Okay, so I need to figure out how many ways 3 people can be assigned to 5 different offices, with each person getting exactly one office and no two people sharing an office. Let me start by understanding the problem. 

First, there are 5 offices, all different, and 3 people. Each person has to get one office, and no two people can share the same office. So essentially, we're looking for the number of ways to assign each person to a unique office from the 5 available. 

Hmm, this seems like a permutation problem because the order in which we assign the offices matters here. Since the offices are different, assigning Person A to Office 1 and Person B to Office 2 is different from assigning Person B to Office 1 and Person A to Office 2. So permutations would be the right approach.

Let me recall the formula for permutations. The number of permutations of n items taken k at a time is given by n! / (n - k)!. In this case, n is 5 offices and k is 3 people. So plugging in the numbers, that would be 5! / (5 - 3)! = 5! / 2!.

Calculating that: 5! is 5 × 4 × 3 × 2 × 1 = 120. Then 2! is 2 × 1 = 2. So 120 divided by 2 is 60. So is the answer 60?

Wait, let me verify that. Another way to think about it is for each person, we choose an office. The first person has 5 choices, the second person has 4 remaining choices (since one office is already taken), and the third person has 3 choices. So multiplying those together: 5 × 4 × 3 = 60. That matches the permutation formula result. 

Alternatively, if we thought of it as combinations, but since the order matters here, combinations wouldn't be sufficient. Combinations would give us the number of ways to choose 3 offices out of 5, which is 5 choose 3, and that's 10. But since each person can be assigned to any of those offices, and the order matters, we have to multiply by the number of ways to arrange the 3 people, which is 3!. So 10 × 6 = 60. That also gives the same answer. 

So all three methods—permutation formula, step-by-step multiplication, and combinations followed by permutations of people—result in 60. That seems solid. 

Wait, could there be a mistake here? Let's check each step again. 

First approach: permutation formula. 5P3 = 5! / (5-3)! = 5!/2! = 120/2 = 60. Correct. 

Second approach: 5 choices for the first person, 4 for the second, 3 for the third. 5×4×3 = 60. Correct. 

Third approach: combinations (5C3) × permutations of people (3!) = 10 × 6 = 60. Correct. 

All methods converge to 60, so I think that's the right answer. 

Is there another way to think about this? Maybe using injective functions. Since each person must be assigned a unique office, it's like counting the number of injective functions from a set of 3 elements to a set of 5 elements. The number of injective functions is given by P(n, k) = n! / (n - k)! which is the same as before. So again, 5! / 2! = 60. 

Therefore, all methods of reasoning lead to the same result. I don't see any alternative interpretations of the problem that would change the answer. The key points are that each person must get exactly one office, no two people share an office, and the offices are distinct. Therefore, the problem is a standard permutation problem. 

I think that's thorough. All angles confirm the answer is 60.
So, the number of ways to assign 3 people to 5 different offices, with each person getting exactly one office and no two people sharing an office, is $\boxed{60}$.
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