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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.

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    $\begingroup$ You have ten spaces, eight filled by "01" and two filled by "1". $\endgroup$ – vadim123 Nov 6 '14 at 23:44
  • $\begingroup$ Which means I have C(10,8) ways to distribute the "01" and 1 way to slot the two "1" into the remaining 2 spaces. By rule of product I have C(10,8)*1 = 45 spaces? $\endgroup$ – Dunka Nov 6 '14 at 23:48
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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.

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  • $\begingroup$ My thinking was definitely off. We start off with 18 spaces to put a eight "0"s and ten (1s). I forgot to adjust how many spaces we had when I turned our bits into eight "01"s and 2 "1"s $\endgroup$ – Dunka Nov 7 '14 at 0:04

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