How many bit strings contain exactly eight $0\,s$ and ten $1\,s$ if every $0$ must be immediately followed by a $1$ Question

How many bit strings contain exactly eight $0\,s$ and ten $1\,s$ if every $0$ must be immediately followed by a $1$
I know a question is already posted here, but i am getting doubt in my approach.

My approach/solution

My arrangement diagram goes like-:

$${\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}$$
I have to fill the gap$(-)$ by remaining $1^{s}$ i.e $2 \,\,1^{s}$

so total number of string possible=
$$\binom{18}{2}$$

but the answer is
$$\binom{10}{2}$$
where am i wrong ?
 A: If you put just one $1$ in a particular gap between $01$s you don't care whether it occupies the left blank or right blank.  All the cases were the $1$s went in different gaps are counted four times.
A: We can also derive the correct result $$\binom{10}{2}$$ based upon your approach. We do so by assuring that each admissible selection is counted precisely once. 

The configuration
  $${\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}01{\color{Red} --}$$
  has $9$ positions starting with the left-most  ${\color{Red} --}$ up to the right-most ${\color{Red} --}$. We can place at each of these $9$ positions one or two $1$s at most a total of two $1$s.

We observe admissible selections belong precisely to one of two different types: 

Either the two $1$s are placed at different positions or the two $1$s are placed at the same position. 
  
  
*
  
*There are $\binom{9}{2}$ ways to place two $1$s at different positions.
  
*There are $\binom{9}{1}$ ways to place two $1$s at the same position.
We conclude there is total of
  \begin{align*}
\binom{9}{2}+\binom{9}{1}\color{blue}{=\binom{10}{2}}
\end{align*}
  admissible selections.

A: Your method does not work, because using your coloring: if the first remaining $1$ goes in a black position, then the second cannot go in a black position. So, you cannot pick any two of the $18$ positions, as that would include two black ones, or two red ones.
Put differently: If you were to put the first $1$ in the first black position, and the second in the second black position, then you get the same string as when you put the first $1$ in the first black position and the second in the second red position. So, you are overcounting.
In fact, you are also counting putting the first in the first black position and the second in the second red position as different frim putting the first in the first red position, and the second in the second black position, but these result in the same string as well, so you are overcounting in that way as well.
Instead, think about it this way: you basically have $10$ 'units' that you are trying to arrange: $8$ units of $01$, and two units of $1$. So the problem has become isomorph to something like: how many strings are there consisting of $8$ $A$'s and $2$ $B$'s?  
