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 \frac{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\frac{2!+1}e\right\rfloor = 300`

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

`(300)/(25!) \approx \boxed{5.17*10^(-22)}`