Why shuffle them up?

Let $\mathbb{P}(n)$ denote the probability of "the required event"

$\mathbb{P}(n)=\sum_{r=0}^{n}\frac{(-1)^{r}}{r!}$

Let us first convince ourselves that the maximum occurs at even $n$

Why? if maxima occurs at a $\mathbb{P}(2n+1)$ , the next term is an addition which makes the maxima lesser than $\mathbb{P}(2n+2)$ which is a contradiction

Now,let us assume that a maximum occurs at a certain $k > 4$

$\mathbb{P}(k)=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-...+\frac{1}{(k-2)!}- \large \frac{1}{(k-1)!}+\frac{1}{k!}$

Clearly, $\mathbb{P}(k-2)> \mathbb{P}(k)$

And so, $\mathbb{P}(2)$ is greatest.But since there have to be atleast 3 students, our maxima is at $\mathbb{P}(4)$ which is 0.375

To find the maximum probability, note that your expression for the probability when there are LaTeX: $n$ students is a partial sum of an alternating series. Use this (along $n\geq 3$ ) to find maximum probability is just $\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} = 0.375$