Question

In a class of 25 students, we want to distribute their exam paper results.

What is the probability that only 2 students DO NOT get their exam paper. (23 get theirs and 2 are swapped).
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Answer 1

See the link in comments (or if it gets deleted, look up "partial derangement"). The number of ways in which exactly $k$ items from a pool of $n$ items can be rearranged such that only these $k$ items are in their original positions is

$R(n,k) = \dbinom nk \, !(n-k)$

where $!(n-k)$ is the so-called sub-factorial that satisfies the relation

$!n = \left\lfloor \dfrac{n!+1}e \right\rfloor$

and $\lfloor x \rfloor$ denotes the floor or greatest integer function of $x$, i.e. the greatest integer smaller than or equal to $x$.

There are 25! ways of handing back exams with no extra constraint.

There are $R(25,23)$ ways of handing back 23 exams to their original owners and 2 exams otherwise, and

$R(25,23) = \dbinom{25}{23} \, !(25-23) = 300\left\lfloor\dfrac{2!+1}e\right\rfloor = 300$

Then the probability of returning 23 exams correctly and 2 exams swapped is

$\dfrac{300}{25!} \approx \boxed{5.17\times10^{-22}}$