The total number of 4-letter combinations formed by the letters of the word "IITIISERISIIIST" can be calculated using combinatorics. Here's the approach:

Step 1: Count the frequencies of each letter

• I: 7 occurrences
• T: 3 occurrences
• S: 3 occurrences
• E: 1 occurrence
• R: 1 occurrence

So, we have the letters $I, T, S, E, R$ with these respective frequencies.

Step 2: Break down the problem by cases

We will calculate the number of ways to form 4-letter combinations based on how many different letters are chosen. There are several cases based on how many distinct letters are used and the corresponding frequencies.

Case 1: All four letters are different.

We need to select 4 different letters from {I, T, S, E, R}. Since we only have 5 distinct letters, the total number of combinations is $\binom{5}{4}$, which gives 1 combination, and for each combination, the arrangement is $4! = 24$.

Total for Case 1: $1 \times 24 = 24$.

Case 2: Three letters are the same, one is different.

Here we have to select the letter that appears three times from {I, T, S} and then pick one more different letter from the remaining options.

• Pick the repeated letter: 3 ways (I, T, S).
• Pick one other letter from {I, T, S, E, R} excluding the chosen repeated letter: 4 ways.
• Arrangement: Since three letters are the same, we only have $\frac{4!}{3!} = 4$ ways to arrange them.

Total for Case 2: $3 \times 4 \times 4 = 48$.

Case 3: Two letters are the same, and two others are different.

Here we select 1 letter to appear twice, and 2 different letters to appear once each.

• Pick the letter to appear twice from {I, T, S}: 3 ways.
• Pick 2 other letters from the remaining {I, T, S, E, R} (but not the chosen letter): $\binom{4}{2} = 6$ ways.
• Arrangement: $\frac{4!}{2!} = 12$ ways.

Total for Case 3: $3 \times 6 \times 12 = 216$.

Case 4: Two letters are the same, two are the same (two pairs).

Here we need to pick 2 different letters that both appear twice.

• Pick 2 letters to appear twice from {I, T, S}: $\binom{3}{2} = 3$.
• Arrangement: $\frac{4!}{2!2!} = 6$ ways.

Total for Case 4: $3 \times 6 = 18$.

Case 5: All four letters are the same.

The only letter that can appear four times is I. There is only 1 combination, and only 1 arrangement.

Total for Case 5: 1.

Step 3: Add up all the cases

Total combinations = $24 + 48 + 216 + 18 + 1 = 307$.

Thus, the total number of 4-letter combinations is 307.

Would you like a detailed breakdown of any specific case?

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Here are 5 related questions:

1. How would the number change if one more letter were added?
2. How many distinct 3-letter combinations can be formed from the word "IITIISERISIIIST"?
3. What if repetitions were not allowed in the combinations? How would you calculate the total?
4. How do the cases change if we need 5-letter combinations instead of 4?
5. Can we apply a similar approach for forming words from larger sets of letters?

Tip: Always verify whether repetition of letters is allowed when forming combinations or permutations.