To solve the problem of finding how many bit-strings of length 9 either start with "1" or end with "00," let's break it down using the principle of inclusion and exclusion.

1. Total number of bit-strings of length 9: There are $2^9 = 512$ total bit-strings of length 9, since each bit can be either 0 or 1.

2. Bit-strings that start with "1": If the bit-string starts with "1," the remaining 8 bits can be anything. So there are $2^8 = 256$ bit-strings that start with "1."

3. Bit-strings that end with "00": If the bit-string ends with "00," the remaining 7 bits can be anything. So there are $2^7 = 128$ bit-strings that end with "00."

4. Bit-strings that start with "1" and end with "00": If the bit-string starts with "1" and ends with "00," the remaining 6 bits can be anything. So there are $2^6 = 64$ bit-strings that start with "1" and end with "00."

5. Apply the principle of inclusion and exclusion:

   • Bit-strings that start with "1" or end with "00" = (bit-strings that start with "1") + (bit-strings that end with "00") - (bit-strings that start with "1" and end with "00")
   • This gives us $256 + 128 - 64 = 320$.

Thus, the correct answer is C. 320.

Would you like more details or have any questions?

Here are 5 related questions:

1. How many bit-strings of length 10 start with "1" or end with "000"?
2. What is the probability that a random bit-string of length 9 starts with "1"?
3. How many bit-strings of length 9 do not start with "1" and do not end with "00"?
4. If a bit-string of length 9 is randomly generated, what is the probability that it ends with "00"?
5. How many bit-strings of length 8 contain exactly two "1"s?

Tip: The principle of inclusion and exclusion is useful for solving problems where you're asked to count elements that satisfy at least one of several conditions.