How many permutations of latter STRESSLESS as not end TRL

To determine how many permutations of the word "STRESSLESS" do **not** end in the letters "TRL," let's break down the problem:

### Step 1: Total Number of Permutations
The word "STRESSLESS" has 10 letters. Among these, the letters "S," "E," and "S" are repeated multiple times:

- S occurs 4 times.
- E occurs 2 times.

The total number of unique permutations is:

\[
\text{Total permutations} = \frac{10!}{4! \times 2!} = \frac{3,628,800}{24 \times 2} = 75,600
\]

### Step 2: Count Permutations That End with "TRL"
We first fix the last three positions as "TRL." We are left with the remaining 7 letters: "S," "T," "R," "E," "S," "S," "S."

The number of permutations of these 7 letters is:

\[
\text{Permutations with "TRL" at the end} = \frac{7!}{4!} = \frac{5,040}{24} = 210
\]

### Step 3: Subtract from the Total
The number of permutations that **do not** end in "TRL" is:

\[
\text{Permutations not ending in "TRL"} = 75,600 - 210 = 75,390
\]

### Conclusion
There are **75,390** permutations of the word "STRESSLESS" that do **not** end in "TRL."

Would you like more details or have any questions?

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Here are 5 related questions you might find interesting:
1. How would the result change if the word had more repeated letters?
2. Can you explain why dividing by factorials accounts for repeated letters?
3. How do we approach similar problems when the restricted sequence is at the beginning?
4. What are the differences between counting restricted and unrestricted permutations?
5. How would you calculate permutations if there were additional constraints (e.g., no consecutive identical letters)?

**Tip:** When solving permutation problems with restrictions, first calculate the unrestricted total, then account for cases that meet the restrictions separately.

Explore how to count permutations of the word STRESSLESS that do not end in TRL. Understand the factorial calculations and combinatorial methods used. Suitable for Grades 10-12 students and beyond.

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