in Combinatory
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How many bit strings contain exactly eight $0s$ and $10\:\: 1s$ if every $0$ must be immediately followed by a $1?$
in Combinatory
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Well, you have 8 01s and then 2 1s left over.

Say a = 01 and b = 1.

You are then basically looking to find the number of unique strings with 8 a's and 2 b's.

For example, aaaaaaaabb

C(10,8)=10*9/2=45

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