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How many bit strings contain exactly eight 0s and 10 1s if every 0 must be immediately followed by a 1 ?
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you can think this way

you have a string of length 10 and you have two letters to use : eight "01"s and two "1"s.

ie. You have ten spaces, eight filled by "01" and two filled by "1"

In other words, there are 10 locations in the string and you're choosing 2 of them to be special,

so the answer is  = 10C2 = 45

selected

After arranging $\color{red}{01}$ ,$8$ times in a line we have created $9$ gaps. In this gaps, we need to fill two $1$'s Now,

1. All $9$ gaps are distinct and $1$'s are obviously indistinguishable.

Therefore we can have \begin{align*} \binom{9+2-1}{2} = \binom{10}{2} = 45 \\ \end{align*} arrangement of $1$'s. And finally $45$ overall bit strings.