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?
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:
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.