In how many ways can the letters of the word 'PERMUTATIONS' be arranged if each word starts with P and ends with S?

Explanation

To solve the problem of how many ways the letters of the word "PERMUTATIONS" can be arranged such that each arrangement starts with 'P' and ends with 'S', we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Letters**: The word "PERMUTATIONS" has 12 letters in total.

2. **Fix the First and Last Letters**: Since we want each arrangement to start with 'P' and end with 'S', we fix 'P' at the beginning and 'S' at the end. This leaves us with the letters in between.

3. **Count the Remaining Letters**: After fixing 'P' and 'S', we have the following letters left: E, R, M, U, T, A, T, I, O, N. This gives us a total of 10 letters to arrange.

4. **Identify Repeated Letters**: In the remaining letters, the letter 'T' appears twice. The other letters (E, R, M, U, A, I, O, N) are all unique.

5. **Calculate the Arrangements**: The number of ways to arrange n items where there are repetitions is given by the formula:
\[ \frac{n!}{p_1! \cdot p_2! \cdots p_k!} \]
where \( n \) is the total number of items, and \( p_1, p_2, \ldots, p_k \) are the frequencies of the repeated items.

In our case:
- Total letters to arrange (n) = 10
- The letter 'T' is repeated 2 times.

Therefore, the number of arrangements is:
\[ \frac{10!}{2!} \]

6. **Calculate Factorials**:
- \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
- \( 2! = 2 \times 1 = 2 \)

7. **Perform the Calculation**:
\[ \frac{10!}{2!} = \frac{3628800}{2} = 1814400 \]

8. **Conclusion**: Therefore, the total number of ways to arrange the letters of the word "PERMUTATIONS" such that it starts with 'P' and ends with 'S' is **1814400**.