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

Let $A$ be the set of bit strings with exactly two $1$-bit among the first $4$ bits, and $B$ be the set of bit strings with exactly two $1$-bit among the last $4$ bits.
\begin{align}
\#A &= \binom{4}{2} 2^4 = 6\cdot2^4 \\
\#B &= 2^4 \binom{4}{2} = 6\cdot2^4 \\
\#A\cap B &= \binom{4}{2}^2 = 6^2 \\
\#A\cup B &= \#A + \#B - \# A \cap B \\
&= 6 (2^4 \cdot 2 - 6) \\
&= 6 \cdot 26 = 156
\end{align}
