N people are to be seated in a row. Four of them, A, B, C, and D cannot sit next to each other*. How many different sitting arrangements are possible? N people are to be seated in a row. A, B, C, and D cannot sit next to each other*. How many arrangements are there?
 * No two of them can be adjacent
I thought I would first calculate all cases in which two of them are adjacent and tried to use the inclusion-exclusion principle to do so.
So my first case was when two of the four are together Case 1: 
$$\binom{4}{2}*2*(n-1)!$$
When 3 of them are together Case 2:
$$\binom{4}{3}*3!*(n-2)!$$
When 4 of them are together Case 3:
$$4!*(n-3)!$$
and then I thought it would be Case 1 - Case 2 + Case 3 but I believe I am overcounting by this.
Apologies for any mistakes, I'm new to StackExchange and still getting used to its workings
 A: Method 1:  Arrange other $n - 4$ people in a row, which can be done in $(n - 4)!$ ways.  This creates $n - 3$ spaces in which we can place $A$, $B$, $C$, and $D$, $n - 5$ between successive people and two at the ends of the row.  
To illustrate what I am saying, suppose $n = 10$.  Then there are $10 - 4 = 6$ people other than $A$, $B$, $C$, and $D$.
$$\square P \square P \square P \square P \square P \square P \square$$
Each $P$ represents the position of one of those six people.  Each of the $10 - 3 = 7$ squares represents a place where person $A$, $B$, $C$, and $D$ could be placed.
To separate $A$, $B$, $C$, and $D$, we must choose four of these $n - 3$ spaces in which to place $A$, $B$, $C$, and $D$.  We can arrange $A$, $B$, $C$, and $D$ in these spaces in $4!$ ways.  Hence, the number of arrangements of the $n$ people in a row so that no two of $A$, $B$, $C$, and $D$ are adjacent is 
$$(n - 4)!\binom{n - 3}{4}4!$$
Method 2: We use the Inclusion-Exclusion Principle.
There are $n!$ arrangements of $n$ people in a row.  From these, we must subtract those arrangements in which two or more of the four people are adjacent.
A pair of the four people are adjacent:  There are $\binom{4}{2}$ ways to select two of the four people to be in the pair.  We then have $n - 1$ objects to arrange, the pair and the other $n - 2$ people.  The objects can be arranged in $(n - 1)!$ ways.  The two people can be arranged in $2!$ ways.  Hence, there are 
$$\binom{4}{2}(n - 1)!2!$$
such arrangements, as you found.
Two pairs of people are adjacent:  This can occur in two ways.  Either the pairs are overlapping or they are separate.
Two overlapping pairs:  This means that three of the four people are consecutive.  Choose which three of the four are consecutive in $\binom{4}{3}$ ways.  We now have $n - 2$ objects to arrange, the block of three people and the remaining $n - 3$ people.  The objects can be arranged in $(n - 2)!$ ways. The three people can be arranged within the block in $3!$ ways, as you found.  Hence, there are 
$$\binom{4}{3}(n - 2)!3!$$
such arrangements, as you found.
Two separate pairs:  There are three ways to choose whether $B$, $C$, or $D$ is paired with $A$.  We then have $n - 2$ objects to arrange, the pair containing $A$, the pair not containing $A$, and the other $n - 4$ people.  The objects can be arranged in $(n - 4)!$ ways.  The pairs can each be arranged in $2!$ ways.  Hence, there are 
$$3(n - 2)!2!2!$$
such arrangements.
This is the case you omitted.
Three pairs of adjacent people:  This can only occur if the four people are all consecutive.  We now have $n - 3$ objects to arrange, the block of four people and the other $n - 4$ people.  The objects can be arranged in $(n - 3)!$ ways. The four people can be arranged within the block in $4!$ ways.  Hence, there are
$$(n - 3)!4!$$
such arrangements, as you found.
By the Inclusion-Exclusion Principle, the number of admissible arrangements is 
$$n! - \binom{4}{2}(n - 1)!2! + 3(n - 2)!2!2! + \binom{4}{3}(n - 2)!3! - (n - 3)!4!$$
