The Question
How many bit strings contain exactly eight 0s and 10 1s
if every 0 must be immediately followed by a 1?
Note: There is a similar but also very different question on this site. Please do not report this as a duplicate because it is not!
My Attempt
I thought could we not also have $8$ 01 strings and $2$ 1 bits. This ensures that 0 is directly followed by a 1. I then have 9 spaces (since the 01 will occupy two of our 18 spaces) to distribute these into, which can be done in 9 ways, then I put the 1s in the remaining spaces, giving us 9 bit strings.
My Problem
My answer was not correct. My textbook gives the answer 45. I'm wondering what was wrong with my approach, and how I would approach this question correctly.
Best Answer
Per vadim123's comment (and your own thoughts), you have a string of length 10 and you have two letters to use: eight "01"s and two "1"s. In other words, there are 10 locations in the string and you're choosing 2 of them to be special, so the answer is $\binom{10}{2}$.
I don't really follow your attempt-- the problem is as if you have 10 spaces, so I don't know why you say 9.