How many 10-digit numbers whose first four digits match either the next four digits or the last four digits? 
If in a 10-digit number, the first four digits are either exactly the same as the next four digits or the last four digits. How many such 10 digit numbers are there?

My attempt:
First I consider the case with first four and next four are same.
For the first four places, we have $9\times 10\times 10\times 10$ ways of choosing digits as zero cant come in first digit.
The digits for next four places are already choosen as they are the same.
The remaining two places can be filled in $10\times 10$ ways.
Therefore total is $9\times 10^5$ ways.
Since the second case first four and last four digits are same, the answer will again be $9\times 10^5$ ways.
Therefore, total numbers are $2\times 9\times 10^5$. However, correct ans is $9\times 10^5-90$ .

Why am I wrong?


If I'm correct, the first digit would be 1-9. Thus, we start the equation with a 9.
The second digit can have 10 different numbers, so the next step of the equation is 10
The third and fourth digit can also have 10 different numbers, which brings us 10*10 numbers.
The first four numbers will have 9*10*10*10= 9000 possibilities
Now, we calculate the remaining 2 digits for both cases, which will be (10*10)*2 =200
Since there are 2 different types of combinations, we multiply 9000 by two and add 200, which will be 18200 different combinations.
I'm a new member so I don't really know where to put my answer if I'm not sure about it, and if I'm wrong, please tell me why and I'll delete my post immediately. :)
Edit: Based on @dhanesh vijay 's answer, we should minus 90 as it is the possibilities with both conditions being true as they are already counted. Thus, the answer would be 18200-90=18110 different combinations.
