Question: is my solution to the following problem correct? I'm also interested in (possibly) more elegant solutions.
I have the following problem: Suppose there are $n$ people of different length and assign them a place in a row on an arbitrary basis. The chance that we can fix the order from small to large by switching two people, equals $1/48$. How many people are there?
My solution The number of ways we can put these people in a line equals $n!$.
The number of sequences where there are 2 people not on the right spot equals $\binom{n}{2}$; starting from the right order, we need to pick two people, without repetition, and switch them. This gives a sequence of people which can be fixed by exactly one switch. The number of ways to do this equals $\binom{n}{2}$.
Hence the chance of a sequence where 2 people are in the wrong spot equals $$\frac{\binom{n}{2}}{n!} = \frac{1}{48}.$$ Writing things out, we find that this equals to $$48\frac{n(n-1)}{2} = n!$$ and hence $$24 = (n-2)!$$ which gives $n = 6$.
community wiki post so that the question can be closed
Your solution is correct.