How many permutations exist of the word "PASCALS"? Leave your answer in factorial form.7!/(2!)(2!)7!/2!2^7 = 1287!/(4!)(2!)

Step 1: Break Down the Problem

To determine the number of unique permutations of the letters in the word "PASCALS," we need to account for repetitions of letters. In "PASCALS," the letters A and S appear twice each.

Step 2: Relevant Concepts

The formula for the number of permutations of a multiset (a set with repetitions) is given by:

$\text{Permutations} = \frac{n!}{n_1! \cdot n_2! \cdots n_k!}$

where $n$ is the total number of letters, and $n_1, n_2, \ldots, n_k$ are the frequencies of the distinct letters.

Step 3: Analysis and Detail

1. Count the total letters: "PASCALS" has 7 letters.

2. Identify repetitions:

   • A appears 2 times
   • S appears 2 times
   • Other letters (P, C, L) appear 1 time each.

3. Apply the formula:

   • Total letters $n = 7$
   • Repetitions: $2!$ for A and $2!$ for S

Thus, the number of permutations is:

$\text{Permutations} = \frac{7!}{2! \cdot 2!}$

Step 4: Verify and Summarize

Calculating it provides:

$\text{Permutations} = \frac{7!}{2! \cdot 2!} = \frac{5040}{2 \cdot 2} = \frac{5040}{4} = 1260$

Final Answer

The total number of unique permutations of the word "PASCALS" is:

$\frac{7!}{2! \cdot 2!} = 1260$