Counting when there are two inclusive conditions. 
How many 8-bit sequences begin with $101$ OR (inclusive) have a $1$ as
  their fourth bit?

For the first condition, we need only to decide the values of the $5$ other bits, so there are
$$2^5$$
sequences starting with $101$.
For the second condition, we have to decide the values of the other $7$ bits, so there are
$$2^7$$
sequences with a $1$ as their fourth bit.

The final answer, however, surely cannot be
$$2^5+2^7$$
Because $2^5$ includes some scenarios with a $1$ as the fourth digit, whereas $2^7$ includes cases with a $101$ at the beginning, so I would be over-counting this. What do I do in this case?

I think the easiest method here is to take the total of all 8-bit sequences and subtract those that don't meet the requirement: those that do not begin with $101$ AND do have a $0$ as their fourth bit.
There are $7$ choices for the first $3$ bits, $1$ choice for the fourth, and $2^4=16$ for the remaining bits.  This gives an answer of
$$2^8-7(2^4)=16(2^4)-7(2^4)=9(2^4)$$ 
