Compute the number of 7-digit sequences that contain a 5-digit consecutive substring 
Let say any 7-digit sequence, even 0000000 is a valid number. Compute the number of sequences that contain a 5-digit consecutive substring. For example “23456” and “12345”?

Approached it as (where X can be (0-9)) :
Which gives me a total as 1800, however the answer is 1700. Can you please point out what I over-counted and share a wise way to approach similar question ?  
 0 1 2 3 4 X X 
 1 2 3 4 5 X X
 2 3 4 5 6 X X
 3 4 5 6 7 X X
 4 5 6 7 8 X X
 5 6 7 8 9 X X 

 X 0 1 2 3 4 X  
 X 1 2 3 4 5 X 
 X 2 3 4 5 6 X 
 X 3 4 5 6 7 X 
 X 4 5 6 7 8 X 
 X 5 6 7 8 9 X

 X X 0 1 2 3 4  
 X X 1 2 3 4 5 
 X X 2 3 4 5 6 
 X X 3 4 5 6 7 
 X X 4 5 6 7 8 
 X X 5 6 7 8 9

 A: Yes, you are overcounting the number of such strings: for example  $0 1 2 3 4 5 6$ is counted three times:
as a member of $0 1 2 3 4 X X$, of  $X 1 2 3 4 5 X$, and of $X X 2 3 4 5 6$.
As explained by the comments the right tool to use is 
the inclusion-exclusion principle.
Hint. Let $S(01234)$ be the set of 7-digit sequences which contain a copy of the string $01234$ then, from your scheme,
$$|S(01234)|=3\cdot 10\cdot 10=300.$$
By the inclusion-exclusion principle, the number of 7-digit sequences such that contain at least a 5-digit consecutive substring in increasing order is
$$\begin{align}&|S(01234)|+|S(12345)|+\dots+|S(56789)|\\&-
|S(012345)|-|S(123456)|-\dots-|S(456789)|\end{align}.$$
P.S. In order to apply the inclusion-exclusion principle you should consider the intersections of the sets $S(\cdot)$. Note that 
$$S(01234)\cap S(12345)=S(012345),\\
S(01234)\cap S(23456)=S(0123456),\\
S(01234)\cap S(34567)=\emptyset,\\
S(01234)\cap S(12345)\cap S(23456)=S(0123456)
.$$
A: You are counting some strings twice.
For example, $$xx56789$$ and $$x45678x$$ both results in $$3456789$$ 
