How many bit strings? 
How many bit strings of length $8$ have either exactly two $1$-bit among the first $4$ bits or exactly two $1$-bit among the last $4$ bits?

My solution: 
A bit only contains $0$ and $1$, so $2$ different numbers, i.e., $0$ and $1$. For the first part we have $2^6=64$ ways. Similar for the other way. Hence there exists $2^4=16$ bit strings. Is my answer true?
Update: I mean $2^6+2^6-2^4=112$ bit strings

Among the first $4$ bits, choose $2$ to set them to one and the other two would be set to $0$ and there are $4$ of them with absolute freedom. 
$$\binom{4}{2}\cdot 2^4$$
Similar when we focus on the last $4$ bits. 
When we focus on intersection. We would pick $2$ from the first $4$ and pick $2$ from the last $4$.
So my overall answer would be
$$2\binom42 \cdot 2^4 - \binom42^2$$
Remark: I think your mistake is thinking that you can set arbitary $6$ bits to anything.
