Key Concepts and Formulas

• Partitions of an integer: A partition of a positive integer $n$ is a way of writing $n$ as a sum of positive integers. The order of the summands does not matter.
• Combinations: The number of ways to choose $k$ objects from a set of $n$ distinct objects is given by the binomial coefficient ${n \choose k} = \frac{n!}{k!(n-k)!}$.
• Indistinguishable Objects: When distributing items into indistinguishable containers, we need to account for overcounting due to permutations of the containers.

Step-by-Step Solution

Step 1: Identify the problem as partitioning 5 into at most 4 parts.

The problem asks for the number of ways to seat 5 different employees into 4 indistinguishable offices. This is equivalent to partitioning the number 5 into at most 4 parts, where each part represents the number of employees in an office. We need to consider all possible distributions of the employees across the offices.

Step 2: List all possible partitions of 5 into at most 4 parts.

The partitions of 5 into at most 4 parts are:

• (5, 0, 0, 0)
• (4, 1, 0, 0)
• (3, 2, 0, 0)
• (3, 1, 1, 0)
• (2, 2, 1, 0)
• (2, 1, 1, 1)

These are all the possible ways to divide the number 5 into at most 4 non-negative integer parts.

Step 3: Calculate the number of ways to arrange the employees for each partition.

Since the employees are distinct, we need to consider the number of ways to choose which employees go into each office for each partition.

• (5, 0, 0, 0): All 5 employees are in one office. There is only 1 way to do this.
• (4, 1, 0, 0): Choose 4 employees to be in one office, and the remaining employee is in another office. The number of ways to do this is ${5 \choose 4} = \frac{5!}{4!1!} = 5$.
• (3, 2, 0, 0): Choose 3 employees to be in one office, and the remaining 2 employees are in another office. The number of ways to do this is ${5 \choose 3} = \frac{5!}{3!2!} = 10$.
• (3, 1, 1, 0): Choose 3 employees to be in one office, and then choose 1 employee from the remaining 2 for another office, and the last employee goes to the remaining office. Since the two offices with 1 employee are indistinguishable, we don't need to worry about ordering. We choose 3 employees for one office. The other two go to the remaining offices. The number of ways is ${5 \choose 3} = \frac{5!}{3!2!} = 10$.
• (2, 2, 1, 0): Choose 2 employees to be in one office, then choose 2 employees from the remaining 3 to be in another office, and the last employee goes to the remaining office. However, since the two groups of 2 are indistinguishable, we must divide by 2 to correct for overcounting. Therefore, the number of ways is $\frac{{5 \choose 2}{3 \choose 2}}{2} = \frac{10 \cdot 3}{2} = 15$.
• (2, 1, 1, 1): Choose 2 employees to be in one office, and the remaining 3 employees are each in their own office. The number of ways to do this is ${5 \choose 2} = \frac{5!}{2!3!} = 10$.

Step 4: Sum the number of ways for each partition.

Add the number of ways for each partition to find the total number of ways: $1 + 5 + 10 + 10 + 15 + 10 = 51$

Common Mistakes & Tips

• Forgetting to account for indistinguishable offices: The key is to list all partitions first. This helps visualize the problem and avoid double-counting.
• Overcounting when groups are of the same size: When distributing into indistinguishable containers and some groups have the same size, divide by the factorial of the number of such groups to avoid overcounting.
• Careful Calculation of Combinations: Ensure the combinations are calculated correctly. A small error can lead to a wrong final answer.

Summary

We solved the problem by first identifying the different ways to partition the number 5 into at most 4 parts. These partitions represent the number of employees in each of the four indistinguishable offices. For each partition, we calculated the number of ways to assign the 5 distinct employees according to the partition. Finally, we summed these values to obtain the total number of ways to seat the employees. The total number of ways is 51.

Final Answer

The final answer is \boxed{51}, which corresponds to option (C).