<Question>
If no one shares an office, in how many ways can 3 people be assigned to 5 different offices? (Each person gets exactly one office).
</Question>

<think>
<Plan_1>
Okay, so I need to figure out how many ways 3 people can be assigned to 5 different offices, with the condition that no two people share the same office. Each person gets exactly one office. Let me start by understanding the problem properly.

First, let's break down the problem.

<prompt_1.1> Break down the problem to understand the task of assigning 3 people to 5 different offices without any two sharing the same office. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
We have 3 individuals, let's call them Person A, Person B, and Person C. There are 5 distinct offices available, and each person must be assigned a unique office. So, the key here is that each person gets exactly one office, and no two people can have the same office. The question is asking for the number of possible ways to do this assignment.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Hmm, okay. Let's think about permutations here because the order might matter. Since each person is getting a different office, the order in which we assign the offices matters. For example, assigning Office 1 to Person A and Office 2 to Person B is different from assigning Office 2 to Person A and Office 1 to Person B.

So, if there are 5 offices and we need to assign 3 of them to the people, where each person gets a unique office, this is similar to permutations of 5 offices taken 3 at a time.

<prompt_2.1> Explain why permutations are the correct approach for assigning 3 people to 5 different offices without any two sharing the same office. </prompt_2.1>

Let me verify this.

<prompt_2.2> Verify the permutation approach by calculating the number of ways for the first person, then the second, and finally the third. </prompt_2.2>

Alternatively, using the permutation formula: P(n, k) = n! / (n - k)! where n is the number of offices (5) and k is the number of people (3).

<prompt_2.3> Use the permutation formula P(n, k) = n! / (n - k)! to verify the solution. </prompt_2.3>

Wait, but let me make sure I haven't missed anything. Is there any possibility that the problem allows for offices to be left empty?
The problem states "assigned to 5 different offices" and each person gets exactly one office. It doesn't specify that each office must be occupied, so some offices might remain empty. But since we are only assigning offices to the 3 people, the total number of assignments is still about selecting 3 distinct offices out of 5 and assigning each to a person.

Alternatively, maybe we can think of it as assigning each person one office, and since the offices are different, each assignment is independent but with the constraint that no two people share the same office.

<prompt_2.4> Consider alternative methods to confirm the solution by thinking of the problem as assigning each person an office with the constraint that no two people share the same office. </prompt_2.4>

Another way to think about it: the first person has 5 choices, the second person has 4 remaining choices, and the third person has 3 remaining choices.
So, 5 × 4 × 3 = 60. This is the same as the permutation formula. So that's consistent.

Is there any chance that the problem is considering the offices as identical except for their labels? For example, if the offices are considered the same regardless of their label, then the number would be different. But the problem states "5 different offices," so each office is distinct. Therefore, the labels matter, and different assignments to different labels count as different.

Alternatively, if the offices were identical, the problem would probably say "5 identical offices," and then it would be a combination problem. But since it's different offices, order matters here.

So, conclusion: the number of ways is 5 × 4 × 3 = 60.
Alternatively, P(5, 3) = 5! / (5 - 3)! = 120 / 2 = 60.

I think that's solid. Let me check if there's any other angle.

<prompt_2.5> Check the solution from a different angle by considering the problem as a permutation of 5 offices taken 3 at a time. </prompt_2.5>

Alternatively, if we thought of it as first choosing the offices and then assigning them to the people. First, choose 3 offices out of 5, which is C(5,3) = 10. Then, assign the 3 distinct offices to the 3 people, which is 3! = 6. So, total ways: 10 × 6 = 60. That's another valid approach, combination followed by permutation.

So, all approaches lead to 60.
Therefore, I think the answer is 60.
</Plan_2>

<Execution_2>
<execution_2.1>
The formula for permutations is P(n, k) = n! / (n - k)! , where n is the total number of items to choose from, and k is the number of items we are choosing. In this case, n = 5 offices and k = 3 people. Plugging in those numbers, we get P(5, 3) = 5! / (5 - 3)! = 5! / 2! . Calculating that, 5! is 5 × 4 × 3 × 2 × 1 = 120, and 2! is 2 × 1 = 2. So, 120 divided by 2 is 60.
</execution_2.1>

<execution_2.2>
The formula for permutations is indeed n! / (n - k)! . Plugging in n = 5 and k = 3, that would be 5! / (5 - 3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 120 / 2 = 60.
</execution_2.2>

<execution_2.3>
Plugging in the numbers, that would be 5! / (5 - 3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 120 / 2 = 60.
</execution_2.3>

<execution_2.4>
So, for each person, we have to choose an office, but without replacement.

So, for the first person, there are 5 offices available. Once we've assigned an office to the first person, there are 4 remaining offices for the second person. Then, for the third person, there are 3 offices left. So, multiplying these together: 5 × 4 × 3. Let me calculate that: 5 times 4 is 20, 20 times 3 is 60. So, 60 possible ways.
</execution_2.4>

<execution_2.5>
Suppose we have 5 offices labeled Office 1 to Office 5. For each person, we need to assign them an office such that no two have the same. So, for the first person, there are 5 possible offices. Once we've assigned an office to the first person, the second person can't have that one, so 4 remaining. The third person can't have either of the first two, so 3 remaining. Multiply them together: 5 × 4 × 3.
</execution_2.5>
</Execution_2>

<Final_answer>
Therefore, the number of ways to assign 3 people to 5 different offices, with no two people sharing the same office, is $\boxed{60}$.
</Final_answer>
</think>