probability of empty box 
Each of 10 balls (numbered from 1 to 10) is placed into one of 10 boxes (numbered from 1 to 10). What is the probability that exactly box of number 1 and box of number 10 are empty?

I think If there are 2 empty boxes, so there are 2 boxes with 2 balls;
(n-4)! ways to arrange (n-4) balls to put in exactly (n−4) holes.
It's have $$n\choose 2$$ choices for the empty box
$$n-1 \choose 2$$ choices left for the boxes with 2 balls
result:
$$\frac{{n-1\choose 2}{n\choose 2} (n-4)! }{n^n}$$
Is this correct?
 A: There are two cases -
$1$) two boxes have two balls each
$2$) one of the boxes has three balls.
Number of arrangements are -
case $1$: $ \displaystyle {8 \choose 2}{10 \choose 4} {4 \choose 2} \cdot6!$
which is we select $2$ boxes out of $8$ for $2$ balls each (as $1$ and $10$ are empty), choose $4$ balls that will go into those two selected boxes and then arrange $2$ balls each in those two boxes. Finally arrange $6$ remaining balls in $6$ boxes.
case $2$: $ \displaystyle {8 \choose 1}{10 \choose 3}  \cdot 7!$
The explanation is similar to case $1$.
To find probability, add both cases and divide by $10^{10}$.
Alternatively you can apply Principle of Inclusion Exclusion or Stirling Number of the second kind.
A: The correct (general) answer for the two-boxes-with-two-balls numerator is
$${n-2\choose2}{n\choose2}{n-2\choose2}(n-4)!$$
That is, you first choose which two of the $n-2$ "interior" boxes (i.e., not box $1$ or box $n$, which are to remain empty) will receive two balls, then you pick two of the $n$ balls to go in the first of these, then two of the remaining $n-2$ balls to go into the second, and then distribute the remaining $n-4$ balls into the remaining $n-4$ boxes.
But as Henry observed in a comment, you need to consider the possibility that one box gets three balls (and the other "interior" boxes get one each). This gives another
$${n-2\choose1}{n\choose3}(n-3)!$$
possibilities. So the overall probability is
$${\displaystyle{n-2\choose2}{n\choose2}{n-2\choose2}(n-4)!+{n-2\choose1}{n\choose3}(n-3)!\over n^n}$$
