How many bit strings of length 10 can the bits of "1" be in pairs? Ie: 0110000110 has the property but 0011100110 does not)

My solution to this was to group the two 1bits into a block and the maximum pairs can only be 3 pairs in separated positions in a string of length 10. Then I choose the middle spaces between a string of zeros like this:

$\displaystyle \binom{9}{1}+\binom{7}{2}+\binom{5}{2}=40$

But The correct answer given was $\displaystyle \binom{11}{0}+\binom{9}{1}+\binom{7}{2}+\binom{5}{ 2}=40$

What I don't understand is why is there $\displaystyle \binom{11}{0}$ ? Why is there a 11 choose 0, which creates an additional 1 more possibility?