If $\mathrm{n}$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $\mathrm{n}$ is equal to :

<p>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.</p> <p>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.</p> <p>The partitions of 5 into at most 4 parts are as follows :</p> <ol> <li>$(5, 0, 0, 0)$ – One office has 5 employees, and the other three have none.</li><br> <li>$(4, 1, 0, 0)$ – One office has 4 employees, another one has 1, and the other two have none.</li><br> <li>$(3, 2, 0, 0)$ – One office has 3 employees, another one has 2, and the other two have none.</li><br> <li>$(3, 1, 1, 0)$ – One office has 3 employees, two have 1 each, and one has none.</li><br> <li>$(2, 2, 1, 0)$ – Two offices have 2 employees each, one has 1, and one has none.</li><br> <li>$(2, 1, 1, 1)$ – One office has 2 employees, and the other three have 1 each.</li> </ol> <p>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.</p> <p>So we end up with 6 distinct partition scenarios.</p> <p>Now, for each partition, we need to find the number of ways to assign the employees according to each partition :</p> <ol> <li>$(5, 0, 0, 0)$ – All employees are in one office. There is 1 way to do this because the offices are indistinguishable.</li><br> <li>$(4, 1, 0, 0)$ - 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.</li><br> <li>$(3, 2, 0, 0)$ - 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.</li><br> <li>$(3, 1, 1, 0)$ - 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.</li><br> <li>$(2, 2, 1, 0)$ – 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.</li><br> <li>$(2, 1, 1, 1)$ - 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.</li> </ol> <p>Let's compute these cases :</p> <ul> <li>$(5, 0, 0, 0)$ corresponds to $1$ way.</li><br> <li>$(4, 1, 0, 0)$ corresponds to ${}^5{C_4} = 5$ ways.</li><br> <li>$(3, 2, 0, 0)$ corresponds to ${}^5{C_3} = 10$ ways.</li><br> <li>$(3, 1, 1, 0)$ corresponds to ${}^5{C_3} = 10$ ways.</li><br> <li>$(2, 2, 1, 0)$ corresponds to $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2} = \frac{10 \cdot 3}{2} = 15$ ways.</li><br> <li>$(2, 1, 1, 1)$ corresponds to ${}^5{C_2} = 10$ ways.</li> </ul> <p>Adding up all the ways we get :</p> <p>$1 + 5 + 10 + 10 + 15 + 10 = 51$</p> <p>Therefore, the number of ways five different employees can sit into four indistinguishable offices ($\mathrm{n}$) is 51, which corresponds to Option C.</p>