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since we can choose the first spot and the last two spots in one way so we have 2^5 for the others Is that right?

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    $\begingroup$ Yup, that's right. $\endgroup$
    – Arthur
    Nov 15 '17 at 15:48
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Break the question down into parts, it always helps me to understand the problem:

How many bit strings of length 8

So we have 8 places to be filled that can either be 1 or 0 (binary)

$ \_ $ $ \_ $ $ \_ $ $ \_ $ $ \_ $ $ \_ $ $ \_ $ $ \_ $

start with a 1 bit and end with two bits 00

These are fixed values and take up the following positions:

$ 1$ $ \_ $ $ \_ $ $ \_ $ $ \_ $ $ \_ $ $ 0 $ $ 0 $

There are 5 remaining positions that can be filled by either a 1 or 0.

Since we have 2 choices for each of the remaining 5 tasks, and repeat elements are aloud (11,00) we can use permutations formula:

$$n^r$$

Where $n$ is the number of possible choices to for each task and $r$ is the number of tasks.

$$\therefore n^r = 2^5 = 32 $$


In short, yes your answer would be correct, I just wanted to break the problem down for you.

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