In how many ways can the letters of the word PERMUTATIONS be arranged such that P comes before S.

Answer

To solve the problem of arranging the letters of the word "PERMUTATIONS" such that P comes before S, we can follow these steps:

### Step 1: Count the total letters and identify repetitions
The word "PERMUTATIONS" consists of 12 letters, where the letter T is repeated twice.

### Step 2: Calculate the total arrangements without restrictions
The total arrangements of the letters can be calculated using the formula for permutations of a multiset:
\[ \text{Total arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots} \]
where \( n \) is the total number of letters, and \( p_1, p_2, \ldots \) are the frequencies of the repeated letters.

In our case:
- Total letters \( n = 12 \)
- The letter T is repeated 2 times.

Thus, the total arrangements are:
\[ \text{Total arrangements} = \frac{12!}{2!} \]

### Step 3: Consider the condition P comes before S
Since we want to count only those arrangements where P comes before S, we can observe that in any arrangement of the letters, P and S can appear in two orders: PS or SP. Since we want P to come before S, we can take half of the total arrangements calculated in Step 2.

Thus, the number of arrangements where P comes before S is:
\[ \text{Arrangements with P before S} = \frac{1}{2} \times \frac{12!}{2!} \]

### Step 4: Simplify the expression
Now we can simplify this expression:
\[ \text{Arrangements with P before S} = \frac{12!}{2 \times 2!} \]
Since \( 2! = 2 \), we can further simplify:
\[ = \frac{12!}{4} \]

### Step 5: Final answer
Thus, the final answer for the number of ways the letters of the word "PERMUTATIONS" can be arranged such that P comes before S is:
\[ \frac{12!}{4} \]