdots \cap A_{i_k}| = (n-k)!$$ (fix $$k$$ elements, permute the rest), and there are $$\binom{n}{k}$$ ways to choose the $$k$$ elements, giving the formula above.

Calculate $$D_5$$.

$$ D_5 = 5! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right) $$

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

Computing the sum inside the parentheses with a common denominator of 120:

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

$$ = 120 \times \frac{44}{120} = 44 $$

The number of ways is 44.