To solve this problem, we need to create an 8-letter string from the first 12 letters of the alphabet, with specific conditions:

1. Each letter in the string must be distinct.
2. The letters "H" and "I" must both be included but cannot be adjacent.
3. We select the 8 letters from the set {A, B, C, D, E, F, G, H, I, J, K, L}.

Let's work through this step-by-step:

Step 1: Choose 8 Letters from 12, Including H and I

Since H and I must be included, we only need to choose 6 additional letters from the remaining 10 letters (A, B, C, D, E, F, G, J, K, L). The number of ways to choose 6 letters out of 10 is calculated by: $\binom{10}{6} = 210$

So, there are 210 ways to select the remaining 6 letters.

Step 2: Arrange the 8 Letters with the Condition on H and I

For each choice of 8 letters, we need to arrange them such that H and I are not next to each other. Let’s calculate the total arrangements and then use a restriction approach to exclude cases where H and I are adjacent.

Total Arrangements without Any Restrictions

With 8 distinct letters, the total number of unrestricted arrangements is: $8! = 40,320$

Arrangements Where H and I Are Adjacent

To count the arrangements where H and I are adjacent, treat "HI" (or "IH") as a single "block." This forms 7 "positions" to fill (the "HI" block plus the remaining 6 letters).

The number of ways to arrange 7 "blocks" is: $7! = 5,040$ Since H and I can appear as "HI" or "IH" within their block, we multiply by 2: $7! \times 2 = 5,040 \times 2 = 10,080$

Valid Arrangements (H and I Not Adjacent)

Now, we subtract the cases where H and I are adjacent from the total arrangements: $8! - 7! \times 2 = 40,320 - 10,080 = 30,240$

Thus, for each selection of 8 letters that includes H and I, there are 30,240 valid arrangements.

Step 3: Combine Selections and Arrangements

Since there are 210 ways to select the 6 other letters, and for each selection, there are 30,240 valid arrangements, we find the total number of valid strings as: $210 \times 30,240 = 6,350,400$

Final Answer

The total number of ways to create the desired 8-letter string such that H and I are included but not adjacent is: $\boxed{6,350,400}$

Would you like more details on any part of this solution?

Here are some related questions you might consider:

1. What if we wanted a string of 7 letters instead of 8?
2. How does the solution change if H and I must be adjacent?
3. How would the solution change if we were limited to only vowels from A to L?
4. What if we added the constraint that the string must start with a consonant?
5. How many arrangements would there be if the letters A and B also could not be adjacent?

Tip: When solving combinatorial problems, breaking down into "selection" and "arrangement" steps can clarify the solution.