Probability of almost correct order Let's say I'm a teacher handing tests back to seven students. 
If I do it with my eyes closed, what's the probability I hand exactly 5 of the tests back to the correct students? 
There are many possible ways this could happen, and if my understanding of probability is correct, the sum of the individual probabilities of all possibilities where I hand exactly 5 back to the correct student yields the answer I'm looking for. 
One such probability (hand out the first two tests incorrectly): 
$$\frac{6}{7}\cdot\frac{5}{6}\cdot\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
Another probability (hand out the last two tests incorrectly): 
$$\frac{1}{7}\cdot\frac{1}{6}\cdot\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
One more: 
$$\frac{1}{7}\cdot\frac{1}{6}\cdot\frac{4}{5}\cdot\frac{1}{4}\cdot\frac{2}{3}
\cdot\frac{1}{2}\cdot\frac{1}{1} $$
We can simplify these probabilities to: 
$$\frac{6\cdot5}{7!}, \frac{1\cdot1}{7!}, \frac{4\cdot2}{7!} $$
More generally, we can think about all possible times when the two incorrect letters might occur, and deduce that the sum of all possible probabilities is: 
$$\frac{6(5+4+3+2+1)+5(4+3+2+1)+4(3+2+1)+3(2+1)+2(1)+1(1)}{7!}$$
Is this really the answer? I really feel like I missed something that would simplify this answer. 

Or as another way way of reaching @Zubin Mukerjee correct answer:
There are $\binom{7}{5}$ ways to pick the 5 students who get their proper papers, and only one way to give the remaining two the wrong papers...
Edit to respond to OP's comment...
Well, consider the first way you enumerate.. Your first two terms give the total probability of getting the first two wrong, including times when the first two got papers from #3 through #7.  But that's not possible, since you then give #3 through #7 their correct papers...
