Inclusion & Exclusion: In how many permutations of the digits $0,...,9$ there's no continuity of 7 digits or more? In how many permutations of  the digits $0,...,9$ there's no continuity of 7 digits or more?
(Ex. the number 203456789 1 should not be counted)
I believe that the basic case, for the inclusion exclusion principle would be how many permutations of 0 to 9 will give us an integer. The answer for that would be: $9$ options for the first digit and $9!$ options for the rest which would give us: $9 \cdot 9!$, right?
What's the next step? Should I use the complementary?
 A: The question appears to be about permutations of digits, not $10$-digit numbers. So $0$ should not be treated as special, there are $10!$ permutations.
Now we need to count the bad permutations, the ones that have $7$ or more consecutive digits. We need to define consecutive digits. We will interpret it as consecutive increasing. And we take $0$ as smallest. 
One could divide into cases. This probably would be the easiest approach, but presumably we want to practice Inclusion/Exclusion.  So the main part of this answer deals with Inclusion/Exclusion. But in a remark at the end, we give a cruder but more practical approach. 
Let $A$ be the collection of strings that are consecutive from the beginning  to the $7th$ place, at least. The first digit of such a string can be any of $0$ to $3$. So there are $(4)(3!)$ such strings.
Let $B$ be the collection of strings that are consecutive from at least the second position to the $8$-th. Again, there are $(4)(3!)$ such strings. 
Similarly, define $C$ as the collection of strings that are consecutive at least from the third position to the $9$-th, and $D$ the collection of strings that are consecutive from at least the fourth position on. Each of $C$ and $D$ has $(4)(3!)$ elements.
Suppose now we add together $4$ copies of $(4)(3!)$. Then we overcount the strings that belong to two or more of our sets. 
How many belong to $A\cap B$? We have to be consecutive from position $1$ to position $8$. There are $3$ choices for the first digit, and for each there are $2!$ ways to permute $2$ numbers that don't belong to the string of $8$, for a total of $(3)(2!)$.  
What about $A\cap C$? There are $2$ choices for how the string begins, and then there are "$1!$" ways to permute the last digit, for a total of $(2)(1!)$. (I am using this slightly silly notation to make things symmetrical. Of course the answer is clearly just $2$.)  
What about $A\cap D$? There is only $1$ string here, but we can call the number $(1)(0!)$.
What about $B\cap C$? You should come up with $(3)(2!)$. There are a few others.
If we subtract the sum of all the numbers we get by counting the above intersections, we will have subtracted too much. So we need to add back the number of strings that belong to various combinations of $3$ of our sets. These are very easy to count.
Add finally, after we have done that, we have added back the string which is consecutive from beginning to end. so we must subtract $1$. 
Remark: There are less sophisticated but far easier ways to count. In this "small" situation, dividing into cases is crude but reliable.
Count the strings of exactly $7$ that start from the first position. If the first digit is $0$, there are $2$ choices for the $8$-th digit (it can't be $8$) and then $2!$ ways to permute the rest, for a total of $4$. If the first digit is $1$, the number is the same, $(2)(2!)$. If the first digit is $2$, again we get $(2)(2!)$. If the first digit is $3$, nothing is forbidden at the end, so we get $3!$. 
Count the strings of length exactly $8$ that start from the first position also exactly $9$, also exactly $10$. These are all pretty easy to count.
Now count the strings of length exactly $7$, $8$, $9$ that start at the second position (so were not already counted). Do the same for third position, fourth position. If one cares to take advantage of it, there is even some symmetry to cut down on the work.   
A: You are essentially there, I think. You have to INCLUDE $9\cdot 9!$ total possibilities. You need to EXCLUDE the cases where the first 7 are a run (noticing the first digit is special), the second 7, ... Then you need to INCLUDE the cases where you have two types of these runs, ...
Main things to be careful of are that


*

*You treat the first digit carefully throughout all calculations.

*You remember to use inclusion-exclusion by considering ALL pairs etc. of possibilities, not just adjacent runs. This means being slightly careful with runs with an overlap of two for instance.

