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!
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 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.