There are four different hotels in a town. Three different men check into hotels in a day, each choosing a hotel independently and at random. What is the probability that each man checks into a different hotel, so that no two men are in the same hotel?

Introduction / Context:
This question involves discrete probability with independent choices among several options. Each of three men chooses one of four hotels independently. We are asked for the probability that all three choose different hotels, meaning that no hotel is chosen by more than one man. This problem illustrates counting outcomes when order matters and then forming a probability ratio.

Given Data / Assumptions:

• There are exactly 4 distinct hotels in the town.
• There are 3 distinct men making choices.
• Each man chooses exactly one hotel.
• Each man chooses a hotel independently and each hotel is equally likely for each man.
• We want the probability that all three men end up in three different hotels.

Concept / Approach:
We treat each man's choice as an independent selection from 4 options, so the total number of ordered outcomes is 4^3. To find favorable outcomes, we need to count the number of ordered triples in which all three chosen hotels are distinct. This is equivalent to counting permutations of 4 hotels taken 3 at a time, because each man must be assigned a different hotel and order matters.

Step-by-Step Solution:
Step 1: For each man, there are 4 choices of hotel, so the total number of ordered outcomes is 4 * 4 * 4 = 4^3 = 64.Step 2: To have all different hotels, the first man can choose any of the 4 hotels.Step 3: Once the first man has chosen, the second man must choose a different hotel, leaving 3 choices.Step 4: The third man must choose a hotel different from both previous choices, leaving 2 choices.Step 5: The number of favorable outcomes is therefore 4 * 3 * 2 = 24.Step 6: The required probability is favorable outcomes divided by total outcomes, giving 24 / 64, which simplifies to 3 / 8 after dividing numerator and denominator by 8.
Verification / Alternative check:
We can cross check by computing the probability directly using conditional reasoning. The first man picks any hotel, which has probability 1. The second man must pick a different hotel, which has probability 3/4. The third man must pick a different hotel from the two already chosen, which has probability 2/4 = 1/2. Multiplying these gives 1 * (3/4) * (1/2) = 3/8, which matches the result obtained from counting permutations. This independently confirms the answer.

Why Other Options Are Wrong:
1/2 would result from incorrectly assuming that half of the outcomes are favorable, which is not supported by the detailed counts. 3/4 would require 48 favorable outcomes, but we only have 24. 4/7 and 9/16 have denominators that do not match 64 when scaled and do not correspond to any simple product of 4 and 3 in this context. Only 3/8 arises naturally from correct counting and conditional probability reasoning.

Common Pitfalls:
Students may mistakenly treat the men as indistinguishable, leading to undercounting. Others may forget that order matters when counting total outcomes, or they might simply think about combinations of hotels without accounting for which man goes to which hotel. Working step by step with conditional probabilities or using permutations of hotels taken three at a time helps avoid these errors.

Final Answer:
The probability that each man checks into a different hotel is 3/8.