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

Question

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 ?

Answer 1

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.