To determine the number of ways in which five different employees can be seated in four indistinguishable offices, we will use the concept of partitioning of an integer.

The problem is equivalent to partitioning the number 5 (representing the 5 different employees) into at most 4 parts (representing the 4 indistinguishable offices) where each part represents the number of employees in each office.

The partitions of 5 into at most 4 parts are as follows :

1. $(5, 0, 0, 0)$ – One office has 5 employees, and the other three have none.


2. $(4, 1, 0, 0)$ – One office has 4 employees, another one has 1, and the other two have none.


3. $(3, 2, 0, 0)$ – One office has 3 employees, another one has 2, and the other two have none.


4. $(3, 1, 1, 0)$ – One office has 3 employees, two have 1 each, and one has none.


5. $(2, 2, 1, 0)$ – Two offices have 2 employees each, one has 1, and one has none.


6. $(2, 1, 1, 1)$ – One office has 2 employees, and the other three have 1 each.

However, since the offices are indistinguishable, we must not count partitions that differ only by the order of the parts. This means that partitions like $(5, 0, 0, 0)$, $(0, 5, 0, 0)$, $(0, 0, 5, 0)$, and $(0, 0, 0, 5)$ all represent the same scenario and thus are counted as one.

So we end up with 6 distinct partition scenarios.

Now, for each partition, we need to find the number of ways to assign the employees according to each partition :

1. $(5, 0, 0, 0)$ – All employees are in one office. There is 1 way to do this because the offices are indistinguishable.


2. $(4, 1, 0, 0)$ - For the group with 4 employees, there are $${}^5{C_4}$$ ways to choose which 4 out of the 5 will be together. Then the remaining employee is automatically assigned to the second office. So, there are $${}^5{C_4}$$ ways for this partition.


3. $(3, 2, 0, 0)$ - For the group with 3 employees, there are ${}^5{C_3}$ ways to choose them, and for the group with 2 employees, there are $${}^2{C_2}$$ ways to choose them (which is essentially 1 way since the last 2 employees are automatically grouped). But the 2 offices these groups can occupy are indistinguishable, so we do not multiply by 2. So, there are $${}^5{C_3}$$ ways for this partition.


4. $(3, 1, 1, 0)$ - For the group with 3 employees, there are $${}^5{C_3}$$ ways. Then we have two indistinguishable offices, each with 1 employee. The order does not matter as offices are indistinguishable. So, there are simply $${}^5{C_3}$$ ways.


5. $(2, 2, 1, 0)$ – For the first group of 2 employees, there are $${}^5{C_2}$$ ways to choose them. For the next group of 2 employees, there are $${}^3{C_2}$$ ways from the remaining 3. Then, the last employee is alone, and since the offices are indistinguishable, there are no further combinations to consider. However, we have double-counted since the two groups of two are indistinguishable. To adjust for this, we divide by 2. This gives us $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2}$ ways.


6. $(2, 1, 1, 1)$ - For the group with 2 employees, there are ${}^5{C_2}$ ways to choose them. The other three employees are each in their own office, with no further combinations since the offices are indistinguishable, so there are ${}^5{C_2}$ ways for this partition.

Let's compute these cases :

• $(5, 0, 0, 0)$ corresponds to $1$ way.


• $(4, 1, 0, 0)$ corresponds to ${}^5{C_4} = 5$ ways.


• $(3, 2, 0, 0)$ corresponds to ${}^5{C_3} = 10$ ways.


• $(3, 1, 1, 0)$ corresponds to ${}^5{C_3} = 10$ ways.


• $(2, 2, 1, 0)$ corresponds to $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2} = \frac{10 \cdot 3}{2} = 15$ ways.


• $(2, 1, 1, 1)$ corresponds to ${}^5{C_2} = 10$ ways.

Adding up all the ways we get :

$$ 1 + 5 + 10 + 10 + 15 + 10 = 51 $$

Therefore, the number of ways five different employees can sit into four indistinguishable offices ($\mathrm{n}$) is 51, which corresponds to Option C.