Problem Breakdown:

We are tasked with finding the number of bit strings that meet the following conditions:

1. Length: The bit string has a total length of 15.
2. Zeros: The string must contain exactly 6 zeros.
3. Prefix: The string starts with 10.

Step 1: Analyze the constraints

• Since the string starts with 10, the first two bits are fixed. This leaves 13 bits remaining.
• The 6 zeros include any that may have already been placed in the prefix. However, there are no zeros in the second bit of the prefix, so 1 zero has already been placed in the prefix.
• This leaves 5 zeros to be distributed among the remaining 13 bits.

Step 2: Determine the number of ways to arrange the bits

The problem reduces to arranging 5 zeros among 13 positions. The remaining positions will be filled with ones, resulting in 8 ones (since $13 - 5 = 8$).

The number of ways to arrange 5 zeros in 13 positions is a combination: $C(13, 5) = \frac{13!}{5!(13 - 5)!} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287$

Final Answer:

The total number of valid bit strings is 1,287.

Would you like a detailed explanation of the steps or additional related problems?

Follow-Up Questions:

1. What if the string needed to start with 101 instead of 10?
2. How does the answer change if the string length increases to 16?
3. Can you calculate the total number of bit strings with 6 zeros but no prefix constraints?
4. What if there are no restrictions on the total number of zeros?
5. How many strings of length 15 start with 10 and have at least 6 zeros?

Tip:

Always confirm the number of fixed bits when dealing with prefix constraints to simplify combinatorial calculations.