To solve this, we analyze the number of 10-bit strings that begin with "101" or "00":

1. Strings beginning with "101":

A 10-bit string starting with "101" fixes the first 3 bits. The remaining $10 - 3 = 7$ bits can be chosen freely. Therefore, there are: $2^7 = 128$ strings that begin with "101."

2. Strings beginning with "00":

A 10-bit string starting with "00" fixes the first 2 bits. The remaining $10 - 2 = 8$ bits can be chosen freely. Therefore, there are: $2^8 = 256$ strings that begin with "00."

3. Total Count (Union of Sets):

Since "101" and "00" are disjoint starting sequences (no overlap), the total number of such strings is: $2^7 + 2^8 = 128 + 256 = 384$

Matching with Choices:

The answer corresponds to: $2^7 + 2^8$ or $27 + 28$.

Final Answer:

$27 + 28$

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Do you want further clarification or examples? Here are five related questions to expand your understanding:

1. How many 10-bit strings start with "1"?
2. What is the total number of 10-bit strings?
3. How many 10-bit strings begin with "111" or "0"?
4. How do overlapping conditions affect such counting problems?
5. How would the solution change if "101" and "00" could overlap?

Tip: Always consider whether the conditions overlap or are disjoint when summing possibilities.