This is the way it was explained to me . . .
We have this set of objects: . . in their original positions.
For the first position, there are choices for the replacement for
Suppose it is . .We have: .
What is placed in the second space?
There are two basic choices:
. .  is not placed in the second space.
. .  is placed in the second space.
 Suppose is not in the second space.
. . .Then we have objects to derange.
. . .There are: . ways.
 If is in the second space, we have: .
. . . and has swapped places; they are already deranged.
. . .We must derange the other objects: . ways.
Hence, if is in the first space,
. . there are: . possible derangements.
Since there are choices for , the number of derangements is:
With Inclusion method you can prove the following: