A local delivery company has three packages to deliver to three different houses. Each house is supposed to receive exactly one specific package. If the driver delivers the three packages at random, in how many possible ways will at least one house receive the wrong package?

Introduction / Context:
This problem is about counting permutations with a condition involving correct and incorrect placements. There are three distinct packages and three distinct houses. Each package has a unique intended destination. We are asked to count the number of ways to assign the packages to the houses so that at least one house gets the wrong package. This is an example of using the complement principle in counting, where it can be easier to count the opposite event and subtract from the total.

Given Data / Assumptions:

• There are 3 distinct packages.
• There are 3 distinct houses.
• Each package must be delivered to exactly one house.
• Every arrangement corresponds to a permutation of the 3 packages among the 3 houses.
• We want the number of permutations where at least one package is misdelivered.

Concept / Approach:
The total number of possible delivery patterns is the number of permutations of 3 distinct objects, which is 3! = 6. Among these, there is exactly 1 arrangement in which every package reaches its correct house. All the other arrangements have at least one package delivered to the wrong house. Instead of counting all the incorrect patterns directly, we use the complement principle: subtract the number of completely correct arrangements from the total number of arrangements.

Step-by-Step Solution:
Step 1: Compute the total number of ways to assign the 3 packages to 3 houses. Total permutations = 3! = 3 * 2 * 1 = 6. Step 2: Count the number of perfectly correct deliveries. There is exactly 1 way in which each package goes to its intended house. Step 3: Use the complement principle. Number of ways with at least one wrong delivery = total ways minus all correct ways. So, required count = 6 - 1 = 5.

Verification / Alternative check:
We can list all permutations of three packages to verify this result. There is 1 arrangement with no mistakes and 2 arrangements where exactly one package is correct and the other two are swapped, plus 3 arrangements where no package is in its correct place. This gives 1 + 2 + 3 = 6 total, with 5 of them having at least one wrong package. This enumeration matches the complement calculation.

Why Other Options Are Wrong:
The value 6 represents the total number of permutations and would be correct only if every arrangement had at least one wrong package, which is not true because there is one perfect arrangement. The values 3 and 2 are too small and would correspond to miscounting specific partial cases. The value 120 is 5!, which is unrelated to this problem since we have only 3 packages and 3 houses.

Common Pitfalls:
A frequent mistake is to forget the existence of the fully correct arrangement and simply assume that all permutations correspond to at least one error. Another mistake is to confuse this with the count of permutations where no package is correct, which is called a derangement. In this small example, there are 2 derangements, but the question asks for at least one wrong package, which includes both partial and complete misdelivery patterns. The complement method keeps the logic clear and avoids such confusion.

Final Answer:
There are 5 different ways in which at least one house receives the wrong package.