Suppose you are going to create a string of 8 distinct letters from the first 12 letters of the
alphabet (ABCDEF GHIJKL). If the letters H and I must be included but cannot be
next to each other, how many ways are there to create such a string?

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.

Suppose you are going to create a string of 8 distinct letters from the first 12 letters of the alphabet (ABCDEF GHIJKL). If the letters H and I must be included but cannot be next to each other, how many ways are there to create such a string?

Learn how to calculate the number of ways to create an 8-letter string from the first 12 letters of the alphabet, ensuring letters H and I are included but not adjacent. This solution covers combinatorics, including the use of combinations, permutations, and the Inclusion-Exclusion Principle for accurate counting. Suitable for Grades 10-12.

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