To solve this problem, we need to find the number of distinct arrangements of the letters in the word INITIATION. The word consists of 10 letters in total, but some of them are repeated.

Step 1: Count the frequency of each letter.

In the word INITIATION, the letter frequencies are:

• I: 4
• N: 2
• T: 2
• A: 1
• O: 1

Step 2: Apply the formula for permutations of multiset.

The formula for the number of distinct permutations of letters in a word, where some letters are repeated, is:

$\text{Number of distinct arrangements} = \frac{n!}{k_1! \times k_2! \times \cdots \times k_r!}$

Where:

• $n$ is the total number of letters (in this case, $n = 10$),
• $k_1, k_2, \dots, k_r$ are the frequencies of the repeated letters.

Thus, the formula for INITIATION is:

$\frac{10!}{4! \times 2! \times 2! \times 1! \times 1!}$

Step 3: Calculate factorial values.

• $10! = 3,628,800$
• $4! = 24$
• $2! = 2$ (for each occurrence of 2 repeated letters)
• $1! = 1$

Step 4: Compute the final value.

$\frac{3,628,800}{24 \times 2 \times 2 \times 1 \times 1} = \frac{3,628,800}{96} = 37,800$

Thus, the number of distinct arrangements of the letters in the word INITIATION is 37,800.

Would you like further details or have any questions?

Related Questions:

1. How would the solution change if one more letter was repeated?
2. What is the total number of distinct arrangements of the letters in a word where all letters are unique?
3. How does the factorial function impact the outcome of such problems?
4. Can you explain the significance of using the multiset permutation formula?
5. How do different frequency distributions affect the complexity of such permutation problems?

Tip:

When solving permutation problems involving repeated elements, always make sure to divide by the factorials of the counts of each repeated element.