Questions: How many distinct permutations can be formed using the letters of the word "PREPARED"? There are distinct permutations. (Type a whole number.)

Solution Steps

To find the number of distinct permutations of the letters in the word "PREPARED", we need to account for the repeated letters. The formula for permutations of a multiset is given by dividing the factorial of the total number of letters by the factorial of the number of occurrences of each repeated letter.

1. Count the total number of letters in the word.
2. Identify and count the occurrences of each repeated letter.
3. Use the formula for permutations of a multiset: \( \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!} \), where \( n \) is the total number of letters, and \( n_1, n_2, \ldots, n_k \) are the counts of each repeated letter.

The word "PREPARED" consists of 8 letters in total.

The counts of each letter in "PREPARED" are as follows:

• \( P: 2 \)
• \( R: 2 \)
• \( E: 2 \)
• \( A: 1 \)
• \( D: 1 \)

To find the number of distinct permutations, we use the formula for permutations of a multiset:

\[ \text{Distinct Permutations} = \frac{n!}{n_1! \times n_2! \times n_3! \times \ldots} \]

Where:

• \( n = 8 \) (total letters)
• \( n_1 = 2 \) (for \( P \))
• \( n_2 = 2 \) (for \( R \))
• \( n_3 = 2 \) (for \( E \))
• \( n_4 = 1 \) (for \( A \))
• \( n_5 = 1 \) (for \( D \))

Substituting the values, we have:

\[ \text{Distinct Permutations} = \frac{8!}{2! \times 2! \times 2! \times 1! \times 1!} \]

Calculating the factorials:

\[ 8! = 40320, \quad 2! = 2, \quad 1! = 1 \]

Thus, the denominator becomes:

\[ 2! \times 2! \times 2! \times 1! \times 1! = 2 \times 2 \times 2 \times 1 \times 1 = 8 \]

Now, substituting back into the formula:

\[ \text{Distinct Permutations} = \frac{40320}{8} = 5040 \]

Final Answer