What is the number of permutations of the letters of the word POSSIBILITY, which contains 3 I's and 2 S's?

Steps to Solve

1. Identify the total number of letters

The word 'POSSIBILITY' has a total of 11 letters.

2. Count the repeating letters

In 'POSSIBILITY', we have:

• 3 I's
• 2 S's

3. Apply the permutation formula for a multiset

The formula for the number of unique arrangements of letters (permutations of a multiset) is given by:

$$ \frac{n!}{n_1! \times n_2! \times \cdots \times n_k!} $$

where:

• $n$ is the total number of letters,
• $n_1, n_2, ..., n_k$ are the frequencies of the repeating letters.

For 'POSSIBILITY', this gives us:

$$ \text{Unique arrangements} = \frac{11!}{3! \times 2!} $$

4. Calculate the factorials

Now let's calculate the necessary factorials:

$$ 11! = 39916800 $$

$$ 3! = 6 $$

$$ 2! = 2 $$

5. Substitute the values into the formula

Now substitute the values into the permutation formula:

$$ \text{Unique arrangements} = \frac{39916800}{6 \times 2} $$

6. Perform the arithmetic

Calculate the product of the denominators:

$$ 6 \times 2 = 12 $$

Now, divide the numerator by this product:

$$ \frac{39916800}{12} = 3326400 $$