The question asks how many ways the letters in the word "school" can be arranged.

The word "school" consists of 6 letters, but not all are unique. The letter "o" is repeated twice. When calculating the number of distinct permutations of a word with repeated letters, we use the following formula:

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

Where:

• $n$ is the total number of letters,
• $k_1, k_2, \dots, k_r$ are the frequencies of the repeated letters.

For "school":

• Total number of letters, $n = 6$,
• The letter "o" is repeated twice, so $k_1 = 2$.

Thus, the formula becomes:

$\text{Number of arrangements} = \frac{6!}{2!}$

First, calculate $6!$ (6 factorial): $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

Now, calculate $2!$ (2 factorial): $2! = 2 \times 1 = 2$

Now, divide the total factorial by the repeated letters' factorial: $\frac{720}{2} = 360$

So, the number of distinct ways to arrange the letters in "school" is 360.

Would you like further details or have any other questions?

Here are 5 related questions to explore:

1. How would the result change if another letter in "school" were repeated?
2. What if all letters were distinct—how many arrangements would there be?
3. Can you calculate the arrangements for the word "bookkeeper" with multiple repeated letters?
4. How does the formula for arrangements change if the word contains no repeated letters?
5. How can this concept be applied to real-world problems, such as arranging people in seats?

Tip: When working with factorials and permutations, it helps to identify any repeated elements early on to adjust the calculations efficiently.