rearrangements of ABCDE having exactly 1 letter in its original position. The question asks to find the no. of rearrangements of $ABCDE$ having exactly $1$ letter in its original position.
(Rearrangement of a set means any arrangement of the set including its original ordering).
Do we first have to fix an element (suppose $A$) and then have to find all the possible ways to arrange other $4$ elements in remaining $4$ positions?
 A: We have to choose an element, then to choose a permutation of the remaining elements without fixed points (a derangement). Since there are $9$ permutations without fixed points in $S_4$ ($3$ elements with the cycle structure of $(1\,2)(3\,4)$ and $6$ elements with the cycle structure of $(1\,2\,3\,4)$), the answer is $5\cdot 9=45$. 
A: Alternative Method (Using Inclusion-Exclusion Principle):
Fix any letter in its proper place, and the remaining $4$ have to be arranged such that none is in its proper place (so they become a derangement). Since the fixed letter is in its proper place, we can ignore it. Then the number of derangements of the other $4$ are:

$4! -  \binom 4 1 3! + \binom 4 2 2! - \binom 4 3 1! + 0! = 9$

where there are:  

$4!$ total permutations of $4$ letters
  $3!$ permutations with a (chosen) fixed letter, and $\binom 4 1$ ways to choose that letter
  $2!$ permutations with two (chosen) fixed letters, and $\binom 4 2$ ways to choose them
  $1!$ permutation with three fixed letters, and $\binom 4 3$ ways to choose them
  $0!$ permutation with all letters fixed

There are $5$ possible letters (out of $ABCDE$) that can be fixed in its proper place, and $9$ derangements of the other $4$ letters, so the total number permutations with exactly one letter in its proper place is $5 \times 9 = \boxed{45}$.
Note: There is no overlap between the permutations with $A$ in its proper place and all other letters deranged, and $B$ in its proper place and all other letters deranged, etc.
One advantage of this method is that it can easily be generalized to $n$ letters.

Principle of Inclusion and Exclusion:
  Out of $N$ objects, the number of objects $N(\overline A_1 \overline A_2 \ldots \overline A_n)$ that do not have any of the $n$ properties $A_i$ ($i = 1, \ldots, n$), is
  $$N(\overline A_1 \ldots \overline A_n) = N - N(A_1) - N(A_2) - \cdots - N(A_n) + N(A_1 A_2) + N(A_1 A_3) + \cdots + N(A_{n-1} A_n) - \cdots + (-1)^n N(A_1 A_2 \ldots A_n)$$
  where $N(A_i)$ is the number of objects with property $A_i$, $N(A_i A_j)$ is the number of those with both properties $A_i$ and $A_j$, etc.

