Unfortunately not. :/
When you say the possible strings are 2^5 you're stand fixed the three 0 at beginning and the two 0 at end. You need to count each case separately:
First case - Three 0 at beginning:
The first, second and third numbers just can take 0 value, so 1^3. Now you have the five middle numbers, which match to 2^5. The two numbers at final...
if you count as 2^2, you are counting the "0001001100" for example. When you count the second case you need to pay attention to don't count this case again. To leave this problem away, we count the cases where the two final numbers are not 0 at same time. So we have 2x1.
For the first case we have 1^3 + 2^5 + 2x1 = 1+32+2 = 35 (wrong)
For the first case we have 1^3 x 2^5 x 2x1 = 1x32x2 = 64
The second case - 00 at final:
Here we have the middle numbers at normal way 2^5, and the final two are just 2^1. As the first case, now we have a similar problem: we can't count 000... again. So let's count the first three number with at least one be 1... 1x2^2. Total: 2^5+2^1+1x2^2 = 32+2+4 = 38 (wrong)
Total: 2^5 x 2^1 x 1x2^2 = 32 x 2 x 4 = 256
Now, all cases: 64+256+32, this +32 at final is because of the cases weren't counted, the cases 000xxxxx00, so total 352
Sorry for the english, I've never explained math in english before.
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My mistake at beginning was sum the possibilities, where the foundamental principal of counting says we need to times them. We need to sum just the separated cases.