We are tasked with finding how many bit strings contain exactly five 0s and 14 1s, with the condition that every 0 must be immediately followed by two 1s.

Step 1: Understand the Structure

The key constraint is that each 0 must be followed by two 1s. So, every "0" forms a triplet "011". This means that each 0 contributes two 1s immediately after it, and these 0s and 1s are tied together.

• Since there are exactly 5 zeros, this condition will create 5 triplets of "011".
• Each "011" triplet contributes 1 zero and 2 ones.

Now, we need to distribute the remaining 9 ones (since 5 * 2 = 10 ones are used up with the 5 zeros) in the string.

Step 2: Calculate the Total Number of Positions

The total number of bits in the string is:

$\text{Total bits} = 5 \times 3 + 9 = 15 \text{ bits}.$

The 15 bits consist of 5 triplets "011" and 9 individual 1s.

Step 3: Distribute the Remaining 1s

We need to place the remaining 9 ones into the 10 possible positions:

• Before the first triplet "011"
• Between the 5 "011" triplets
• After the last triplet "011"

This means there are 10 positions where the remaining 9 ones can be placed.

Step 4: Calculate the Number of Ways to Distribute the Remaining 1s

This is a problem of distributing 9 indistinguishable ones into 10 distinguishable positions, which can be solved using the stars and bars formula. The number of ways to distribute $n$ indistinguishable objects into $k$ distinguishable bins is given by:

$\binom{n + k - 1}{k - 1}$

For our case, $n = 9$ (remaining 1s) and $k = 10$ (available positions):

$\binom{9 + 10 - 1}{10 - 1} = \binom{18}{9}$

Step 5: Compute the Binomial Coefficient

Now, we compute $\binom{18}{9}$:

$\binom{18}{9} = \frac{18!}{9! \cdot 9!} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 48620$

Final Answer:

The number of bit strings that contain exactly five 0s and 14 1s, with each 0 immediately followed by two 1s, is $\boxed{48620}$.

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Would you like to go into more detail on any part of this explanation, or do you have additional questions?

Here are 5 related questions you might find interesting:

1. How can we calculate the number of bit strings with exactly 5 zeros and 14 ones without the given constraint (i.e., without requiring that every 0 be followed by two 1s)?
2. What is the stars and bars method, and how can it be applied to other problems of distributing objects?
3. How does the concept of "combinations with repetition" relate to distributing indistinguishable items into distinguishable bins?
4. What would change in this problem if the number of zeros were different, or if the condition was modified (e.g., if each 0 were followed by 3 1s)?
5. How does the binomial coefficient formula generalize to other combinatorial problems, such as selecting groups from a set?

Tip: When tackling combinatorial problems, break down the constraints first to understand the structure, then apply methods like stars and bars or binomial coefficients to solve.