In how many ways can the letters of the word ' PERMUTATIONS' be arranged if the
words start with P and end with S

Answer

To solve the problem of arranging the letters of the word "PERMUTATIONS" such that it starts with 'P' and ends with 'S', we can follow these steps:

### Step-by-Step Solution:

1. **Identify the total letters in the word**: The word "PERMUTATIONS" consists of 12 letters.

2. **Fix the first and last letters**: According to the problem, the first letter must be 'P' and the last letter must be 'S'. This means we are left with the letters in between.

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

4. **Account for repetitions**: In the letters we have left (E, R, M, U, T, A, T, I, O, N), the letter 'T' appears twice.

5. **Calculate the arrangements**: The formula for arranging n items where there are repetitions is given by:
\[ \text{Number of arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots} \]
where \( n \) is the total number of items, and \( p_1, p_2, \ldots \) are the frequencies of the repeated items.

In our case:
- Total letters to arrange = 10 (E, R, M, U, T, A, T, I, O, N)
- The letter 'T' is repeated 2 times.

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

6. **Calculate 10! and 2!**:
- \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3628800 \)
- \( 2! = 2 \times 1 = 2 \)

7. **Final calculation**:
\[ \text{Number of arrangements} = \frac{3628800}{2} = 1814400 \]

Thus, 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**.