A postman has to deliver ten different letters to ten different people. That morning, however, he forgot his glasses so he delivered them randomly. What is the probability that he got right at least six of them?
How the situation does change if he is delivering generally n letters to n persons and at least m are delivered to right persons?
Right after I posted the problem I thing the solution goes like this ... n! general ordering. Fix the people which are correct, there are nCm of them. For each there are (n-m)! how to deliver the others. How many of them are incorrect? I would guess based on a different problem that the answer is $x = (n-m)!(1-1/1! + 1/2! - \cdots +(-1)^{n-m}/(n-m)!)$ So I suspect $P[\text{m is correct}] = nCm*x/n! = (1-1/1! + 1/2! - \cdots +(-1)^{n-m}/(n-m)!)/m!$ so
$$P[\text{at least m are correct}] = \sum_{k=m}^{n} \sum_{l=0}^{n-k} \frac{(-1)^l }{l!k!}$$
Which is for $n=10$ and $m=6$ is 17/28350.
Is it correct?