Find the number of rearrangements of the string 12345 in which none of the sequences 12, 23, 34, 45 occur By using the principle of inclusion/exclusion, I wanted to start with the total number of arrangements.
$S_0 = 5!$
Now I need to find the ones that are being counted more than once.
letting $S_1 = |12| + |23| + |34| + |45|$ and $S_2 = |123| + |234| + |345|$, $S_3 = |1234| + |2345|$, $S_4 = |12345|$ 
I think the general way to go about it would be $S_0 - S_1 + S_2 - S_3 + S_4$.
Is this the correct way to go about this problem? If so, I'm unsure of how to calculate the $S_x$
EDIT: progress update:
$S_0 = 5! = 120$
$S_1 = 4 * \binom{4}{1} * 3!  = 96$
$S_2 = 3 * \binom{3}{1} * 2! = 18$
$S_3 = 4$
$S_4 = 1$
Therefore, I have $120 - 96 + 18 - 4 +1  = 39$ arrangements
 A: Call $A_k$ the set of permutations of $[n]$ where $k+1$ is right adjacent to $k$ then out of all $n!$ total permutations
$$|A_k|= (n-1)!$$
but also it is true for any intersection
$$|A_k\cap A_l|=(n-2)!$$
Because we may have, for example, 1-2 3 4-5 (3 objects to permute) or 1-2-3 4 5 (3 objects to permute). For every adjacency condition we add there is always 1 less object. Hence
$$|A_k\cap A_l\cap A_m|=(n-3)!$$
etc.
In general there are $n-1$ sets $A_k$ hence
$$S_1=\binom{n-1}{1}|A_k|\\S_2=\binom{n-1}{2}|A_k\cap A_l|\\\vdots$$
Therefore the desired count is, in general
$$\text{valid permutations}=n!-\binom{n-1}{1}(n-1)!+\binom{n-1}{2}(n-2)!-\ldots +(-1)^{n-1}\binom{n-1}{n-1}1!$$
In your case $n=5$ so
$$\text{valid permutations}=5!-\binom{4}{1}4!+\binom{4}{2}3!-\binom{4}{3}2! +\binom{4}{4}1!=53\qquad\blacksquare$$
A different method is to develop a recurrence for permutations of length $n$ with no cases where $k+1$ is right-adjacent to $k$, the required recurrence is
$$a_n=(n-1)a_{n-1}+(n-2)a_{n-2}$$
with $a_1=1$ and $a_n=0$ for $n\le 0$. 
This can be seen by noting that all valid permutations of $[n]$ can be formed by placing $n$ after all numbers except $n-1$ or at the start in all valid permutations of $[n-1]$. Or $n$ can go between two adjacent numbers $k$ and $k+1$ in all permutations of $[n-1]$ with exactly one right adjacency (there are clearly $(n-2)a_{n-2}$ of these because we treat the two adjacent numbers k-(k+1) as a single character and there are no more right adjacencies).
The sequence $a_n$ begins
$$1,1,3,11,53,309,\ldots$$
It is possible to show that this sequence has an exponential generating function
$$f(x)=\frac{e^{-x}}{(1-x)^2}$$
