1. Interpretation of the problem

We have 5 different employees and 4 indistinguishable offices.

• Offices are indistinguishable $ we only care about how the 5 employees are grouped.
• Any office may have any number of persons, including $0$.

So the problem is equivalent to finding the number of ways to partition 5 distinct employees into at most 4 nonempty unlabeled groups.

That means we need:

where $S(n,k)$ is the Stirling number of the second kind.

2. Compute each Stirling number

For $n=5$:

• $S(5,1)=1$

  (all 5 employees in one office)

• $S(5,2)=15$

• $S(5,3)=25$

• $S(5,4)=10$

Now add them:

3. Alternative direct partition-count check

Let us verify by possible occupancy patterns of nonempty offices.

Since offices are indistinguishable, only group sizes matter.

Possible partitions of 5 into at most 4 positive parts are:

1. $5$

   Number of ways: $1$

2. $4+1$

   Choose the singleton:

   $$\binom{5}{1}=5$$

3. $3+2$

   Choose 2 people for the pair:

   $$\binom{5}{2}=10$$

   (remaining 3 form the other group)

4. $3+1+1$

   Choose 3 people for the triple:

   $$\binom{5}{3}=10$$

5. $2+2+1$

   Choose the singleton first, then split remaining 4 into two unlabeled pairs:

   $$\binom{5}{1}\cdot \frac{1}{2}\binom{4}{2}=5\cdot 3=15$$

6. $2+1+1+1$

   Choose the pair:

   $$\binom{5}{2}=10$$

Total:

So,

4. Check options

• A: $47$  incorrect
• B: $53$  incorrect
• C: $51$  correct
• D: $43$  incorrect

Therefore, the correct option is C.