Four digit-number that cannot contain sequence $45$

Question

I keep coming up with the result $8720$ for this, but it's not correct.

I calculate like this:

Total number of four-digit numbers: $9000$

Combinations beginning with $45$ ($45xx$): $100$

Combinations with $45$ in the middle ($x45x$): $90$

Combinations with $45$ at the end ($xx45$): $90$

So: $9000-100-90-90=8720$. Where am I wrong?